text:
Larry Fields
It was 30 some years ago.
Larry Fields was back in Pittsburg for the holidays.
He and his wife Sooki were driving down 4th Street.
"I pointed over at that brick building that was in disrepair and told Sooki that I'm going to buy it someday," Larry said. "I just felt like it needed to be taken care of."
That building was the former Frisco Freight Depot at 4th and Elm. A career railroad man, Fields spent much of his adult life in hundreds of depots around the world. This particular one in Pittsburg always occupied a special place in his heart.
Regret is a one way street.
"It's a beautiful building, and it has such a rich history," he said. "I didn't want to see it torn down or just sit there and waste away. It took a little work to finally purchase it, but I was determined to make it happen."
Pittsburg today knows the building as the Frisco Event Center, a beautifully-restored venue for weddings, family reunions, concerts, and more. The depot's story, which began in 1876, lives on in the laughter and tears and memories made.
"I just love seeing people inside there having a good time, enjoying the space," Fields said. "Pittsburg needed a venue like this, and it meant a lot to Sooki and me to fix up the old depot and give it new life."
Due in large part to the Fields' hard work and determination, the Frisco now sits firmly on the National Register of Historic Places.
"I enjoy working on things."
"I didn't really know if we could get it on there, but I wanted to try," Fields said. "I thought the building deserved to be on there, but it's not an easy thing to get done. Our history is our history, and these buildings tell that story.
"I am not a rabid historian that thinks every old building can or should be saved," he said. "Sometimes tearing it down is the right or only choice. But if there is a chance to preserve these buildings and bring them back to life, we should try. Regret is a one-way street. Once you tear them down, there's no going back."
Sooki probably saw it coming.
Shortly after completing work on the Frisco, Fields started thinking about taking on another project. With so many historic, beautiful buildings in Pittsburg, he felt a calling to restore at least one more.
"I guess it becomes a little bit like an addiction," he said. "It's really a lot of fun to get your hands on these buildings. Plus, I enjoy working on things."
In 2019, Fields purchased and began renovating what's commonly known as the Pittcraft building on Locust Street. Built in 1879, the building's original use was production of equipment and goods to be used by miners.
Today, it's The Family Fun Depot, a place where families and children can gather to participate in a variety of games and activities. Fields said such a venue was needed in Pittsburg.
"We have plenty of bars and restaurants," he said. "Kids have sports. We have a bowling alley. We have dance clubs. What we didn't have, however, was a place indoors where families can come and just play together. We needed a place for kids to come and just be kids."
The Fun Depot features a variety of activities, including bounce houses, ball pits, arcade games, trampoline, pop-a-shot basketball, and more.
"It will change over time," Fields said. "We'll keep bringing new things in, give families new activities and challenges. We have a great manager named Darcie Shultz, and she's really doing a wonderful job. We're excited for what this place is becoming."


John McGoldrick with a punch card at the museum.
It's a much-loved Leeds landmark with roots that lie firmly in the past.
Now, though, Leeds Industrial Museum is preparing to set the girls of today on an inspirational path to the careers of the future.
The Armley Mills-based visitor attraction will be playing host later this month to its first ever STEAM Hack event.
Organisers say the day has been designed to encourage girls aged between seven and 14 to consider pursuing STEAM (Science, Technology, Engineering, Arts and Maths) careers.
Its programme will include Minecraft and Patterncraft workshops, with participants being given the chance to get to grips with equipment such as Micro:Bits and Raspberry Pi computers.
Looking ahead to the event, Leeds City Council leader Coun Judith Blake said she hoped it would inspire those attending to go on and take leading roles in the "modern, digital workplace".
Coun Blake said: "Leeds Industrial Museum has played a huge part in the story of our city and once helped establish Leeds as a global standard-bearer in the textile industry.
Expert advice during the day will come courtesy of professional mentors from Leeds-based software developers BJSS.
Attendees will also be able to run the rule over old punch cards which paved the way for the development of some of the world's earliest computer software.
John McGoldrick, curator of industrial history at Leeds Museums and Galleries, said: "The museum is filled with hundreds of fascinating relics of the city's industrial heyday and it's hard to stand in front of these impressive machines and not be in awe of the skill, dexterity and bravery it must have taken to operate them day in, day out.
"Some of the technology which was used here was at the absolute cutting edge of production, and women factory workers in particular were a cornerstone of keeping it moving, even through the unprecedented demands of the First World War.
The STEAM Hack at Leeds Industrial Museum is taking place next Saturday, January 26, and runs from 12.30pm to 4pm.
The £3 fee to attend includes child entry to the museum.
For further information about the event, call the museum on 0113 378 3173.
According to a report released last year by the Tech Nation group, just 19 per cent of the country's digital tech workforce is female.
Across all sectors and jobs, that figure rises to 49 per cent.
Recent research also found that boys are nearly four times more likely than girls to study physics at A-level.


text:
JOEBILL MUÑOZ
director | producer | filmmaker
"Tonny Trocha-Morelos -- en español."
Born in a low-income neighborhood in Bogatå, Colombia, Tony had a long and difficult journey before he first stepped foot on the basketball court at Texas A&M. Because Tony is a native Spanish-speaker and a student in Texas A&M's English Language Institute, JoeBill believed it was best for Tony to share his story in his own words and his native language. "Tony Trocha Morelos - en español," was Texas A&M Athletics' first feature in Spanish-only with English subtitles. (Translation by JoeBill).
"Autism Speaks, Texas A&M Basketball Listens"
Texas A&M Men's Basketball Strength & Conditioning Coach Darby Rich shares his story about his son with Autism--and how the guys on the team have treated him as a member of the Aggie Basketball Family.
"The Beat: Tyler Davis"
Weighing 325 pounds as a freshman in high school, Tyler Davis changed his diet, lost fifty pounds his sophomore year, and three years later worked himself into an all-freshmen collegiate athlete.
"The Khaalia Hillsman Story."
Born and raised in Chicago and a star student-athlete at the historic Whitney Young High School, Khaalia Hillsman chose to play college basketball more than one-thousand miles away at Texas A&M. However, in her first-ever college game, Khaalia returned to Chicago to defeat the hometown DePaul Blue Demons. JoeBill traveled with the team to Chicago to produce this feature.
"The Beat: Chris Collins."
Chris Collins began his career at Texas A&M as a walk-on, but after his play on the court as a sophomore, he earned himself a full-ride scholarship. On this feature, JoeBill served as a cinematographer and gaffer.
FEATURE PRODUCER
©2016 BY JOEBILL MUÑOZ. PROUDLY CREATED WITH WIX.COM

redpajama_set_name: RedPajamaCommonCrawl


Election Day is here, and New Yorkers can look forward to an innovative way to stay informed about who gets elected. Through a collaboration with Instagram, CA Technologies, Obscura Digital, and the Empire State Building itself, CNN is projecting the election results on the Empire State Building, starting at 8 p.m. EST. The projection on the south facade of the famous skyscraper will help voters keep track of how many electoral votes their candidate of choice has.
The display includes "photography, real-time election results, maps, voter mosaics, and animations," according to CNN. It also includes a countdown clock that tracks when polls close throughout the United States. And when the final election results come in, the face of America's next president will be broadcast across the New York City skyline.
And if you've ever used the hashtag #MyVote to share an election-themed image, your creation may well be featured on the side of the Empire State Building, CNN reported. "We wanted to break the fourth wall," Ed O'Keefe, a CNN senior vice president told the Washington Post. "We wanted to make a connection between the television program, which is very much speaking to people, and this election, which has fostered so many emotions and conversations among people."
This isn't the first time the Empire State Building has played host to projections, but it is possibly the most ambitious project yet, the Post reported. The projection is immense; it covers 70,000 square feet and has been created with 40 projectors, O'Keefe told the Post. This makes sense, seeing as the projection is the finale of CNN's #MyVote nationwide tour, during which CNN correspondents traveled across the United States and interviewed voters. Fittingly enough, both Hillary Clinton and Donald Trump's election night gatherings, including the candidates themselves, are within walking distance of the Empire State Building. Both candidates have ties to New York; Clinton was the state's senator from 2001 until 2009, and Trump was born in New York City.
If you're not in New York City, don't worry. You can still watch the Empire State Building either on CNN's television channel, online at CNN's website, or through CNN's Facebook page, which is live-streaming the event. And while it may be the biggest projection screen in the United States tonight, the Empire State Building isn't the only facade to display CNN's election-related news. Washington, D.C.'s Newseum, as well as Las Vegas' Miracle Mile shopping mall will be displaying election results as well, according to the Washington Post.


text:
\section{Generalization and extension}
We assume that thermodynamic forces $D^{-1}\boldsymbol{v}^\alpha(\alpha=1, \cdots, n)$ satisfy the following two conditions, (i) the constraint condition: $\sum_{\alpha=1}^n\boldsymbol{v}^\alpha=\boldsymbol{v}$ and (ii) the orthogonal condition $\bkt{D^{-1}\boldsymbol{v}^\alpha,D^{-1}\boldsymbol{v}^\beta}\propto\delta_{\alpha,\beta}$. The corresponding current and EP are defined as $j_d^\alpha=\bkt{\boldsymbol{d},D^{-1}\boldsymbol{v}^\alpha}$ and $\sigma^\alpha:=||D^{-1}\boldsymbol{v}^\alpha||^2$ respectively.

Including the normalized vector fields $\boldsymbol{e}^\alpha=D^{-1}\boldsymbol{v}^\alpha/||D^{-1}\boldsymbol{v}^\alpha||$ in the basis, \eref{projective TUR} can be generalized to
\begin{align}
    \sum_{\alpha=1}^n \frac{(j_d^\alpha)^2}{\sigma^\alpha}\leq \calD_d
\end{align}
in the same way as \eref{derivation of projective TUR}. In \eref{projective TUR}, the chosen basis includes the housekeeping and excess vector fields $(\alpha=\mathrm{hk},\mathrm{ex})$, and the thermodynamic forces already satisfy the constraint condition. Therefore, even though the lower bound can be tightened by adding other components of the basis, $\boldsymbol{e}^\alpha$, the components have no information about the norm of the corresponding thermodynamic force, making $j_d^\alpha$ and $\sigma^\alpha,$ undefined.

For the thermodynamic forces and arbitrary signs $\chi^\alpha=\pm1\ (\alpha=1, \cdots, n)$, the following inequality holds:
\begin{align}
    &\left\langle\boldsymbol{d},D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha\right\rangle\nonumber\\
    &\leq\bkt{\boldsymbol{d},\boldsymbol{d}}\left\langle D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha,D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha\right\rangle\nonumber\\
    &=\bkt{\boldsymbol{d},\boldsymbol{d}}\bkt{D^{-1}\boldsymbol{v}, D^{-1}\boldsymbol{v}},
\end{align}
where the Cauchy-Schwartz inequality is used in the first line. From this inequality, \eref{max TUR} can be generalized  to
\begin{align}
    \sigma^\mathrm{tot}\geq\frac{1}{\calD_d}
    \max_{\chi^\alpha}\left(\sum_{\alpha=1}^n \chi^\alpha j_d^\alpha\right)^2.
\end{align}
The equality is achieved if and only if the coefficient satisfies $\boldsymbol{d}\propto D^{-1}\sum_{\alpha=1}^n \chi^\alpha \boldsymbol{v}^\alpha$. For the fixed coefficient, the best estimation of the dissipation is obtained for $\chi^\alpha$ which makes the signs of $\chi^\alpha j_d^\alpha$ aligned.
{The equality condition of \eref{max TUR} is obtained by setting $\alpha=\mathrm{hk},\mathrm{ex}$. For the case of $j_d^\mathrm{hk} j_d^\mathrm{ex}>0$, the equality condition is $\boldsymbol{d}={j_d^\mathrm{hk}}D^{-1}\boldsymbol{v}/{\sigma^\mathrm{hk}}\propto D^{-1}\boldsymbol{v}$ and the current coefficient is proportional to the thermodynamic force \cite{Gingrich-Horowitz2016}. By contrast, for the case of $j_d^\mathrm{hk} j_d^\mathrm{ex}<0$, the above relation is replaced by  $\boldsymbol{d}=-{j_d^\mathrm{hk}}D^{-1}(\boldsymbol{v}-2\boldsymbol{v}^\mathrm{ss})/{\sigma^\mathrm{hk}}\propto D^{-1}\boldsymbol{v}^\dagger$ (with $\boldsymbol{v}^\dagger:=\boldsymbol{v}-2\boldsymbol{v}^\mathrm{ss}$), which means that the current coefficient is proportional to the thermodynamic force in the dual dynamics \cite{Dechant-Sasa-continuous}.}

{While the single coefficient $d$ is used in inequalities \eqref{projective TUR} and \eqref{respective TUR}, two coefficients $d$ and $d'$ can be utilized to assess $\sigma^\mathrm{hk}$ and $\sigma^\mathrm{ex}$ respectively.}
\eref{respective TUR} can be modified for two coefficients as
\begin{align}
    \sigma^\mathrm{tot}\geq
    \frac{\left(j_{d}^\mathrm{hk}\right)^2}{D_{d}}
    +\frac{\left(j_{d'}^\mathrm{ex}\right)^2}{D_{d'}}.
\end{align}
This TUR can be applied for the case when the kind of coefficient is restricted by the experimental condition. The equality is obtained if and only if $\boldsymbol{d}\propto\boldsymbol{e}^\mathrm{hk},\boldsymbol{d}'\propto\boldsymbol{e}^\mathrm{ex}$ is satisfied. {Therefore, if we set $\boldsymbol{d}=\boldsymbol{f}\propto D^{-1}\boldsymbol{v}^\mathrm{ss}$ and $\boldsymbol{d}'=-\boldsymbol{\nabla} U\propto D^{-1}(\boldsymbol{v}-\boldsymbol{v}^\mathrm{ss})$ in the example of the two-dimensional Brownian motion, we can completely estimate the housekeeping and excess EP rate. Therefore, we can estimate all the EP rates by taking measurements for these particular two currents.}
Note that \eref{projective TUR} is still tighter as long as we focus on the estimation with the single coefficient.
Moreover, adding small coefficient $\delta d$ such that $\delta d\propto\boldsymbol{e}^\mathrm{ex},\delta d'\propto\boldsymbol{e}^\mathrm{hk}$, \eref{projective TUR} can be extended to
\begin{align}
    \sigma^\mathrm{tot}\geq
    \frac{\left(j_{d}^\mathrm{hk}\right)^2}{D_{d+\delta d}}
    +\frac{\left(j_{d'}^\mathrm{ex}\right)^2}{D_{d'+\delta d'}}.
\end{align}

\section{Markov jump processes}
For the Langevin dynamics, the orthogonal condition \eqref{orthogonal condition} is satisfied because the surface term vanishes due to its boundary condition. By contrast, this condition does not hold for Markov jump processes, and neither \eref{projective TUR} nor \eqref{max TUR} can be derived. On the other hand, as the counterpart of \eref{respective TUR}, the following TURs hold:
\begin{align}
    \label{Markov jump: respective TUR}
    \sigma^\mathrm{hk}\geq c_0\frac{\left(j_d^\mathrm{hk}\right)^2}{\tilde{D}_d},\
    \sigma^\mathrm{ex}\geq c_0\frac{\left(j_d^\mathrm{ex}\right)^2}{\tilde{D}_d}.
\end{align}
Contrary to \eref{respective TUR}, the constant $c_0=0.896\dots$ is required and the variance of the current $\tilde{D}_d$ is also different. It is nontrivial whether the average and variance of the current are directly measurable or not. Nonetheless, we see that these take a simple form (\eref{QD: hk current}\eqref{QD: ex current}\eqref{QD: HS variance}) for the example of the quantum dot mentioned below.

Compared to the conventional TUR, \eref{TUR},
\begin{align}
    \label{Markov jump: sign condition}
    \frac{(j_d)^2}{\calD_d}
    =\frac{1}{c_0}\frac{\tilde{D}_d}{\calD_d}
    \left(c_0\frac{\left(j_d^\mathrm{hk}\right)^2}{\tilde{D}_d}+c_0\frac{\left(j_d^\mathrm{ex}\right)^2}{\tilde{D}_d}\right)
    +\frac{2j_d^\mathrm{hk} j_d^\mathrm{ex}}{\calD_d}
\end{align}
means that, roughly speaking, it is better to estimate the total dissipation by decomposing it into the housekeeping and excess parts when the product of the currents $j_d^\mathrm{hk} j_d^\mathrm{ex}$ is negative.

\subsection{Derivation}
We assume that the system is coupled to multiple reservoirs and its discrete state $x$ obeys the master equation:
\begin{align}
    \dot{p}_{x}
    =\sum_\nu\sum_{x'}R_{xx'}^\nu p_{x'}
    =\sum_\nu\sum_{x'}j_{xx'}^\nu,
\end{align}
where $j_{xx'}^\nu:=R_{xx'}^\nu p_{x'}-R_{x'x}^\nu p_{x}=K_{xx'}^\nu-K_{x'x}^\nu$ is the probability flow from the state $x'$ to $x$ driven by the reservoir $\nu$ $(K_{xx'}^\nu:=R_{xx'}^\nu p_{x'})$.
In the steady state, this satisfies the balanced condition $\sum_\nu\sum_{x'}{j^\mathrm{ss}}_{xx'}^\nu=0$.
The EP rates are given by \cite{Three-faces-Master-equation,3DFT-Esposito2010}
\begin{align}
    \label{Markov jump: total EP}
    \sigma^\mathrm{tot}&=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{K_{x'x}^\nu},\\
    \label{Markov jump: hk EP}
    \sigma^\mathrm{hk}&=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{\tilde{K}_{xx'}^\nu},
\end{align}
and
\begin{align}
    \label{Markov jump: ex EP}
    \sigma^\mathrm{ex}=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{\tilde{K}_{x'x}^\nu}.
\end{align}
We define $\tilde{K}$ as $\tilde{K}_{xx'}^\nu:=\tilde{R}_{xx'}^\nu p_{x'}$ with the dual transition rate $\tilde{R}_{xx'}^\nu:=R_{x'x}^\nu p^\mathrm{ss}_{x}/p^\mathrm{ss}_{x'}$.

Alhough the mean local velocity is not defined conventionally for Markov jump processes, we introduce it with the definition $v_{xx'}^\nu:=j_{xx'}^\nu/p_{x'}$ in analogy with the Langevin dynamics. Dividing the probability flow $j_{xx'}$ by the probability $p_{x'}$, the mean local velocity captures the net current of the transition. On the contrary to the probability flow, the mean local velocity does not satisfy the skewed symmetry, i.e., $v_{xx'}^\nu\neq v_{x'x}^\nu$.

The total current average is expressed as
\begin{align}
    \label{Markov jump: def of current}
    j_d(t)
    &=\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu K_{xx'}^\nu
    =\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu j_{xx'}^\nu\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu v_{xx'}^\nu p_{x'}.
\end{align}
We define the housekeeping and excess current averages as
\begin{align}
    \label{Markov jump: def of hk current}
    j_d^\mathrm{hk}(t)
    :&=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    (K_{xx'}^\nu-\tilde{K}_{xx'})\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu {v^\mathrm{ss}}_{xx'}^{\nu} p_{x'}
\end{align}
and
\begin{align}
    \label{Markov jump: def of ex current}
    j_d^\mathrm{ex}(t)
    :&=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    (K_{xx'}^\nu+\tilde{K}_{xx'})\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    ({v^\mathrm{ss}}_{xx'}^{\nu}-v_{xx'}^\nu)p_{x'}
\end{align}
respectively. Then, the current decomposition $j_d=j_d^{\textrm{hx}}+j_d^\mathrm{ex}$ holds by definition. These current averages have almost the same form as their counterparts in Langevin dynamics and a simple calculation confirms that these coincide in the continuous limit.
In addition, we define both the housekeeping and excess current variances as
\begin{subequations}
\begin{align}
    \tilde{D}_d:&=\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)\\
    &=\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{x'x}^\nu).
\end{align}
\end{subequations}

We are now ready to derive \eref{Markov jump: respective TUR}.
\begin{align}
    \sigma^\mathrm{hk}\tilde{D}_d
    =\sum_\nu&\sum_{x\neq x'}\left(K_{xx'}^\nu
    \ln\frac{K_{xx'}^\nu}{\tilde{K}_{xx'}^\nu}
    +\tilde{K}_{xx'}^\nu-K_{xx'}^\nu\right)\nonumber\\
    &\cdot\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)\nonumber\\
    \geq c_0 &\sum_\nu\sum_{x\neq x'}
    \frac{(K_{xx'}^\nu-\tilde{K}_{xx'}^\nu)^2}
    {K_{xx'}^\nu+\tilde{K}_{xx'}^\nu}\nonumber\\
    &\cdot\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)
    \nonumber\\
    \geq c_0 &\left(\frac{1}{2}\sum_\nu\sum_{x\neq x'}
    d_{xx'}^\nu(K_{xx'}^\nu-\tilde{K}_{xx'}^\nu)\right)^2\nonumber\\
    =c_0&\left(j_d^\mathrm{hk}\right)^2
\end{align}
The derivation for the excess part follows in exactly the same way. We used an equality $\sum_x K_{xx'}^\nu=\sum_x \tilde{K}_{xx'}^\nu=0$, an inequality
\begin{align}
    \label{basic ineq c=0.896}
    a\ln\frac{a}{b}+b-a\geq \frac{c_0(a-b)^2}{a+b}
\end{align}
($a,b>0, c_0=0.896\dots$) \cite{Shiraishi-Funo-Saito,Shiraishi-Saito-Tasaki2016}, and the Cauchy-Schwaltz inequality in each transformation.

\subsection{The necessity of $c_0$}
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{MJ_RW.eps}
    \caption{\label{fig: MJ_RW}
    (a) The sketch of the random walk. A single particle is hopping among sites driven by a heat reservoir.
    (b)(c)The test of the short-time TUR for each EP rate in the $1$-dimensional random walk (\eref{1dRW: tot TUR}\eqref{1dRW: HS TUR}). The parameters of the model are chosen at random from $A\in[-4, 4],\ R_x\in[0, 1],\ d_x\in[-1, 1]$. The probability distribution of the system $p_x$ is also at random. The sample size is $10^5$. The system size is $M=3$ for (b), and $M=5$ for (c).
    }
\end{figure}
We see the constant $c_0=0.896\dots$ is required for Markov jump processes in the simple one-dimensional random walk model. There are $M$ sites in the one direction with the periodic boundary condition, and a single particle is hopping among them, being driven by a heat reservoir (see \fref{fig: MJ_RW} (a)). We denote the transition rate to the forward (backward) direction at site $x(=1,2,\dots,M)$ as $R_{x,+}\ (R_{x,-})$. We assume the local detailed balance condition for this system, $\frac{R_{x,+}}{R_{x+1,-}}=e^A$, with the thermodynamic force $A$. Thus, the transition rate is written by $R_{x,+}:=R_xe^{A/2},\ R_{x,-}:=R_{x-1}e^{-A/2}$.
The current coefficient obeys the skewed symmetry:
\begin{align}
    d_{xx'}=\begin{cases}
            d_{x-1} & (x'=x-1)\\
            -d_x & (x'=x+1)\\
            0 & (\textrm{otherwise})
            \end{cases}.
\end{align}

In this setup, we examine the TURs
\begin{align}
    \label{1dRW: tot TUR}
    q:&=\frac{\sigma^\mathrm{tot} {D}_d}
    {\left(j_d\right)^2}\geq1\\
    \label{1dRW: HS TUR}
    q:&=\frac{\sigma^l \tilde{D}_d}
    {\left(j_d^l\right)^2}\geq c_0
    \ (l=hk,\mathrm{ex})
\end{align}
by plotting $q$ for many systems with different probability distributions, transition rates, and current coefficients.
In \fref{fig: MJ_RW} (b), the lower bound $q\geq1$ is violated for some blue and red points, meaning that \eref{Markov jump: respective TUR} does not hold without the constant $c_0$. At the same time, there are few points near the bound $q\geq c_0$, which suggests that the quality condition is hard to satisfy.

Since the conventional TUR \eqref{TUR} is satisfied in the equilibrium limit, the lower bound $q\geq1$ gets tighter for small thermodynamics forces. The blue plot behaves similarly to the black plot because the housekeeping EP captures the violation of the detailed balance condition. On the other hand, the excess EP comes from the deviation from the steady state. Therefore, the red points are widely distributed and the tightness of the lower bound is almost independent of $A$.

The samples close to the lower bound get scarce when we increase the size of the system (see \fref{fig: MJ_RW} (c)). This suggests that achieving equality requires the state and coefficients to be tuned commensurately and this tuning gets extremely hard for large systems. Therefore, the lower bound $q\geq$ is practically never broken for sufficiently large systems, and the same TUR as the Langevin dynamics, \eref{respective TUR}, can be applied.

\subsection{Example: quantum dot}
The EP is estimated using TUR \eqref{Markov jump: respective TUR} in the model of the quantum dot. The number of levels is one, and we denote state 1 when the particle is occupied and state 0 when it is empty. The transition rate of the heat reservoir $\nu(=\textrm{h,c})$ is given by the Fermi golden rule as $R_{10}^\nu=\Gamma_0 f_\nu,\ R_{01}^\nu=\Gamma_0(1-f_\nu)\ (f_\nu:=1/(1+e^{\beta_\nu(E-\mu_\nu)}))$ \cite{Benenti-Casati-Saito-Whitney2017, Esposito-Lindenberg-Broeck2009-QD}. $E$ is the energy of the level and $\mu_\nu$ is the chemical potential of the heat reservoir $\nu$. The coupling strength to the heat reservoir is assumed to be the same $\Gamma_0$, which is sufficiently small that the sequential tunneling approximation holds.

The steady state is easily calculated:
\begin{align}
    p_0^\mathrm{ss}=\frac{R_{01}}{R_{01}+R_{10}},\
    p_1^\mathrm{ss}=\frac{R_{10}}{R_{01}+R_{10}}
    =\frac{1}{2}(f_{\textrm{h}}+f_{\textrm{c}}).
\end{align}
The distinguished feature of 2-state systems is that the overall dual transition rate matches the original transition rate, that is,
\begin{align}
    \tilde{R}_{xx'}=\sum_\nu\tilde{R}_{xx'}^\nu=\sum_\nu\frac{{R}_{x'x}^\nu p_{x}^\mathrm{ss}}{p_{x'}^\mathrm{ss}}={R}_{xx'}.
\end{align}

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{MJ_QD.eps}
    \caption{\label{fig: MJ_QD}
    (a) The averages of the particle current in the quantum dot model. All of the averages are linear in terms of $p_1$. The vertical dot line depicts the steady state, $p_1=p_1^\mathrm{ss}$. (b) Each EP rate. In the steady state, the excess EP rate vanishes and the housekeeping EP rate coincides with the total one. (c) The estimation of each EP rate based on \eref{TUR}\eqref{Markov jump: respective TUR} using the particle current. $\widehat{\sigma^l}$ denotes the estimation of $\sigma^l\ (l=\mathrm{tot},\mathrm{hk},\mathrm{ex})$. The gray line represents the estimation with $\widehat{\sigma^\mathrm{tot}}=c_0[(j_d^\mathrm{hk})^2+(j_d^\mathrm{ex})^2]/\tilde{D}_d$. The parameters used in calculation are $\beta_\textrm{h}(E-\mu_\textrm{h})=1.0,\beta_\textrm{c}(E-\mu_\textrm{c})=3.0$.
    }
\end{figure}
As a current, consider a particle current through the system in the direction from the high-temperature heat reservoir to the low-temperature heat reservoir. The coefficient is given by $d_{xx'}^\nu=M_{xx'}^\nu$ with
\begin{align}
    M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}=\frac{(N_x-N_{x'})}{2}.
\end{align}
$N_x$ is the number of particles of the system in state $x$, where $N_0=0,N_1=1$. Since we are dealing with non-stationary conditions here, we need to consider the exchange of particles with both heat reservoirs.
Computing each current average based on \eref{Markov jump: def of current}-\eqref{Markov jump: def of ex current},
\begin{align}
    \label{QD: tot current}
    j_d&=\sum_\nu M_{10}^\nu(R_{10}^\nu p_0-R_{01}^\nu p_1)=\frac{\Gamma_0}{2}(f_h-f_c)\\
    \label{QD: hk current}
    j_d^\mathrm{hk}
    &=\frac{1}{2}\sum_\nu M_{10}^\nu
    \left[(R_{10}^\nu p_0-\tilde{R}_{10}^\nu p_0)
    -(R_{01}^\nu p_1-\tilde{R}_{01}^\nu p_1)\right]\nonumber\\
    &=\frac{\Gamma_0}{4}\left(\frac{p_0}{p_0^\mathrm{ss}}
    +\frac{p_1}{p_1^\mathrm{ss}}\right)(f_h-f_c)\\
    \label{QD: ex current}
    j_d^\mathrm{ex}
    &=j_d^\mathrm{tot}-j_d^\mathrm{hk}\nonumber\\
    &=\frac{\Gamma_0}{4}\left(\left(1-\frac{p_0}{p_0^\mathrm{ss}}\right)
    +\left(1-\frac{p_1}{p_1^\mathrm{ss}}\right)\right)(f_h-f_c)
\end{align}
These are plotted in \fref{fig: MJ_QD} (a). It can be seen that the current mean of the total system is always constant, independent of the state probability $p_x$. This is due to the two-state nature and the property $M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}$. The housekeeping and excess current averages are the current averages of the total system multiplied by $({p_0}/{p_0^\mathrm{ss}}+{p_1}/{p_1^\mathrm{ss}})/2$ and $((1-{p_0}/{p_0^\mathrm{ss}})+(1-{p_1}/{p_1^\mathrm{ss})})/2$, respectively. The housekeeping current average is always positive and the excess current average changes its sign at $p_1=p_1^\mathrm{ss}$.

On the other hand, each current variance is computed as
\begin{align}
    \label{QD: variance}
    \calD_d
    &=\frac{1}{8}\sum_\nu(R_{10}^\nu p_0+R_{01}^\nu p_1)\\
    \label{QD: HS variance}
    \tilde{D}_d&=
    \frac{1}{16}\sum_\nu
    \left[(R_{10}^\nu p_0+\tilde{R}_{10}^\nu p_0)
    +(R_{01}^\nu p_1+\tilde{R}_{01}^\nu p_1)\right]\nonumber\\
    &=\frac{1}{8}\sum_\nu(R_{10}^\nu p_0+R_{01}^\nu p_1)
    =\calD_d
\end{align}
and is the same value. This is also due to the special properties of the two states and $M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}$, where the variance generally takes different values.

In this model, the EP rates calculated by \eref{Markov jump: total EP}-\eqref{Markov jump: ex EP} are shown in \fref{fig: MJ_QD} (b). The excess EP rate takes a minimum value of $0$ at the steady state $p_1=p_1^\mathrm{ss}$ and increases rapidly away from it. The housekeeping EP rate is linear and monotonically increasing because $\ln(K_{x'x}^\nu/\mathcal{K}_{x'x}^\nu)$ is independent of $p_1$. Also, from a simple calculation, it can be shown that this slope increases as the thermal force $A=\beta_\textrm{c}(E-\mu_\textrm{c})-\beta_\textrm{h}(E-\mu_\textrm{h})$ is larger. Reflecting these behaviors, the EP rate for the total system reaches a finite minimum at $p_1$ which is slightly smaller than $p_1^\mathrm{ss}$.

Finally, the EP rate estimations given by the TUR \eqref{Markov jump: respective TUR} are plotted in \fref{fig: MJ_QD} (c). The accuracy of the EP rate for the total system tends to have a maximum value of about $90\%$ near the minimum of the EP rate, and the accuracy decreases as one moves away from the minimum. For the excess EP rate, the overall accuracy is less than $10\%$, but the accuracy improves as the steady state is approached. On the other hand, for the housekeeping EP rate, the overall accuracy is high, and as $p_1$ increases, the accuracy gradually decreases. The gray line in the figure shows the estimated EP rate for the entire system from the lower bounds of the housekeeping and excess EP rates. From \eref{Markov jump: sign condition}, it is better to decompose and estimate the housekeeping and excess components approximately when the product of the currents $j_d^\mathrm{hk} j_d^\mathrm{ex}$ becomes negative. In fact, for $p_1>p_1^\mathrm{ss}$ where the product of the currents is negative (see \fref{fig: MJ_QD}(a)), the gray line is located above the black line except near its boundary, and the EP rate for the total system is estimated more accurately.

\section{Details of the example: 2-dimensional flow}
In this section, we present some calculations and figures to supplement the first example in the main text.
Since $D$ is proportional to the identity matrix, the coefficient proportional to the nonconservative force, $\boldsymbol{d}=\boldsymbol{d}_\theta:=d_\theta[-x_2, x_1]^T$, satisfies \eref{hk coeff}, and the coefficient proportional to the conservative force, $\boldsymbol{d}=\boldsymbol{d}_r:=-d_r[x_1, x_2]^T$, satisfies \eref{ex coeff}. Here, we consider coefficients including only $\boldsymbol{d}_\theta,\boldsymbol{d}_r$, but note that this does not mean that all coefficients can be expanded by them.

The current averages with $\boldsymbol{d}=\boldsymbol{d}_\theta+\boldsymbol{d}_r$ are given by
\begin{align}
    \label{2Dflow: hk coeff}
    j_d^\mathrm{hk}&=j_{d_\theta}^\mathrm{hk}=j_{d_\theta}
    =\frac{2\kappa T'd_\theta}{k}
\end{align}
and
\begin{align}
    \label{2Dflow: ex coeff}
    j_d^\mathrm{ex}&=j_{d_r}^\mathrm{ex}=j_{d_r}
    =2 T'\left(1-\frac{T}{T'}\right)d_r.
\end{align}
While $J_{D}^\mathrm{hk}$ has the same sign regardless of $T'$, $J_{D}^\mathrm{ex}$ switches positive and negative at $T'=T$. The current variances with $\boldsymbol{d}=\boldsymbol{d}_\theta,\ \boldsymbol{d}_r$ are computed as
\begin{align}
    D_{d_\theta}=\frac{2 TT'{d_\theta}^2}{k},\ \textrm{and }
    D_{d_r}=\frac{2 TT'{d_r}^2}{k}
\end{align}
respectively. Since $\boldsymbol{d}_\theta^T\boldsymbol{d}_r=0$ (and $D$ is proportional to the identity matrix),
the current with $\boldsymbol{d}=\boldsymbol{d}_\theta+\boldsymbol{d}_r$ satisfies the decomposition of the variance as well:
\begin{align}
    \label{2Dflow: decomp of fluctuation}
    \calD_d=D_{d_\theta}+D_{d_r}.
\end{align}
For the coefficient $\boldsymbol{d}=\boldsymbol{F}$, the averages are plotted in \fref{fig: 2dflow} (b) and the variance is plotted in \fref{fig: 2dflow_variance}.

\begin{figure}
    \centering
    \includegraphics[width=0.8\linewidth]{2dflow_variance.eps}
    \caption{\label{fig: 2dflow_variance}
    The variance of the power exerted on the system in the 2-dimensional flow model. Due to the decomposition of the coefficient, \eref{2Dflow: decomp of fluctuation}, the black line agrees with the sum of the blue and red lines. The parameters used in calculation are $\mu=1.0, k=1.0, \kappa=2.0, T=1.5$.
    }
\end{figure}

Next, we calculate the EP rates. For the current coefficient,
\begin{align}
    \boldsymbol{d}=D^{-1}\boldsymbol{v}
    &=\frac{1}{T}(-\boldsymbol{\nabla} U+\boldsymbol{f}-T\boldsymbol{\nabla}\ln p)\nonumber\\
    &=\frac{\kappa}{T}\boldsymbol{d}_\theta+\frac{k}{T}
    \left(1-\frac{T}{T'}\right)\boldsymbol{d}_r,
\end{align}
the housekeeping (excess) EP rate is given as its housekeeping (excess) current due to the orthogonal condition \eqref{orthogonal condition}:
\begin{align}
    \sigma^\mathrm{hk}&=\frac{2\kappa^2}{k}\frac{T'}{T},
\end{align}
and
\begin{align}
    \sigma^\mathrm{ex}&=\frac{2\kappa^2}{k}\frac{T'}{T}\left(\frac{k}{\kappa}\right)^2
    \left(1-\frac{T}{T'}\right)^2.
\end{align}
The total EP rate is just the sum of them:
\begin{align}
    \sigma^\mathrm{tot}
    =\frac{2\kappa^2}{k}\frac{T'}{T}
    \left[1+\left(\frac{k}{\kappa}\right)^2
    \left(1-\frac{T}{T'}\right)^2\right].
\end{align}
These are plotted in \fref{fig: 2dflow} (c). {$\sigma^\mathrm{ex}$ and $\sigma^\mathrm{tot}$ diverges at $T'\rightarrow0$ because the system state takes the form of the delta-function distribution which is an extremely nonequilibrium state.}

\section{Details of the example: rocking ratchet}
In this section, we present some calculations and figures to supplement the second example in the main text.
\subsection{Model}
The Langevin equation of the rocking ratchet is
\begin{align}
    \dot{x}=F(x,t)+\xi(t)=-\partial_x U(x)+R(t)+\xi(t),
\end{align}
where the potentials are periodic in space and time: $U(x+L)=U(x)$ and $R(t\calT)=R(t)$.
The thermal noise $\xi(t)$ is the white Gaussian noise, i.e., $\bkt{\xi(t)}=0$ and $\xi(t)\xi(t')=\delta(t-t')$. The corresponding Fokker-Planck equation is
\begin{align}
    \label{rocking: Fokker-Planck equation}
    \partial_t p(x,t)=-\partial_x j(x,t)=-\partial_x[F(x,t)-D\partial_x]p(x,t),
\end{align}
and the mean local velocity is expressed as $v(x,t):=j(x,t)/p(x,t)=F-D\partial_x \ln p$. Since the system is periodic in space, the probability $p$ and the probability current $j$ are punishingly small. We assume that these quantities are also periodic, $p(x+L,t)=p(x,t), j(x,t)=j(x+L,t)$, and take the summation $\hat{p}(x, t):=\sum_{n=-\infty}^\infty p(x+nL,t),\hat{j}(x, t):=\sum_{n=-\infty}^\infty j(x+nL,t)$. Then, the reduced quantities $\hat{p}$ and $\hat{j}$ also get periodic in space and have finite values, satisfying the same form of the Fokker-Planck equation as \eref{rocking: Fokker-Planck equation}. This assumption is plausible if we consider the periodic steady state or prepare the initial state periodic in space.
{The instantaneous steady state is given by a modified canonical distribution:
\begin{align}
    p^\mathrm{ss}(x, t)=\frac{\int^{x+L}_x dye^{[V(y,t)-V(x,t)]/D}}{\int_0^L dx\int^{x+L}_x dye^{[V(y,t)-V(x,t)]/D}},
\end{align}
where $V(x, t):=U(x)-R(t)x$ and $\int^{x+L}_x dye^{V(y, t)/D}$ is required to satisfy the periodicity $p^\mathrm{ss}(x+L)=p^\mathrm{ss}(x)$.}
We define the reduced local mean velocity as $\hat{v}(x, t):=\hat{j}(x,t)/\hat{p}(x,t)$ which coincides with the usual definition $v$ as long as the probability has the periodicity $p(x+L,t)=p(x,t)$. These reduced quantities are denoted as $p,j,v$ in the main text. The currents are EP rates can be calculated in the usual way, but the integration range is replaced by $[0,L]$.

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{rocking_mlv.eps}
    \caption{\label{fig: rocking_mlv}
    The steady-state mean local velocity in the rocking ratchet model. Since the nonconservative force vanishes at $t/T=0,0.5$, the mean local velocity vanishes as well at this time. The plot has kinks at $x/L=0, 0.8$ in accordance with the indifferentiability of the potential $U(x)$. Apart from these points, the plot is approximately flat. The parameters used in calculation are $D = 0.1, T=0.7, a=0.8, R_0 = 6.0$.
    }
\end{figure}

\subsection{The coefficient choice}
In the main text, we used the particle current with $d(x)=1$. Here, we consider another choice of coefficient. We plot the mean local velocity for the instantaneous steady state in \fref{fig: rocking_mlv}. $v^\mathrm{ss}$ tends to be constant except for the top and valley of the potential $U(x)$, which suggests that the coefficient $d(x)=1$ approximately satisfies the condition $d\propto e^\mathrm{hk}$ (\eref{hk coeff}).
Consequently,$j_d^\mathrm{ex}$ is small compared to $j_d^\mathrm{hk}$ and the estimation of the excess EP rate does not have high accuracy. (see \fref{fig: rocking} (a) and (d)).

Since the mean local velocity is expected to consist of low-frequency Fourier components, we use the coefficient $d(x)=\cos(2\pi x/L),\sin(2\pi x/L)$ instead here. We plot the current averages in \fref{fig: rocking_cossin}(a)(b). The housekeeping current is decreased in comparison with the case $d(x)=1$, and is now comparable to the excess current. We can see the improvement of the estimation in the first half of the period (see \fref{fig: rocking_cossin}(c)(d)).
Although $\cos(2\pi x/L)$ and $\sin(2\pi x/L)$ are not orthogonal to each other in terms of the inner product $\bkt{\cdot,\cdot}$, the currents and estimations tend to compensate each other. Combining these results, we can estimate roughly over $50\%$ of the excess EP rate.
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{rocking_cossin.eps}
    \caption{\label{fig: rocking_cossin}
    (a)(b) Each current average with $d(x)=\cos(2\pi x/L)$ for (a) and $d(x)=\sin(2\pi x/L)$ for (b).  (c)(d) The estimation of the excess EP rate based on \eref{projective TUR} (green line) and \eref{respective TUR} (red line) with $d(x)=\cos(2\pi x/L)$ for (c) and $d(x)=\sin(2\pi x/L)$ for (d).
    The parameter used in calculation are $D = 0.1, T=0.7, a=0.8, R_0 = 6.0$.
    }
\end{figure}

\section{2-beads model}
In the first example in the main text, the direction of $\boldsymbol{e}^\mathrm{hk},\boldsymbol{e}^\mathrm{ex}$ or $\boldsymbol{v}^\mathrm{ss},\boldsymbol{v}-\boldsymbol{v}^\mathrm{ss}$ was obvious. Also, as noted in the discussion, the direction of $\boldsymbol{v}^\mathrm{ss}$ is considered experimentally estimable. In this section, we consider a two-dimensional heat conduction system in the absence of prior knowledge of the steady state.

\subsection{Setup}
Consider two coupled Brownian particles governed by the Langevin equation
\begin{align}
    \label{2-beads: Langevin eq}
    \dot{\boldsymbol{x}}&=\mu A\boldsymbol{x}+\sqrt{2}G\boldsymbol{\xi}(t)\\
    &\textrm{with  }
    A=\mu k\begin{bmatrix}
                -2 & 1\\
                1 & -2
            \end{bmatrix}\
    \textrm{and } G=\begin{bmatrix}
                \sqrt{\mu T_1} & 0\\
                0 & \sqrt{\mu T_2}
            \end{bmatrix}
\end{align}
and the Fokker-Planck equation
\begin{align}
    \label{2-beads: Fokker-Planck eq}
    \partial_t &p(\boldsymbol{x}, t)=-\boldsymbol{\nabla}^T\boldsymbol{j}(\boldsymbol{x}, t)\\
    \label{2-beads: probability flow}
    &\textrm{with } \boldsymbol{j}(\boldsymbol{x}, t)=(A\boldsymbol{x}-D\boldsymbol{\nabla})p(\boldsymbol{x}, t).
\end{align}
The particles undergo thermal fluctuations interacting with a heat bath at temperatures $T_1, T_2\ (T_1 > T_2)$, respectively (see \fref{fig: 2-beads_model}(a)). This model has also been investigated in the literature \cite{Li-Horowitz-Gingrich-Fakhri2019,Otsubo-Ito-Dechant-Sagawa2020} and others. Here, we obtain the distribution of states according to the literature \cite{Li-Horowitz-Gingrich-Fakhri2019}.
Provided that the steady-state distribution is expressed as the Gaussian distribution
\begin{align}
    p^\mathrm{ss}(\boldsymbol{x})=\frac{1}{2\pi\sqrt{\det \bar{C}}}
    \exp \left[-\frac{1}{2}\boldsymbol{x}^T\bar{C}^{-1}\boldsymbol{x}\right]
\end{align}
with a symmetric correlation matrix $\bar{C}$, the steady-state probability distribution is given by
\begin{align}
    \boldsymbol{j}^\mathrm{ss}=(A+D\bar{C}^{-1})\boldsymbol{x} p^\mathrm{ss}.
\end{align}
At this time, $\bar{C}$ satisfies the Lyapunov equation \cite{lyapunov1992general}
\begin{align}
    \label{2-beads: Lyapunov eq}
    A\bar{C}+\bar{C}A+2D=0.
\end{align}
The steady-state mean local velocity is plotted in \fref{fig: 2-beads_model}(b). Unlike in the case of a 2-dimensional vortex system, the flow is elliptical and its shape depends on the temperature of the heat bath. Therefore, it is difficult to determine the direction of $\boldsymbol{v}^\mathrm{ss}$ from physical considerations. The current defined below does not satisfy the condition of \eref{hk coeff}\eqref{ex coeff} and is not reduced to the total current average, and hence it is difficult to measure them experimentally.

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{2-beads_model.eps}
    \caption{\label{fig: 2-beads_model}
    (a) The sketch of the 2-beads model. The two beads (particles) are coupled to each other. The left (right) bead is influenced by thermal fluctuation from the hot (cold) reservoir. (b) The steady-state mean local velocity, $\boldsymbol{v}^\mathrm{ss}$. $\boldsymbol{v}^\mathrm{ss}$ is divided by $\mu k$ to be dimensionless, and the scale is multiplied by $1/10$ for illustration. (c) The time-evolution of the covariance matrix of the state, \eref{2-beads: state distribution}. The parameter used in calculation are $\mu=1.0, k=1.0, T_1=1.5, T_2=0.5$.
    }
\end{figure}

The solution of the Langevin equation (\eref{2-beads: Langevin eq}) is
\begin{align}
    \boldsymbol{x}(t)=\int^t_0 ds e^{A(t-s)}\sqrt{2}G\boldsymbol{\xi}(s)+e^{At}\boldsymbol{x}(0)\ (t\geq0).
\end{align}
Assuming $\bkt{\boldsymbol{x}(0)}=0$, the correlation matrix at time $t$ is calculated as
\begin{align}
    \label{2-beads: corr mat}
    C(t):&=\bkt{\boldsymbol{x}(t)\boldsymbol{x}(t)^T}\nonumber\\
        &=2\int^t_0 ds e^{A(t-s)}D^{A(t-s)}+e^{At}C(0)e^{At}.
\end{align}
This is plotted in \fref{fig: 2-beads_model}(c).
If we adopt the Gaussian distribution as the initial state, the state at time $t$ is given by the Gaussian distribution whose covariance matrix is $C(t)$:
\begin{align}
    \label{2-beads: state distribution}
    p(\boldsymbol{x}, t)=\frac{1}{2\pi\sqrt{\det C(t)}}
    \exp \left[-\frac{1}{2}\boldsymbol{x}^T{C(t)}^{-1}\boldsymbol{x}\right].
\end{align}
Since this is the Gaussian distribution, the average of the quantity in the shape of $\boldsymbol{x}^T H\boldsymbol{x}$ is computed as ${\boldsymbol{x}^T H\boldsymbol{x}}=\textrm{tr}(C^{1/2}HC^{1/2})=\textrm{tr}(HC)$. The second term on the right-hand side of \eref{2-beads: corr mat} vanishes in the long-time limit because the all eigenvalues of $A$ are negative. By contrast, the first term, denoted as $C_0(t)$, satisfies \eref{2-beads: Lyapunov eq} due to
\begin{align}
    \frac{d}{dt}C_0(t)=AC_0+C_0A+2D\xrightarrow{t\rightarrow\infty}0.
\end{align}
Therefore, the correlation matrix in the steady state is
\begin{align}
    \bar{C}=\lim_{t\rightarrow\infty}C_0(t)
    =\frac{1}{12k}
    \begin{bmatrix}
        7T_1+T_2 & 2(T_1+T_2)\\
        2(T_1+T_2) & T_1+7T_2
    \end{bmatrix}.
\end{align}

\subsection{Entropy production and current}
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{2-beads_TUR.eps}
    \caption{\label{fig: 2-beads_TUR}
    (a) The current averages in the 2-beads model. The current coefficient is $\boldsymbol{d}=D^{-1}A\boldsymbol{x}$ so that the total current agrees with the EP rate of the reservoirs. The initial state is the delta function localized at the origin. Therefore, at first, the total current average is negative because the energy is received as heat from the heat reservoir. Eventually, when it reaches the steady state, the excess current average vanishes and the housekeeping current average agrees with that of the total current. Since there is a steady heat flow from the high-temperature heat reservoir $T_1$ to the low-temperature heat reservoir $T_2$, the steady-state value of the current average is positive. (b) The current variance. This is initially zero because they are localized at the origin and increases as the particles diffuse due to thermal fluctuations, eventually reaching the steady-state value. (c) The EP rates. The total EP rate diverges at the initial time and decreases monotonically over time, saturating to a finite steady-state value. The excess EP rate is similar but has a steady-state value of $0$. On the other hand, the housekeeping EP rate monotonically increases from $0$ to reach the same steady-state value as the total system. (d) The estimation of each EP rate based on \eref{respective TUR}\eqref{max TUR}. The black and green lines correspond to the estimation $\widehat{\sigma^\mathrm{tot}}=(j_d^\mathrm{hk}+j_d^\mathrm{ex})^2/\calD_d, (j_d^\mathrm{hk}-j_d^\mathrm{ex})^2/\calD_d$ respectively. The gray line represents the sum of the estimation $\widehat{\sigma^\mathrm{tot}}=[(j_d^\mathrm{hk})^2+(j_d^\mathrm{ex})^2]/\calD_d$. The purple line shows the EP rate of the reservoirs $\widehat{\sigma^\mathrm{tot}}=j_d=\sigma^m$. The parameters used in calculation are $\mu=1.0, k=1.0, T_1=1.5, T_2=0.5$.
    }
\end{figure}
Now that we are ready, we calculate the currents and the entropy productions. The initial state is given by a delta function localized at the origin. This gives $C(0)=0$. Choosing $\boldsymbol{d}=D^{-1}A\boldsymbol{x}$ as the coefficient, the current is the EP rate in the heat bath. Each current average is
\begin{align}
    j_d&=
    \textrm{tr}\left(AD^{-1}(A+DC^{-1})C\right),\\
    j_d^\mathrm{hk}&=
    \textrm{tr}\left(AD^{-1}(A+D\bar{C}^{-1})C\right),
\end{align}
and
\begin{align}
    j_d^\mathrm{ex}=
    j_d-j_d^\mathrm{hk}=
    \textrm{tr}\left(A(C^{-1}-\bar{C}^{-1})C\right).
\end{align}
The current variance is computed as
\begin{align}
    \calD_d=\textrm{tr}(AD^{-1}AC).
\end{align}
These are plotted in \fref{fig: 2-beads_TUR}(a) and (b). Since the minimum energy of the system is $0$ at $t=0$, the energy flows into the system as heat, leading to $j_d<0$. For the housekeeping current, $j_d^\mathrm{hk}$ vanishes in the limit $t\rightarrow0$ because of $C\rightarrow0$. At $t\rightarrow\infty$, there is a steady heat flow and  $j_d^\mathrm{hk}\rightarrow j_d>0$. Due to this combination, the housekeeping current is always above $0$ and the excess current is always below $0$.
In the steady state $(t\rightarrow\infty)$, $j_d^ex$ vanishes because of $C\rightarrow\bar{C}$. On the other hand, when the detailed balance condition holds $(T_1=T_2)$, $D\propto I$ leads to $\bar{C}=-DA^{-1}=-A^{-1}D$ and $j_d^\mathrm{hk}$ vanishes.
As for fluctuations, $\calD_d$ increases monotonically and saturates to a stationary value as time passes and particles diffuse from near the origin.

The EP rates are computed as
\begin{align}
    \label{2-beads: tot EP rate}
    {\sigma^\mathrm{tot}}&=
    \textrm{tr}\left((A+DC^{-1})^TD^{-1}(A+DC^{-1})C\right),\\
    \label{2-beads: hk EP rate}
    {\sigma^\mathrm{hk}}&=
    \textrm{tr}\left((A+D\bar{C}^{-1})^TD^{-1}(A+D\bar{C}^{-1})C\right),
\end{align}
and
\begin{align}
    \label{2-beads: ex EP rate}
    {\sigma^\mathrm{ex}}&=
    {\sigma^\mathrm{tot}}-{\sigma^\mathrm{hk}}\nonumber\\
    &=\textrm{tr}\left((C^{-1}-\bar{C}^{-1})^TD(C^{-1}-\bar{C}^{-1})C\right).
\end{align}
We plot them in \fref{fig: 2-beads_TUR} (c). First, the total and excess EP rates diverge in the limit $t\rightarrow0$ because they include $C^{-1}$ which does not cancel out with $C\rightarrow0$, and the housekeeping EP rate with only $C$ terms vanishes in this limit. As in the previous example (2-dimensional flow), the excess EP rate diverges for the localized nonequilibrium state of the delta function and the housekeeping EP rate becomes $0$.
Over time, the housekeeping EP rate monotonically increases and saturates, and the excess EP rate decays to zero.
In the steady state, since the EP rate of the system is zero, the steady-state value reached by the total and housekeeping EP rates is the same as the corresponding steady-state value of the current average (see \fref{fig: 2-beads_TUR}(a)).

Finally, the estimation of the EP rates by TUR is plotted in \fref{fig: 2-beads_TUR} (d). For the EP rate of the total system, the green line is above the black line because the product of the currents is always negative in this setting (see \fref{fig: 2-beads_TUR}(a)). Around the initial time, the accuracy of the estimation using \eref{max TUR} is relatively high. As it approaches the steady state, the accuracy becomes worse, indicating that it is better to use the current as it is, i.e., the EP rate of the heat bath (purple line) for the estimation. This is because the mean of the current asymptotically approaches the mean of the EP rate for the total system, but the variance does not, making the estimation with TUR less accurate.

Since $j_d^\mathrm{hk}$ vanishes in the limit $t\rightarrow0$, all but the red line (and the purple line) converge to the same point. By contrast, $j_d^\mathrm{ex}$ vanishes in the limit $t\rightarrow\infty$, and all but the red line (and the purple line) intersect at the single point.

Around the initial time, the estimation of the excess EP rate has a similar trend to that of the total system. The accuracy deteriorates as it approaches the steady state, but it is obvious that it relaxes to zero in this case. On the other hand, the estimation of the housekeeping EP rate tends to be similar to that of the total system when the system is close to the steady state and is not accurate in any time region.

\section{Generalization and extension}
We assume that thermodynamic forces $D^{-1}\boldsymbol{v}^\alpha(\alpha=1, \cdots, n)$ satisfy the following two conditions, (i) the constraint condition: $\sum_{\alpha=1}^n\boldsymbol{v}^\alpha=\boldsymbol{v}$ and (ii) the orthogonal condition $\bkt{D^{-1}\boldsymbol{v}^\alpha,D^{-1}\boldsymbol{v}^\beta}\propto\delta_{\alpha,\beta}$. The corresponding current and EP are defined as $j_d^\alpha=\bkt{\boldsymbol{d},D^{-1}\boldsymbol{v}^\alpha}$ and $\sigma^\alpha:=||D^{-1}\boldsymbol{v}^\alpha||^2$ respectively.

Including the normalized vector fields $\boldsymbol{e}^\alpha=D^{-1}\boldsymbol{v}^\alpha/||D^{-1}\boldsymbol{v}^\alpha||$ in the basis, \eref{projective TUR} can be generalized to
\begin{align}
    \sum_{\alpha=1}^n \frac{(j_d^\alpha)^2}{\sigma^\alpha}\leq \calD_d
\end{align}
in the same way as \eref{derivation of projective TUR}. In \eref{projective TUR}, the chosen basis includes the housekeeping and excess vector fields $(\alpha=\mathrm{hk},\mathrm{ex})$, and the thermodynamic forces already satisfy the constraint condition. Therefore, even though the lower bound can be tightened by adding other components of the basis, $\boldsymbol{e}^\alpha$, the components have no information about the norm of the corresponding thermodynamic force, making $j_d^\alpha$ and $\sigma^\alpha,$ undefined.

For the thermodynamic forces and arbitrary signs $\chi^\alpha=\pm1\ (\alpha=1, \cdots, n)$, the following inequality holds:
\begin{align}
    &\left\langle\boldsymbol{d},D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha\right\rangle\nonumber\\
    &\leq\bkt{\boldsymbol{d},\boldsymbol{d}}\left\langle D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha,D^{-1}\sum_{\alpha=1}^n\chi^\alpha\boldsymbol{v}^\alpha\right\rangle\nonumber\\
    &=\bkt{\boldsymbol{d},\boldsymbol{d}}\bkt{D^{-1}\boldsymbol{v}, D^{-1}\boldsymbol{v}},
\end{align}
where the Cauchy-Schwartz inequality is used in the first line. From this inequality, \eref{max TUR} can be generalized  to
\begin{align}
    \sigma^\mathrm{tot}\geq\frac{1}{\calD_d}
    \max_{\chi^\alpha}\left(\sum_{\alpha=1}^n \chi^\alpha j_d^\alpha\right)^2.
\end{align}
The equality is achieved if and only if the coefficient satisfies $\boldsymbol{d}\propto D^{-1}\sum_{\alpha=1}^n \chi^\alpha \boldsymbol{v}^\alpha$. For the fixed coefficient, the best estimation of the dissipation is obtained for $\chi^\alpha$ which makes the signs of $\chi^\alpha j_d^\alpha$ aligned.
{The equality condition of \eref{max TUR} is obtained by setting $\alpha=\mathrm{hk},\mathrm{ex}$. For the case of $j_d^\mathrm{hk} j_d^\mathrm{ex}>0$, the equality condition is $\boldsymbol{d}={j_d^\mathrm{hk}}D^{-1}\boldsymbol{v}/{\sigma^\mathrm{hk}}\propto D^{-1}\boldsymbol{v}$ and the current coefficient is proportional to the thermodynamic force \cite{Gingrich-Horowitz2016}. By contrast, for the case of $j_d^\mathrm{hk} j_d^\mathrm{ex}<0$, the above relation is replaced by  $\boldsymbol{d}=-{j_d^\mathrm{hk}}D^{-1}(\boldsymbol{v}-2\boldsymbol{v}^\mathrm{ss})/{\sigma^\mathrm{hk}}\propto D^{-1}\boldsymbol{v}^\dagger$ (with $\boldsymbol{v}^\dagger:=\boldsymbol{v}-2\boldsymbol{v}^\mathrm{ss}$), which means that the current coefficient is proportional to the thermodynamic force in the dual dynamics \cite{Dechant-Sasa-continuous}.}

{While the single coefficient $d$ is used in inequalities \eqref{projective TUR} and \eqref{respective TUR}, two coefficients $d$ and $d'$ can be utilized to assess $\sigma^\mathrm{hk}$ and $\sigma^\mathrm{ex}$ respectively.}
\eref{respective TUR} can be modified for two coefficients as
\begin{align}
    \sigma^\mathrm{tot}\geq
    \frac{\left(j_{d}^\mathrm{hk}\right)^2}{D_{d}}
    +\frac{\left(j_{d'}^\mathrm{ex}\right)^2}{D_{d'}}.
\end{align}
This TUR can be applied for the case when the kind of coefficient is restricted by the experimental condition. The equality is obtained if and only if $\boldsymbol{d}\propto\boldsymbol{e}^\mathrm{hk},\boldsymbol{d}'\propto\boldsymbol{e}^\mathrm{ex}$ is satisfied. {Therefore, if we set $\boldsymbol{d}=\boldsymbol{f}\propto D^{-1}\boldsymbol{v}^\mathrm{ss}$ and $\boldsymbol{d}'=-\boldsymbol{\nabla} U\propto D^{-1}(\boldsymbol{v}-\boldsymbol{v}^\mathrm{ss})$ in the example of the two-dimensional Brownian motion, we can completely estimate the housekeeping and excess EP rate. Therefore, we can estimate all the EP rates by taking measurements for these particular two currents.}
Note that \eref{projective TUR} is still tighter as long as we focus on the estimation with the single coefficient.
Moreover, adding small coefficient $\delta d$ such that $\delta d\propto\boldsymbol{e}^\mathrm{ex},\delta d'\propto\boldsymbol{e}^\mathrm{hk}$, \eref{projective TUR} can be extended to
\begin{align}
    \sigma^\mathrm{tot}\geq
    \frac{\left(j_{d}^\mathrm{hk}\right)^2}{D_{d+\delta d}}
    +\frac{\left(j_{d'}^\mathrm{ex}\right)^2}{D_{d'+\delta d'}}.
\end{align}

\section{Markov jump processes}
For the Langevin dynamics, the orthogonal condition \eqref{orthogonal condition} is satisfied because the surface term vanishes due to its boundary condition. By contrast, this condition does not hold for Markov jump processes, and neither \eref{projective TUR} nor \eqref{max TUR} can be derived. On the other hand, as the counterpart of \eref{respective TUR}, the following TURs hold:
\begin{align}
    \label{Markov jump: respective TUR}
    \sigma^\mathrm{hk}\geq c_0\frac{\left(j_d^\mathrm{hk}\right)^2}{\tilde{D}_d},\
    \sigma^\mathrm{ex}\geq c_0\frac{\left(j_d^\mathrm{ex}\right)^2}{\tilde{D}_d}.
\end{align}
Contrary to \eref{respective TUR}, the constant $c_0=0.896\dots$ is required and the variance of the current $\tilde{D}_d$ is also different. It is nontrivial whether the average and variance of the current are directly measurable or not. Nonetheless, we see that these take a simple form (\eref{QD: hk current}\eqref{QD: ex current}\eqref{QD: HS variance}) for the example of the quantum dot mentioned below.

Compared to the conventional TUR, \eref{TUR},
\begin{align}
    \label{Markov jump: sign condition}
    \frac{(j_d)^2}{\calD_d}
    =\frac{1}{c_0}\frac{\tilde{D}_d}{\calD_d}
    \left(c_0\frac{\left(j_d^\mathrm{hk}\right)^2}{\tilde{D}_d}+c_0\frac{\left(j_d^\mathrm{ex}\right)^2}{\tilde{D}_d}\right)
    +\frac{2j_d^\mathrm{hk} j_d^\mathrm{ex}}{\calD_d}
\end{align}
means that, roughly speaking, it is better to estimate the total dissipation by decomposing it into the housekeeping and excess parts when the product of the currents $j_d^\mathrm{hk} j_d^\mathrm{ex}$ is negative.

\subsection{Derivation}
We assume that the system is coupled to multiple reservoirs and its discrete state $x$ obeys the master equation:
\begin{align}
    \dot{p}_{x}
    =\sum_\nu\sum_{x'}R_{xx'}^\nu p_{x'}
    =\sum_\nu\sum_{x'}j_{xx'}^\nu,
\end{align}
where $j_{xx'}^\nu:=R_{xx'}^\nu p_{x'}-R_{x'x}^\nu p_{x}=K_{xx'}^\nu-K_{x'x}^\nu$ is the probability flow from the state $x'$ to $x$ driven by the reservoir $\nu$ $(K_{xx'}^\nu:=R_{xx'}^\nu p_{x'})$.
In the steady state, this satisfies the balanced condition $\sum_\nu\sum_{x'}{j^\mathrm{ss}}_{xx'}^\nu=0$.
The EP rates are given by \cite{Three-faces-Master-equation,3DFT-Esposito2010}
\begin{align}
    \label{Markov jump: total EP}
    \sigma^\mathrm{tot}&=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{K_{x'x}^\nu},\\
    \label{Markov jump: hk EP}
    \sigma^\mathrm{hk}&=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{\tilde{K}_{xx'}^\nu},
\end{align}
and
\begin{align}
    \label{Markov jump: ex EP}
    \sigma^\mathrm{ex}=\sum_\nu\sum_{x\neq x'}
    K_{xx'}^\nu\ln\frac{K_{xx'}^\nu}{\tilde{K}_{x'x}^\nu}.
\end{align}
We define $\tilde{K}$ as $\tilde{K}_{xx'}^\nu:=\tilde{R}_{xx'}^\nu p_{x'}$ with the dual transition rate $\tilde{R}_{xx'}^\nu:=R_{x'x}^\nu p^\mathrm{ss}_{x}/p^\mathrm{ss}_{x'}$.

Alhough the mean local velocity is not defined conventionally for Markov jump processes, we introduce it with the definition $v_{xx'}^\nu:=j_{xx'}^\nu/p_{x'}$ in analogy with the Langevin dynamics. Dividing the probability flow $j_{xx'}$ by the probability $p_{x'}$, the mean local velocity captures the net current of the transition. On the contrary to the probability flow, the mean local velocity does not satisfy the skewed symmetry, i.e., $v_{xx'}^\nu\neq v_{x'x}^\nu$.

The total current average is expressed as
\begin{align}
    \label{Markov jump: def of current}
    j_d(t)
    &=\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu K_{xx'}^\nu
    =\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu j_{xx'}^\nu\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu v_{xx'}^\nu p_{x'}.
\end{align}
We define the housekeeping and excess current averages as
\begin{align}
    \label{Markov jump: def of hk current}
    j_d^\mathrm{hk}(t)
    :&=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    (K_{xx'}^\nu-\tilde{K}_{xx'})\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu {v^\mathrm{ss}}_{xx'}^{\nu} p_{x'}
\end{align}
and
\begin{align}
    \label{Markov jump: def of ex current}
    j_d^\mathrm{ex}(t)
    :&=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    (K_{xx'}^\nu+\tilde{K}_{xx'})\nonumber\\
    &=\frac{1}{2}\sum_\nu\sum_{x\neq x'}d_{xx'}^\nu
    ({v^\mathrm{ss}}_{xx'}^{\nu}-v_{xx'}^\nu)p_{x'}
\end{align}
respectively. Then, the current decomposition $j_d=j_d^{\textrm{hx}}+j_d^\mathrm{ex}$ holds by definition. These current averages have almost the same form as their counterparts in Langevin dynamics and a simple calculation confirms that these coincide in the continuous limit.
In addition, we define both the housekeeping and excess current variances as
\begin{subequations}
\begin{align}
    \tilde{D}_d:&=\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)\\
    &=\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{x'x}^\nu).
\end{align}
\end{subequations}

We are now ready to derive \eref{Markov jump: respective TUR}.
\begin{align}
    \sigma^\mathrm{hk}\tilde{D}_d
    =\sum_\nu&\sum_{x\neq x'}\left(K_{xx'}^\nu
    \ln\frac{K_{xx'}^\nu}{\tilde{K}_{xx'}^\nu}
    +\tilde{K}_{xx'}^\nu-K_{xx'}^\nu\right)\nonumber\\
    &\cdot\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)\nonumber\\
    \geq c_0 &\sum_\nu\sum_{x\neq x'}
    \frac{(K_{xx'}^\nu-\tilde{K}_{xx'}^\nu)^2}
    {K_{xx'}^\nu+\tilde{K}_{xx'}^\nu}\nonumber\\
    &\cdot\frac{1}{4}\sum_\nu\sum_{x\neq x'}(d_{xx'}^\nu)^2
    (K_{xx'}^\nu+\tilde{K}_{xx'}^\nu)
    \nonumber\\
    \geq c_0 &\left(\frac{1}{2}\sum_\nu\sum_{x\neq x'}
    d_{xx'}^\nu(K_{xx'}^\nu-\tilde{K}_{xx'}^\nu)\right)^2\nonumber\\
    =c_0&\left(j_d^\mathrm{hk}\right)^2
\end{align}
The derivation for the excess part follows in exactly the same way. We used an equality $\sum_x K_{xx'}^\nu=\sum_x \tilde{K}_{xx'}^\nu=0$, an inequality
\begin{align}
    \label{basic ineq c=0.896}
    a\ln\frac{a}{b}+b-a\geq \frac{c_0(a-b)^2}{a+b}
\end{align}
($a,b>0, c_0=0.896\dots$) \cite{Shiraishi-Funo-Saito,Shiraishi-Saito-Tasaki2016}, and the Cauchy-Schwaltz inequality in each transformation.

\subsection{The necessity of $c_0$}
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{MJ_RW.eps}
    \caption{\label{fig: MJ_RW}
    (a) The sketch of the random walk. A single particle is hopping among sites driven by a heat reservoir.
    (b)(c)The test of the short-time TUR for each EP rate in the $1$-dimensional random walk (\eref{1dRW: tot TUR}\eqref{1dRW: HS TUR}). The parameters of the model are chosen at random from $A\in[-4, 4],\ R_x\in[0, 1],\ d_x\in[-1, 1]$. The probability distribution of the system $p_x$ is also at random. The sample size is $10^5$. The system size is $M=3$ for (b), and $M=5$ for (c).
    }
\end{figure}
We see the constant $c_0=0.896\dots$ is required for Markov jump processes in the simple one-dimensional random walk model. There are $M$ sites in the one direction with the periodic boundary condition, and a single particle is hopping among them, being driven by a heat reservoir (see \fref{fig: MJ_RW} (a)). We denote the transition rate to the forward (backward) direction at site $x(=1,2,\dots,M)$ as $R_{x,+}\ (R_{x,-})$. We assume the local detailed balance condition for this system, $\frac{R_{x,+}}{R_{x+1,-}}=e^A$, with the thermodynamic force $A$. Thus, the transition rate is written by $R_{x,+}:=R_xe^{A/2},\ R_{x,-}:=R_{x-1}e^{-A/2}$.
The current coefficient obeys the skewed symmetry:
\begin{align}
    d_{xx'}=\begin{cases}
            d_{x-1} & (x'=x-1)\\
            -d_x & (x'=x+1)\\
            0 & (\textrm{otherwise})
            \end{cases}.
\end{align}

In this setup, we examine the TURs
\begin{align}
    \label{1dRW: tot TUR}
    q:&=\frac{\sigma^\mathrm{tot} {D}_d}
    {\left(j_d\right)^2}\geq1\\
    \label{1dRW: HS TUR}
    q:&=\frac{\sigma^l \tilde{D}_d}
    {\left(j_d^l\right)^2}\geq c_0
    \ (l=hk,\mathrm{ex})
\end{align}
by plotting $q$ for many systems with different probability distributions, transition rates, and current coefficients.
In \fref{fig: MJ_RW} (b), the lower bound $q\geq1$ is violated for some blue and red points, meaning that \eref{Markov jump: respective TUR} does not hold without the constant $c_0$. At the same time, there are few points near the bound $q\geq c_0$, which suggests that the quality condition is hard to satisfy.

Since the conventional TUR \eqref{TUR} is satisfied in the equilibrium limit, the lower bound $q\geq1$ gets tighter for small thermodynamics forces. The blue plot behaves similarly to the black plot because the housekeeping EP captures the violation of the detailed balance condition. On the other hand, the excess EP comes from the deviation from the steady state. Therefore, the red points are widely distributed and the tightness of the lower bound is almost independent of $A$.

The samples close to the lower bound get scarce when we increase the size of the system (see \fref{fig: MJ_RW} (c)). This suggests that achieving equality requires the state and coefficients to be tuned commensurately and this tuning gets extremely hard for large systems. Therefore, the lower bound $q\geq$ is practically never broken for sufficiently large systems, and the same TUR as the Langevin dynamics, \eref{respective TUR}, can be applied.

\subsection{Example: quantum dot}
The EP is estimated using TUR \eqref{Markov jump: respective TUR} in the model of the quantum dot. The number of levels is one, and we denote state 1 when the particle is occupied and state 0 when it is empty. The transition rate of the heat reservoir $\nu(=\textrm{h,c})$ is given by the Fermi golden rule as $R_{10}^\nu=\Gamma_0 f_\nu,\ R_{01}^\nu=\Gamma_0(1-f_\nu)\ (f_\nu:=1/(1+e^{\beta_\nu(E-\mu_\nu)}))$ \cite{Benenti-Casati-Saito-Whitney2017, Esposito-Lindenberg-Broeck2009-QD}. $E$ is the energy of the level and $\mu_\nu$ is the chemical potential of the heat reservoir $\nu$. The coupling strength to the heat reservoir is assumed to be the same $\Gamma_0$, which is sufficiently small that the sequential tunneling approximation holds.

The steady state is easily calculated:
\begin{align}
    p_0^\mathrm{ss}=\frac{R_{01}}{R_{01}+R_{10}},\
    p_1^\mathrm{ss}=\frac{R_{10}}{R_{01}+R_{10}}
    =\frac{1}{2}(f_{\textrm{h}}+f_{\textrm{c}}).
\end{align}
The distinguished feature of 2-state systems is that the overall dual transition rate matches the original transition rate, that is,
\begin{align}
    \tilde{R}_{xx'}=\sum_\nu\tilde{R}_{xx'}^\nu=\sum_\nu\frac{{R}_{x'x}^\nu p_{x}^\mathrm{ss}}{p_{x'}^\mathrm{ss}}={R}_{xx'}.
\end{align}

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{MJ_QD.eps}
    \caption{\label{fig: MJ_QD}
    (a) The averages of the particle current in the quantum dot model. All of the averages are linear in terms of $p_1$. The vertical dot line depicts the steady state, $p_1=p_1^\mathrm{ss}$. (b) Each EP rate. In the steady state, the excess EP rate vanishes and the housekeeping EP rate coincides with the total one. (c) The estimation of each EP rate based on \eref{TUR}\eqref{Markov jump: respective TUR} using the particle current. $\widehat{\sigma^l}$ denotes the estimation of $\sigma^l\ (l=\mathrm{tot},\mathrm{hk},\mathrm{ex})$. The gray line represents the estimation with $\widehat{\sigma^\mathrm{tot}}=c_0[(j_d^\mathrm{hk})^2+(j_d^\mathrm{ex})^2]/\tilde{D}_d$. The parameters used in calculation are $\beta_\textrm{h}(E-\mu_\textrm{h})=1.0,\beta_\textrm{c}(E-\mu_\textrm{c})=3.0$.
    }
\end{figure}
As a current, consider a particle current through the system in the direction from the high-temperature heat reservoir to the low-temperature heat reservoir. The coefficient is given by $d_{xx'}^\nu=M_{xx'}^\nu$ with
\begin{align}
    M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}=\frac{(N_x-N_{x'})}{2}.
\end{align}
$N_x$ is the number of particles of the system in state $x$, where $N_0=0,N_1=1$. Since we are dealing with non-stationary conditions here, we need to consider the exchange of particles with both heat reservoirs.
Computing each current average based on \eref{Markov jump: def of current}-\eqref{Markov jump: def of ex current},
\begin{align}
    \label{QD: tot current}
    j_d&=\sum_\nu M_{10}^\nu(R_{10}^\nu p_0-R_{01}^\nu p_1)=\frac{\Gamma_0}{2}(f_h-f_c)\\
    \label{QD: hk current}
    j_d^\mathrm{hk}
    &=\frac{1}{2}\sum_\nu M_{10}^\nu
    \left[(R_{10}^\nu p_0-\tilde{R}_{10}^\nu p_0)
    -(R_{01}^\nu p_1-\tilde{R}_{01}^\nu p_1)\right]\nonumber\\
    &=\frac{\Gamma_0}{4}\left(\frac{p_0}{p_0^\mathrm{ss}}
    +\frac{p_1}{p_1^\mathrm{ss}}\right)(f_h-f_c)\\
    \label{QD: ex current}
    j_d^\mathrm{ex}
    &=j_d^\mathrm{tot}-j_d^\mathrm{hk}\nonumber\\
    &=\frac{\Gamma_0}{4}\left(\left(1-\frac{p_0}{p_0^\mathrm{ss}}\right)
    +\left(1-\frac{p_1}{p_1^\mathrm{ss}}\right)\right)(f_h-f_c)
\end{align}
These are plotted in \fref{fig: MJ_QD} (a). It can be seen that the current mean of the total system is always constant, independent of the state probability $p_x$. This is due to the two-state nature and the property $M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}$. The housekeeping and excess current averages are the current averages of the total system multiplied by $({p_0}/{p_0^\mathrm{ss}}+{p_1}/{p_1^\mathrm{ss}})/2$ and $((1-{p_0}/{p_0^\mathrm{ss}})+(1-{p_1}/{p_1^\mathrm{ss})})/2$, respectively. The housekeeping current average is always positive and the excess current average changes its sign at $p_1=p_1^\mathrm{ss}$.

On the other hand, each current variance is computed as
\begin{align}
    \label{QD: variance}
    \calD_d
    &=\frac{1}{8}\sum_\nu(R_{10}^\nu p_0+R_{01}^\nu p_1)\\
    \label{QD: HS variance}
    \tilde{D}_d&=
    \frac{1}{16}\sum_\nu
    \left[(R_{10}^\nu p_0+\tilde{R}_{10}^\nu p_0)
    +(R_{01}^\nu p_1+\tilde{R}_{01}^\nu p_1)\right]\nonumber\\
    &=\frac{1}{8}\sum_\nu(R_{10}^\nu p_0+R_{01}^\nu p_1)
    =\calD_d
\end{align}
and is the same value. This is also due to the special properties of the two states and $M^{\textrm{h}}_{xx'}=-M^{\textrm{c}}_{xx'}$, where the variance generally takes different values.

In this model, the EP rates calculated by \eref{Markov jump: total EP}-\eqref{Markov jump: ex EP} are shown in \fref{fig: MJ_QD} (b). The excess EP rate takes a minimum value of $0$ at the steady state $p_1=p_1^\mathrm{ss}$ and increases rapidly away from it. The housekeeping EP rate is linear and monotonically increasing because $\ln(K_{x'x}^\nu/\mathcal{K}_{x'x}^\nu)$ is independent of $p_1$. Also, from a simple calculation, it can be shown that this slope increases as the thermal force $A=\beta_\textrm{c}(E-\mu_\textrm{c})-\beta_\textrm{h}(E-\mu_\textrm{h})$ is larger. Reflecting these behaviors, the EP rate for the total system reaches a finite minimum at $p_1$ which is slightly smaller than $p_1^\mathrm{ss}$.

Finally, the EP rate estimations given by the TUR \eqref{Markov jump: respective TUR} are plotted in \fref{fig: MJ_QD} (c). The accuracy of the EP rate for the total system tends to have a maximum value of about $90\%$ near the minimum of the EP rate, and the accuracy decreases as one moves away from the minimum. For the excess EP rate, the overall accuracy is less than $10\%$, but the accuracy improves as the steady state is approached. On the other hand, for the housekeeping EP rate, the overall accuracy is high, and as $p_1$ increases, the accuracy gradually decreases. The gray line in the figure shows the estimated EP rate for the entire system from the lower bounds of the housekeeping and excess EP rates. From \eref{Markov jump: sign condition}, it is better to decompose and estimate the housekeeping and excess components approximately when the product of the currents $j_d^\mathrm{hk} j_d^\mathrm{ex}$ becomes negative. In fact, for $p_1>p_1^\mathrm{ss}$ where the product of the currents is negative (see \fref{fig: MJ_QD}(a)), the gray line is located above the black line except near its boundary, and the EP rate for the total system is estimated more accurately.

\section{Details of the example: 2-dimensional flow}
In this section, we present some calculations and figures to supplement the first example in the main text.
Since $D$ is proportional to the identity matrix, the coefficient proportional to the nonconservative force, $\boldsymbol{d}=\boldsymbol{d}_\theta:=d_\theta[-x_2, x_1]^T$, satisfies \eref{hk coeff}, and the coefficient proportional to the conservative force, $\boldsymbol{d}=\boldsymbol{d}_r:=-d_r[x_1, x_2]^T$, satisfies \eref{ex coeff}. Here, we consider coefficients including only $\boldsymbol{d}_\theta,\boldsymbol{d}_r$, but note that this does not mean that all coefficients can be expanded by them.

The current averages with $\boldsymbol{d}=\boldsymbol{d}_\theta+\boldsymbol{d}_r$ are given by
\begin{align}
    \label{2Dflow: hk coeff}
    j_d^\mathrm{hk}&=j_{d_\theta}^\mathrm{hk}=j_{d_\theta}
    =\frac{2\kappa T'd_\theta}{k}
\end{align}
and
\begin{align}
    \label{2Dflow: ex coeff}
    j_d^\mathrm{ex}&=j_{d_r}^\mathrm{ex}=j_{d_r}
    =2 T'\left(1-\frac{T}{T'}\right)d_r.
\end{align}
While $J_{D}^\mathrm{hk}$ has the same sign regardless of $T'$, $J_{D}^\mathrm{ex}$ switches positive and negative at $T'=T$. The current variances with $\boldsymbol{d}=\boldsymbol{d}_\theta,\ \boldsymbol{d}_r$ are computed as
\begin{align}
    D_{d_\theta}=\frac{2 TT'{d_\theta}^2}{k},\ \textrm{and }
    D_{d_r}=\frac{2 TT'{d_r}^2}{k}
\end{align}
respectively. Since $\boldsymbol{d}_\theta^T\boldsymbol{d}_r=0$ (and $D$ is proportional to the identity matrix),
the current with $\boldsymbol{d}=\boldsymbol{d}_\theta+\boldsymbol{d}_r$ satisfies the decomposition of the variance as well:
\begin{align}
    \label{2Dflow: decomp of fluctuation}
    \calD_d=D_{d_\theta}+D_{d_r}.
\end{align}
For the coefficient $\boldsymbol{d}=\boldsymbol{F}$, the averages are plotted in \fref{fig: 2dflow} (b) and the variance is plotted in \fref{fig: 2dflow_variance}.

\begin{figure}
    \centering
    \includegraphics[width=0.8\linewidth]{2dflow_variance.eps}
    \caption{\label{fig: 2dflow_variance}
    The variance of the power exerted on the system in the 2-dimensional flow model. Due to the decomposition of the coefficient, \eref{2Dflow: decomp of fluctuation}, the black line agrees with the sum of the blue and red lines. The parameters used in calculation are $\mu=1.0, k=1.0, \kappa=2.0, T=1.5$.
    }
\end{figure}

Next, we calculate the EP rates. For the current coefficient,
\begin{align}
    \boldsymbol{d}=D^{-1}\boldsymbol{v}
    &=\frac{1}{T}(-\boldsymbol{\nabla} U+\boldsymbol{f}-T\boldsymbol{\nabla}\ln p)\nonumber\\
    &=\frac{\kappa}{T}\boldsymbol{d}_\theta+\frac{k}{T}
    \left(1-\frac{T}{T'}\right)\boldsymbol{d}_r,
\end{align}
the housekeeping (excess) EP rate is given as its housekeeping (excess) current due to the orthogonal condition \eqref{orthogonal condition}:
\begin{align}
    \sigma^\mathrm{hk}&=\frac{2\kappa^2}{k}\frac{T'}{T},
\end{align}
and
\begin{align}
    \sigma^\mathrm{ex}&=\frac{2\kappa^2}{k}\frac{T'}{T}\left(\frac{k}{\kappa}\right)^2
    \left(1-\frac{T}{T'}\right)^2.
\end{align}
The total EP rate is just the sum of them:
\begin{align}
    \sigma^\mathrm{tot}
    =\frac{2\kappa^2}{k}\frac{T'}{T}
    \left[1+\left(\frac{k}{\kappa}\right)^2
    \left(1-\frac{T}{T'}\right)^2\right].
\end{align}
These are plotted in \fref{fig: 2dflow} (c). {$\sigma^\mathrm{ex}$ and $\sigma^\mathrm{tot}$ diverges at $T'\rightarrow0$ because the system state takes the form of the delta-function distribution which is an extremely nonequilibrium state.}

\section{Details of the example: rocking ratchet}
In this section, we present some calculations and figures to supplement the second example in the main text.
\subsection{Model}
The Langevin equation of the rocking ratchet is
\begin{align}
    \dot{x}=F(x,t)+\xi(t)=-\partial_x U(x)+R(t)+\xi(t),
\end{align}
where the potentials are periodic in space and time: $U(x+L)=U(x)$ and $R(t\calT)=R(t)$.
The thermal noise $\xi(t)$ is the white Gaussian noise, i.e., $\bkt{\xi(t)}=0$ and $\xi(t)\xi(t')=\delta(t-t')$. The corresponding Fokker-Planck equation is
\begin{align}
    \label{rocking: Fokker-Planck equation}
    \partial_t p(x,t)=-\partial_x j(x,t)=-\partial_x[F(x,t)-D\partial_x]p(x,t),
\end{align}
and the mean local velocity is expressed as $v(x,t):=j(x,t)/p(x,t)=F-D\partial_x \ln p$. Since the system is periodic in space, the probability $p$ and the probability current $j$ are punishingly small. We assume that these quantities are also periodic, $p(x+L,t)=p(x,t), j(x,t)=j(x+L,t)$, and take the summation $\hat{p}(x, t):=\sum_{n=-\infty}^\infty p(x+nL,t),\hat{j}(x, t):=\sum_{n=-\infty}^\infty j(x+nL,t)$. Then, the reduced quantities $\hat{p}$ and $\hat{j}$ also get periodic in space and have finite values, satisfying the same form of the Fokker-Planck equation as \eref{rocking: Fokker-Planck equation}. This assumption is plausible if we consider the periodic steady state or prepare the initial state periodic in space.
{The instantaneous steady state is given by a modified canonical distribution:
\begin{align}
    p^\mathrm{ss}(x, t)=\frac{\int^{x+L}_x dye^{[V(y,t)-V(x,t)]/D}}{\int_0^L dx\int^{x+L}_x dye^{[V(y,t)-V(x,t)]/D}},
\end{align}
where $V(x, t):=U(x)-R(t)x$ and $\int^{x+L}_x dye^{V(y, t)/D}$ is required to satisfy the periodicity $p^\mathrm{ss}(x+L)=p^\mathrm{ss}(x)$.}
We define the reduced local mean velocity as $\hat{v}(x, t):=\hat{j}(x,t)/\hat{p}(x,t)$ which coincides with the usual definition $v$ as long as the probability has the periodicity $p(x+L,t)=p(x,t)$. These reduced quantities are denoted as $p,j,v$ in the main text. The currents are EP rates can be calculated in the usual way, but the integration range is replaced by $[0,L]$.

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{rocking_mlv.eps}
    \caption{\label{fig: rocking_mlv}
    The steady-state mean local velocity in the rocking ratchet model. Since the nonconservative force vanishes at $t/T=0,0.5$, the mean local velocity vanishes as well at this time. The plot has kinks at $x/L=0, 0.8$ in accordance with the indifferentiability of the potential $U(x)$. Apart from these points, the plot is approximately flat. The parameters used in calculation are $D = 0.1, T=0.7, a=0.8, R_0 = 6.0$.
    }
\end{figure}

\subsection{The coefficient choice}
In the main text, we used the particle current with $d(x)=1$. Here, we consider another choice of coefficient. We plot the mean local velocity for the instantaneous steady state in \fref{fig: rocking_mlv}. $v^\mathrm{ss}$ tends to be constant except for the top and valley of the potential $U(x)$, which suggests that the coefficient $d(x)=1$ approximately satisfies the condition $d\propto e^\mathrm{hk}$ (\eref{hk coeff}).
Consequently,$j_d^\mathrm{ex}$ is small compared to $j_d^\mathrm{hk}$ and the estimation of the excess EP rate does not have high accuracy. (see \fref{fig: rocking} (a) and (d)).

Since the mean local velocity is expected to consist of low-frequency Fourier components, we use the coefficient $d(x)=\cos(2\pi x/L),\sin(2\pi x/L)$ instead here. We plot the current averages in \fref{fig: rocking_cossin}(a)(b). The housekeeping current is decreased in comparison with the case $d(x)=1$, and is now comparable to the excess current. We can see the improvement of the estimation in the first half of the period (see \fref{fig: rocking_cossin}(c)(d)).
Although $\cos(2\pi x/L)$ and $\sin(2\pi x/L)$ are not orthogonal to each other in terms of the inner product $\bkt{\cdot,\cdot}$, the currents and estimations tend to compensate each other. Combining these results, we can estimate roughly over $50\%$ of the excess EP rate.
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{rocking_cossin.eps}
    \caption{\label{fig: rocking_cossin}
    (a)(b) Each current average with $d(x)=\cos(2\pi x/L)$ for (a) and $d(x)=\sin(2\pi x/L)$ for (b).  (c)(d) The estimation of the excess EP rate based on \eref{projective TUR} (green line) and \eref{respective TUR} (red line) with $d(x)=\cos(2\pi x/L)$ for (c) and $d(x)=\sin(2\pi x/L)$ for (d).
    The parameter used in calculation are $D = 0.1, T=0.7, a=0.8, R_0 = 6.0$.
    }
\end{figure}

\section{2-beads model}
In the first example in the main text, the direction of $\boldsymbol{e}^\mathrm{hk},\boldsymbol{e}^\mathrm{ex}$ or $\boldsymbol{v}^\mathrm{ss},\boldsymbol{v}-\boldsymbol{v}^\mathrm{ss}$ was obvious. Also, as noted in the discussion, the direction of $\boldsymbol{v}^\mathrm{ss}$ is considered experimentally estimable. In this section, we consider a two-dimensional heat conduction system in the absence of prior knowledge of the steady state.

\subsection{Setup}
Consider two coupled Brownian particles governed by the Langevin equation
\begin{align}
    \label{2-beads: Langevin eq}
    \dot{\boldsymbol{x}}&=\mu A\boldsymbol{x}+\sqrt{2}G\boldsymbol{\xi}(t)\\
    &\textrm{with  }
    A=\mu k\begin{bmatrix}
                -2 & 1\\
                1 & -2
            \end{bmatrix}\
    \textrm{and } G=\begin{bmatrix}
                \sqrt{\mu T_1} & 0\\
                0 & \sqrt{\mu T_2}
            \end{bmatrix}
\end{align}
and the Fokker-Planck equation
\begin{align}
    \label{2-beads: Fokker-Planck eq}
    \partial_t &p(\boldsymbol{x}, t)=-\boldsymbol{\nabla}^T\boldsymbol{j}(\boldsymbol{x}, t)\\
    \label{2-beads: probability flow}
    &\textrm{with } \boldsymbol{j}(\boldsymbol{x}, t)=(A\boldsymbol{x}-D\boldsymbol{\nabla})p(\boldsymbol{x}, t).
\end{align}
The particles undergo thermal fluctuations interacting with a heat bath at temperatures $T_1, T_2\ (T_1 > T_2)$, respectively (see \fref{fig: 2-beads_model}(a)). This model has also been investigated in the literature \cite{Li-Horowitz-Gingrich-Fakhri2019,Otsubo-Ito-Dechant-Sagawa2020} and others. Here, we obtain the distribution of states according to the literature \cite{Li-Horowitz-Gingrich-Fakhri2019}.
Provided that the steady-state distribution is expressed as the Gaussian distribution
\begin{align}
    p^\mathrm{ss}(\boldsymbol{x})=\frac{1}{2\pi\sqrt{\det \bar{C}}}
    \exp \left[-\frac{1}{2}\boldsymbol{x}^T\bar{C}^{-1}\boldsymbol{x}\right]
\end{align}
with a symmetric correlation matrix $\bar{C}$, the steady-state probability distribution is given by
\begin{align}
    \boldsymbol{j}^\mathrm{ss}=(A+D\bar{C}^{-1})\boldsymbol{x} p^\mathrm{ss}.
\end{align}
At this time, $\bar{C}$ satisfies the Lyapunov equation \cite{lyapunov1992general}
\begin{align}
    \label{2-beads: Lyapunov eq}
    A\bar{C}+\bar{C}A+2D=0.
\end{align}
The steady-state mean local velocity is plotted in \fref{fig: 2-beads_model}(b). Unlike in the case of a 2-dimensional vortex system, the flow is elliptical and its shape depends on the temperature of the heat bath. Therefore, it is difficult to determine the direction of $\boldsymbol{v}^\mathrm{ss}$ from physical considerations. The current defined below does not satisfy the condition of \eref{hk coeff}\eqref{ex coeff} and is not reduced to the total current average, and hence it is difficult to measure them experimentally.

\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{2-beads_model.eps}
    \caption{\label{fig: 2-beads_model}
    (a) The sketch of the 2-beads model. The two beads (particles) are coupled to each other. The left (right) bead is influenced by thermal fluctuation from the hot (cold) reservoir. (b) The steady-state mean local velocity, $\boldsymbol{v}^\mathrm{ss}$. $\boldsymbol{v}^\mathrm{ss}$ is divided by $\mu k$ to be dimensionless, and the scale is multiplied by $1/10$ for illustration. (c) The time-evolution of the covariance matrix of the state, \eref{2-beads: state distribution}. The parameter used in calculation are $\mu=1.0, k=1.0, T_1=1.5, T_2=0.5$.
    }
\end{figure}

The solution of the Langevin equation (\eref{2-beads: Langevin eq}) is
\begin{align}
    \boldsymbol{x}(t)=\int^t_0 ds e^{A(t-s)}\sqrt{2}G\boldsymbol{\xi}(s)+e^{At}\boldsymbol{x}(0)\ (t\geq0).
\end{align}
Assuming $\bkt{\boldsymbol{x}(0)}=0$, the correlation matrix at time $t$ is calculated as
\begin{align}
    \label{2-beads: corr mat}
    C(t):&=\bkt{\boldsymbol{x}(t)\boldsymbol{x}(t)^T}\nonumber\\
        &=2\int^t_0 ds e^{A(t-s)}D^{A(t-s)}+e^{At}C(0)e^{At}.
\end{align}
This is plotted in \fref{fig: 2-beads_model}(c).
If we adopt the Gaussian distribution as the initial state, the state at time $t$ is given by the Gaussian distribution whose covariance matrix is $C(t)$:
\begin{align}
    \label{2-beads: state distribution}
    p(\boldsymbol{x}, t)=\frac{1}{2\pi\sqrt{\det C(t)}}
    \exp \left[-\frac{1}{2}\boldsymbol{x}^T{C(t)}^{-1}\boldsymbol{x}\right].
\end{align}
Since this is the Gaussian distribution, the average of the quantity in the shape of $\boldsymbol{x}^T H\boldsymbol{x}$ is computed as ${\boldsymbol{x}^T H\boldsymbol{x}}=\textrm{tr}(C^{1/2}HC^{1/2})=\textrm{tr}(HC)$. The second term on the right-hand side of \eref{2-beads: corr mat} vanishes in the long-time limit because the all eigenvalues of $A$ are negative. By contrast, the first term, denoted as $C_0(t)$, satisfies \eref{2-beads: Lyapunov eq} due to
\begin{align}
    \frac{d}{dt}C_0(t)=AC_0+C_0A+2D\xrightarrow{t\rightarrow\infty}0.
\end{align}
Therefore, the correlation matrix in the steady state is
\begin{align}
    \bar{C}=\lim_{t\rightarrow\infty}C_0(t)
    =\frac{1}{12k}
    \begin{bmatrix}
        7T_1+T_2 & 2(T_1+T_2)\\
        2(T_1+T_2) & T_1+7T_2
    \end{bmatrix}.
\end{align}

\subsection{Entropy production and current}
\begin{figure}
    \centering
    \includegraphics[width=1.0\linewidth]{2-beads_TUR.eps}
    \caption{\label{fig: 2-beads_TUR}
    (a) The current averages in the 2-beads model. The current coefficient is $\boldsymbol{d}=D^{-1}A\boldsymbol{x}$ so that the total current agrees with the EP rate of the reservoirs. The initial state is the delta function localized at the origin. Therefore, at first, the total current average is negative because the energy is received as heat from the heat reservoir. Eventually, when it reaches the steady state, the excess current average vanishes and the housekeeping current average agrees with that of the total current. Since there is a steady heat flow from the high-temperature heat reservoir $T_1$ to the low-temperature heat reservoir $T_2$, the steady-state value of the current average is positive. (b) The current variance. This is initially zero because they are localized at the origin and increases as the particles diffuse due to thermal fluctuations, eventually reaching the steady-state value. (c) The EP rates. The total EP rate diverges at the initial time and decreases monotonically over time, saturating to a finite steady-state value. The excess EP rate is similar but has a steady-state value of $0$. On the other hand, the housekeeping EP rate monotonically increases from $0$ to reach the same steady-state value as the total system. (d) The estimation of each EP rate based on \eref{respective TUR}\eqref{max TUR}. The black and green lines correspond to the estimation $\widehat{\sigma^\mathrm{tot}}=(j_d^\mathrm{hk}+j_d^\mathrm{ex})^2/\calD_d, (j_d^\mathrm{hk}-j_d^\mathrm{ex})^2/\calD_d$ respectively. The gray line represents the sum of the estimation $\widehat{\sigma^\mathrm{tot}}=[(j_d^\mathrm{hk})^2+(j_d^\mathrm{ex})^2]/\calD_d$. The purple line shows the EP rate of the reservoirs $\widehat{\sigma^\mathrm{tot}}=j_d=\sigma^m$. The parameters used in calculation are $\mu=1.0, k=1.0, T_1=1.5, T_2=0.5$.
    }
\end{figure}
Now that we are ready, we calculate the currents and the entropy productions. The initial state is given by a delta function localized at the origin. This gives $C(0)=0$. Choosing $\boldsymbol{d}=D^{-1}A\boldsymbol{x}$ as the coefficient, the current is the EP rate in the heat bath. Each current average is
\begin{align}
    j_d&=
    \textrm{tr}\left(AD^{-1}(A+DC^{-1})C\right),\\
    j_d^\mathrm{hk}&=
    \textrm{tr}\left(AD^{-1}(A+D\bar{C}^{-1})C\right),
\end{align}
and
\begin{align}
    j_d^\mathrm{ex}=
    j_d-j_d^\mathrm{hk}=
    \textrm{tr}\left(A(C^{-1}-\bar{C}^{-1})C\right).
\end{align}
The current variance is computed as
\begin{align}
    \calD_d=\textrm{tr}(AD^{-1}AC).
\end{align}
These are plotted in \fref{fig: 2-beads_TUR}(a) and (b). Since the minimum energy of the system is $0$ at $t=0$, the energy flows into the system as heat, leading to $j_d<0$. For the housekeeping current, $j_d^\mathrm{hk}$ vanishes in the limit $t\rightarrow0$ because of $C\rightarrow0$. At $t\rightarrow\infty$, there is a steady heat flow and  $j_d^\mathrm{hk}\rightarrow j_d>0$. Due to this combination, the housekeeping current is always above $0$ and the excess current is always below $0$.
In the steady state $(t\rightarrow\infty)$, $j_d^ex$ vanishes because of $C\rightarrow\bar{C}$. On the other hand, when the detailed balance condition holds $(T_1=T_2)$, $D\propto I$ leads to $\bar{C}=-DA^{-1}=-A^{-1}D$ and $j_d^\mathrm{hk}$ vanishes.
As for fluctuations, $\calD_d$ increases monotonically and saturates to a stationary value as time passes and particles diffuse from near the origin.

The EP rates are computed as
\begin{align}
    \label{2-beads: tot EP rate}
    {\sigma^\mathrm{tot}}&=
    \textrm{tr}\left((A+DC^{-1})^TD^{-1}(A+DC^{-1})C\right),\\
    \label{2-beads: hk EP rate}
    {\sigma^\mathrm{hk}}&=
    \textrm{tr}\left((A+D\bar{C}^{-1})^TD^{-1}(A+D\bar{C}^{-1})C\right),
\end{align}
and
\begin{align}
    \label{2-beads: ex EP rate}
    {\sigma^\mathrm{ex}}&=
    {\sigma^\mathrm{tot}}-{\sigma^\mathrm{hk}}\nonumber\\
    &=\textrm{tr}\left((C^{-1}-\bar{C}^{-1})^TD(C^{-1}-\bar{C}^{-1})C\right).
\end{align}
We plot them in \fref{fig: 2-beads_TUR} (c). First, the total and excess EP rates diverge in the limit $t\rightarrow0$ because they include $C^{-1}$ which does not cancel out with $C\rightarrow0$, and the housekeeping EP rate with only $C$ terms vanishes in this limit. As in the previous example (2-dimensional flow), the excess EP rate diverges for the localized nonequilibrium state of the delta function and the housekeeping EP rate becomes $0$.
Over time, the housekeeping EP rate monotonically increases and saturates, and the excess EP rate decays to zero.
In the steady state, since the EP rate of the system is zero, the steady-state value reached by the total and housekeeping EP rates is the same as the corresponding steady-state value of the current average (see \fref{fig: 2-beads_TUR}(a)).

Finally, the estimation of the EP rates by TUR is plotted in \fref{fig: 2-beads_TUR} (d). For the EP rate of the total system, the green line is above the black line because the product of the currents is always negative in this setting (see \fref{fig: 2-beads_TUR}(a)). Around the initial time, the accuracy of the estimation using \eref{max TUR} is relatively high. As it approaches the steady state, the accuracy becomes worse, indicating that it is better to use the current as it is, i.e., the EP rate of the heat bath (purple line) for the estimation. This is because the mean of the current asymptotically approaches the mean of the EP rate for the total system, but the variance does not, making the estimation with TUR less accurate.

Since $j_d^\mathrm{hk}$ vanishes in the limit $t\rightarrow0$, all but the red line (and the purple line) converge to the same point. By contrast, $j_d^\mathrm{ex}$ vanishes in the limit $t\rightarrow\infty$, and all but the red line (and the purple line) intersect at the single point.

Around the initial time, the estimation of the excess EP rate has a similar trend to that of the total system. The accuracy deteriorates as it approaches the steady state, but it is obvious that it relaxes to zero in this case. On the other hand, the estimation of the housekeeping EP rate tends to be similar to that of the total system when the system is close to the steady state and is not accurate in any time region.

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