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Reporting on a visit by a high-level Omani delegation to Oxea's headquarters in Houston recently, the newsletter quoted a high-level official as saying that the company will also provide the technology and specialist expertise in the operation of the project, as well as the marketing of its products. Hilal Ali al Kharusi, Executive Managing Director of Oman Oil Duqm Development Company — one of the verticals of the restructured Oman Oil Group - described Oxea as a key building block in the downstream value chain that the parent group is eager to develop and sustain as part of its investments in the Sultanate. Al Kharusi is also a Director on the board of Oxea.
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text:
Q: Are the sets subspaces? We have the following subsets:
\begin{align*}&U_1:=\left \{\begin{pmatrix}x \\ y\end{pmatrix} \mid x^2+y^2\leq 4\right \} \subseteq \mathbb{R}^2\\ &U_2:=\left \{\begin{pmatrix}2a \\ -a\end{pmatrix} \mid a\in \mathbb{R}\right \} \subseteq \mathbb{R}^2 \\ &U_3:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=0\right \}\subseteq \mathbb{R}^3 \\ &U_4:=\left \{\begin{pmatrix}x \\ y \\ z\end{pmatrix} \mid y=1\right \}\subseteq \mathbb{R}^3\end{align*}
I want to check if these are subspaces.
$$$$
I have done the following:

*

*$U_1$ :
The set is non-empty, $(0,0)^T\in U_1$.
Let $(x_1, y_1)^T, (x_2, y_2)^T\in U_1$. That means that $x_1^2+y_1^2\leq 4$ and $x_2^2+y_2^2\leq 4$. For the sum $(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$ we have \begin{align*}(x_1+x_2)^2+(y_1+y_2)^2&=x_1^2+2x_1x_2+x_2^2+y_1^2+2y_1y_2+y_2^2 \\ & =\left (x_1^2+y_1^2\right )+\left (x_2^2+y_2^2\right )+2x_1x_2+2y_1y_2 \\ & \leq 4+4+2x_1x_2+2y_1y_2\end{align*} This is not necessarily less than $4$ and so this is not a subspace. Is that correct?

*$U_2$ :
The set is non-empty, $(0,0)^T\in U_2$.
Let $(x_1, y_1)^T, (x_2, y_2)^T\in U_2$. That means that $y_1=-2x_1$ and $y_2=-2x_2$. For the sum $(x_1, y_1)^T+ (x_2, y_2)^T=(x_1+x_2, y_1+y_2)^T$ we have $y_1+y_2=-2x_1-2x_2=-2(x_1+x_2)$ and that means that the sum is also in $U_2$.
Let $r\in \mathbb{R}$ and $(x_1, y_1)^T\in U_2$. That means that $y_1=-2x_1$. For the scalar multiplication $r(x_1, y_1)^T=(rx_1, ry_1)^T=(rx_1, ry_1)^T$ we have $ry_1=r\cdot (-2x_1)=-2(rx_1)$ and that means that the result is also in $U_2$.
Therefore $U_2$ is a subspace.

*$U_3$ :
The set is non-empty, $(0,0,0)^T\in U_3$.
Let $(x_1, y_1, z_1)^T, (x_2, y_2, z_2)^T\in U_3$. That means that $y_1=y_2=0$. For the sum $(x_1, y_1, z_1)^T+ (x_2, y_2, z_2)^T=(x_1+x_2, y_1+y_2, z_1+z_2)^T$ we have $y_1+y_2=0+0=0$ and that means that the sum is also in $U_3$.
Let $r\in \mathbb{R}$ and $(x_1, y_1, z_1)^T\in U_3$. That means that $y_1=0$. For the scalar multiplication $r(x_1, y_1, z_1)^T=(rx_1, ry_1, rz_1)^T$ we have $ry_1=r\cdot 0=0$ and that means that the result is also in $U_3$.
Therefore $U_3$ is a subspace.

*$U_4$ :
Since the zero vector is nott included, $U_4$ is not a subspace.
$$$$
Is everything correct and complete?

A: For every one want to know:
$U_1: \qquad $ There is no bounded subspace but $\{0\}$
$U_2 = \langle \binom{2}{-1} \rangle: \qquad $ Multiples of any vector produce a subspace of degree $1$.
$U_2' = \langle v_i \rangle_i: \qquad $ linear combinations of any set of vectors produce a subspace of degree equals to the max number of independent vectors in generating set.
$U_{3,4} = y=y_0: \qquad $ Any linear equation produces a subspace of degree $n-1$ where your space is of degree $n$.
$U'_{3,4} : \qquad $ A system of linear equations produce a subspace of degree $n-k$ where you have $k$ independent equations.

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Apparently, dogs regularly board buses and even occupy seats in Seattle, and while small dogs ride for free, pet owners have to pay a fare for larger dogs.
The rules for bringing dogs on Pierce Transit appear to be slightly more strict: "Regular" dogs (meaning those that aren't service animals) have to be kept in a container, like a pet carrier or pet stroller. If you've never seen a pet stroller, yes, that's a real thing.
Service dogs don't need to be kept in a container, but they can't occupy a seat. There's no fare for regular or service animals.
There was one exception recently, a Pierce Transit spokeswoman said. During the Downtown to Defiance and Gig Harbor summertime weekend trolley rides, well-behaved and leashed dogs have been allowed to ride at no extra charge. (Let's be honest, much like children, everyone thinks their dog is well-behaved. But I digress.)
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Washington, B.C. want to make connection via high-speed rail
How do ramp meters work?
Washington drivers might be charged by the mile to make up for lagging gas-tax revenue
By James Drew
Washington state would move toward replacing the gas tax with a pay-per-mile system under a proposal the state Transportation Commission is expected to vote on later this year.
MORE TRAFFIC
Man who died in crash on SR 16 is identified
Traffic metering begins Tuesday at I-5 ramps in Tacoma, Fife
Lakewood man killed in crash near Tacoma Mall identified
Northbound I-5 lanes reopened after oil tanker rollover north of Centralia
Expect higher travel times but then fewer loose rocks on state Route 7 near La Grande
Woman on way to funeral in Kitsap County dies in crash; family witnessed from another car

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Everton have signed defender Cuco Martina on a Bosman after his contract expired at Southampton.
The right-back, 27, worked with Everton boss Ronald Koeman for one season when his countryman was in charge at St Mary's.
Martina, who is captain of the Caribbean nation of Curacao, has signed a three-year contract.
Newcastle United's French forward Yoan Gouffran has joined Goztepe in Turkey. The 31-year-old made more than 140 appearances for the Magpies after joining from Bordeaux in 2013, scoring 19 goals.
Championship side Bolton Wanderers have signed Newcastle's 20-year-old forward Adam Armstrong on loan until January.
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Newcastle United have agreed a £12.5m fee with Norwich City for 22-year-old forward Jacob Murphy, according to the Daily Telegraph.
Have a watch as Britain's Chijindu Ujah wins the men's 100m in 9.98 seconds at the Diamond League meeting in Rabat, Morocco.
Britain's Grace Reid and Katherine Torrance have finished fifth in the 3m synchro springboard final at the Diving World Championships in Budapest.
They finished with a total score of 294.60 after five dives while China's Yani Chang and Tingmao Shi took gold with 333.30.
Canada's Jennifer Abel and Melissa Citrini Beaulieu (323.43) claimed the silver while Russia's Nadezhda Bazhina and Kristina Ilinykh (312.60) got the bronze.
Tom Daley and Daniel Goodfellow will compete in the men's 10m synchro platform final at 17:30 BST, which you'll be able to watch live on this website.
Michu, the Spanish midfielder who helped Swansea win the League Cup in 2013, has been forced to retire.
The 31-year-old was the Swans' top scorer in 2012-13 with 22 goals in a spectacular first season after a £2m move from Rayo Vallecano.
Within two years of helping Swansea finish ninth in the Premier League a recurring ankle injury saw him playing fourth-flight football in Spain.
Michu spent the 2016-17 season with Oviedo and plans to move into coaching.
Celtic striker Leigh Griffiths faces further disciplinary action on top of his bizarre Belfast booking after being accused of "provoking" Linfield supporters during a Champions League qualifying match.
Uefa has opened a case against Griffiths over his actions at the end of Celtic's 2-0 second qualifying round first-leg win at Windsor Park last Friday.
The Scotland international, who was earlier pelted with missiles including a glass bottle, tied a Celtic scarf to a goalpost. The European governing body has also instigated disciplinary action against Linfield over their supporters' behaviour, and Celtic for their yellow card count.
HK Joyce: Carrying an egg on a spoon from one side of the field to another in primary school.
Ruth Sharville: Coming second in the long jump at junior school - a total surprise to me!
Dan James: Played U16 fast pitch softball for Guernsey in a tournament in the UK in mid-80s. WON 37-0 in the final, still a record.
Tweet #bbcsportsday, text 81111 or head to the BBC Sport Facebook page.
Middlesbrough are close to breaking their transfer record by signing Nottingham Forest strikerBritt Assombalonga for £14m.
Assombalonga, 24, is currently in talks with Boro at their Rockliffe Park training ground after the two clubs agreed the broad outlines of the transfer fee.
Burnley made a bid for the Zaire-born forward earlier this summer but Middlesbrough have outbid their rivals and look set to sign a man who scored 14 goals in 33 games last season.
If the deal is finalised then the signing will eclipse the £12.7 million that Middlesbrough paid for Brazilian striker Afonso Alves in 2008.
Hands up who knew it was World Emoji Day?
To celebrate, Italian side Roma have come up with this of Francesco Totti.
Ipswich Town will not be offering former Newcastle United defender Steven Taylor, 31, a new contract, manager Mick McCarthy says.
Taylor featured three times for Ipswich after joining them in January.
His time at Portman Road was hampered by a hamstring injury, but he was invited back for pre-season.
Former England opener and TMS commentator Geoff Boycott is dissecting England's 340-run defeat against South Africa at Trent Bridge.
Crystal Palace are on the verge of signing Ajax defender Jairo Riedewald in a deal worth around £8m, reports the Evening Standard.
New Palace boss Frank de Boer previously managed the 20-year-old at Ajax.
Manchester City flew to United States today for the start of their pre-season tour. A few players, including Yaya Toure and Leroy Sane, weren't there but the club is adamant there is no issue and they will be going.


Ballghazi: The New England Patriots lose an insanely low number of fumbles. Coincidence?
The New England Patriots' prevention of fumbles is nearly impossible.
LeGarrette Blount of the New England Patriots runs with the ball against LaRon Landry of the Indianapolis Colts on Jan. 18, 2015, in Foxboro, Massachusetts.
In light of the ball deflation scandal currently consuming the New England Patriots, football data analyst Warren Sharp looked at the Patriots' recent fumble statistics and found some startling results. A slightly different version of this post first appeared on Sharp's own site. It is reprinted with his permission.
I looked at the last five years of data and examined total fumbles lost in all games (as well as fumbles per game) and, more importantly, total offensive plays run. Thus, I was able to determine average plays per fumble lost. The results are displayed in the chart above. Keep in mind, this is for all games since 2010.
One can clearly see the Patriots, visually, are off the chart. There is no other team even close to being near to their rate of 187 offensive plays per fumble lost. The league average is 105 plays per fumble lost. Most teams are within 21 plays of that number.
Based on the assumption that fumbles per play follow a normal distribution, you'd expect to see, according to random fluctuation, the results that the Patriots have gotten over this period, once in 16,233.77 instances.
Which in layman's terms means that this result only being a coincidence, is like winning a raffle where you have a 0.0000616 probability to win. [In] other words, it's very unlikely that it's a coincidence.
There are a few key takeaways. First and foremost, the 187 plays per fumble lost dwarfs even the rest of the best seasons of the last 25 years. Second, the Patriots have been at the top of the NFL since 2007.
Ironically, as my study on Wednesday showed, the Patriots' performance in wet weather home games mysteriously turned ridiculous starting in 2007. In 2006, they went 0–2. From 2007 onward, they went 14–1.
As you can see, the Patriots won their Super Bowls having a below-average rate of fumbles lost given today's average of 105 plays per fumble lost. But in 2007, something happened to propel them to a much better rate. (You'll remember, that just so happened to be the same year they went 16–0 in the regular season.) But even looking at these numbers, it's clear how insane the 187 number is: They are almost running 100 more plays without a single fumble as compared with the 2002–2006 period when they won two of their three Super Bowls.
Looking at all fumbles, and not just fumbles lost, we come up with similar findings.
First, it should be noted that teams playing indoors fumble the ball less frequently, the foremost reason being that the ball won't be wet from precipitation or damp from late-night condensation. The below graphic looks at all fumbles over five-year periods during the last 25 years. As you can see, of the top 25 team-periods, 17 are dome teams, including 11 of the top 15.
As is apparent, the Patriots are the only outdoor NFL team in the last 25 years to average 70 plays per fumble or better. It's simply uncanny, as the statistics above similarly showed.
The league average from 2010 to 2014 was 50 plays per fumble. For indoor teams, the average was 55 plays per fumble. For outdoor teams, excluding the Patriots, the average was 46 plays per fumble. The Patriots averaged 73 plays per fumble, almost 60 percent more than outdoor teams and almost 50 percent more than the league average the past five years.
As we can clearly see in both near-term and long-term data that dome-based teams (that play at least eight games out of the elements) have an advantage in the fumble department, we can exclude them from comparisons to the Patriots.
Could the Patriots be so good that they just defy the numbers? As my friend theorized: Perhaps they've invented a revolutionary in-house way to protect the ball, or perhaps they've intentionally stocked their skill positions with players who don't have a propensity to fumble. Or perhaps, still, they call plays that intentionally result in a lower percentage of fumbles. Or maybe it's just that they play with deflated footballs on offense. It could be any combination of the above.


text:
\section{Introduction} \label{sec:intro}

There are several combinatorial structures whose enumeration is given by the Catalan sequence: $1, 2, 5, 14, 42, \ldots$, where the $n$th term (for $n\ge 1$) is given by the Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$. Some examples include the set of all triangulations of a regular polygon of $n+2$ sides, the set of non-crossing partitions of an $n$-set (the lattice on which is of much interest to researchers in free probability), and the set of balanced parentheses, with $n$ left and $n$ right parentheses---or equivalently, described as {\em Catalan strings} $x\in \{\pm 1\}^{2n}$ of $n$ $1$'s and $n$ $-1$'s with non-negative partial sums, $\sum_{i=1}^j x_i \ge 0$, for all $j$. This last structure (of Catalan strings) is also known as the set of Dyck paths, and is visualized as lattice paths of $n$ up-steps and $n$ down-steps, representing the $1$'s and the $-1$'s respectively.

For any given $n\ge 1$, generating a Catalan structure (such as triangulation or a Dyck path) uniformly at random from the set of $C_n$ many, is a straightforward task, and can be done in time linear in $n$, not too unlike generating a permutation uniformly at random  (out of the $n!$ many permutations) of $n$ distinct letters.
However, the study (of convergence to equilibrium) of particular Markov chain Monte Carlo algorithms which yield a random Catalan structure is seemingly much more interesting. The inspiration to this endeavor in part stems from  an open problem of David Aldous~\cite{Aldous94}, who conjectured that the random walk on triangulations of a polygon on $n$ sides, performed using uniform diagonal flips, ought to take time roughly  $n^{3/2}$ (up to factors logarithmic in $n$).
Despite much effort by various researchers in this topic over several years, the best known bounds for the mixing time of the chain on triangulations remain those of  McShine-Tetali \cite{McShine1999}, who proved an upper bound of $O(n^4)$ on the relaxation time (also known as the inverse spectral gap), and Molloy-Reed-Steiger \cite{MRS98} who showed  $\Omega(n^{3/2})$ as a lower bound.

In the work, we explore a (different) Markov chain on a different realization of the Catalan number: we consider the set of Dyck paths, described using binary strings $x \in \{\pm 1\}^{2n}$, as mentioned above, and consider the natural local move of exchanging $x_i$ and $x_j$, for a pair $(i,j)$ chosen uniformly at random from among all possible pairs, {\em provided} that the resulting string is a (valid) Catalan string; else we reject such a move. Note that, without the rejection step, we would have a random transposition chain on the set of $\binom{2n}{n}$ many binary strings with an equal number of $1$'s and $-1$'s. It is well-known (see e.g., \cite{Diaconis1987})---sometimes under the name \emph{Bernoulli-Laplace model}---that $O(n\log n)$ such random transpositions are necessary and sufficient to reach (close to) equilibrium. It is also well-known that the so-called random transposition shuffle of $n$ distinct cards mixes in $O(n\log n)$ time. However, the ``sea-level'' constraint of having the partial sums of the (Catalan) string stay non-negative seems to thwart any type of straight forward (or otherwise) analyses, despite the best efforts of several experts in the field.

Our main result is an $O(n^2\log n)$ upper bound on the mixing time of the random transposition walk on Dyck paths of length $n$.  The proof relies on a rephrasing of the walk as a basis exchange walk on a balanced matroid.  In Section \ref{sec:matroids} we describe the Catalan matroid (due to Ardila \cite{Ardila2003})  and the prior work \cite{Feder1992, Jerrum2004, Jerrum-Son} regarding mixing times for the basis exchange walk on balanced matroids.  All that remains to bound the mixing time of our Dyck path walk is the proof in Section \ref{sec:main-result} that the Catalan matroid is balanced.  In Section \ref{sec:lower-bounds} we give a short proof of a mixing time lower bound for Dyck random transpositions and, finally, in Section \ref{sec:open-questions} we explore the possible consequences of this result for other walks on Catalan structures and pose some open questions.

\subsection{Random walks on lattice paths}
Throughout, for integers $a\leq b$ we let $[a,b] \vcentcolon= \set{a, a+1,\dots\,b}$ denote the discrete closed interval and $[a] \vcentcolon= [1,a]$.

A \emph{lattice path} of length $m$ is a string $P\in\set{\pm1}^m$ of \emph{up-steps} ($+$) and \emph{down-steps} ($-$).  The \emph{height} of $P$ at index $i$ is $h_i(P) = \sum_{j=1}^i P_i$, and we can draw $P$ on the grid as the graph of $f_P(i) = h_i(P)$.  That is, we draw a path starting at $(0,0)$ and taking up-steps $(1,1)$ and down-steps $(1,-1)$.
We will refer to a lattice path $P$ of length $m$ as a lattice path \emph{from $(0,0)$ to $(m, h_m(P))$}.

From this we can define a partial order on the set of lattice paths from $(0,0)$ to $(m,2r-m)$ by letting $P \leq Q$ whenever $h_i(P) \leq h_i(Q)$ for all $i\in [m]$.  Note that $P\leq Q$ if and only if $q_i \leq p_i$ for all $i\in[r]$, where $p_i$ (resp. $q_i$) is the index of the $i$th up-step in $P$ (resp. $Q$).

\begin{definition}
    A \emph{Dyck path} of length $2n$ is a lattice path $P$ from $(0,0)$ to $(2n,0)$ with $h_i(P) \geq 0$ for all $i\in[2n]$.
\end{definition}

That is, the Dyck paths are precisely the paths $P$ from $(0,0)$ to $(0,2n)$ with $P \geq (+-)^n$. It is a standard result that the number of Dyck paths of length $2n$ is the Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$.

A natural class of random walks on lattice paths from $(0,0)$ to $(m,h)$ is the \emph{transposition walk}, which at each step picks random indices $i,j\in[m]$ and swaps the steps of $P$ at those indices.
Indeed, if $i$ and $j$ are chosen uniformly independently at random this is just the Bernoulli-Laplace model on $\binom{[m]}{r}$, where $r = (h+m)/2$ is the number of up-steps in every such path, and has been studied extensively.  If we pick $i,j$ to be uniform among \emph{adjacent} pairs of indices, we get a different walk, whose mixing properties were determined in \cite{Wilson2004}.

But what if we consider lattice paths from $(0,0)$ to $(2n, 0)$ and restrict the walk to Dyck paths?  That is, from any Dyck path we pick $i,j$ according to one of these rules but only perform the move if the resulting path is again a Dyck path. (We will call these walks the \emph{Dyck random transpositions walk} and \emph{Dyck adjacent transpositions walk}, respectively.)

Indeed, the analysis of \cite{Wilson2004} extends also to this case, yielding the same upper bound for the mixing time of the Dyck adjacent transpositions walk.  On the other hand, the Dyck random transpositions walk has evaded such precise analysis.

\section{Lattice path matroids} \label{sec:matroids}

\subsection{Matroids and the basis exchange walk}

Recall that a nonempty set $\mathcal{B}\subseteq 2^U$ is the set of \emph{bases} of a \emph{matroid} $M = (U, \mathcal{B})$ if the following \emph{basis exchange axiom} holds:
\begin{quote}
    \textbf{Matroid Basis Exchange Axiom.} For any bases $A,B\in \mathcal{B}$ and every $e\in A\setminus B$ there exists $f\in B\setminus A$ such that $A\setminus\set{e}\cup\set{f}\in \mathcal{B}$ is a basis.
\end{quote}
Among other things, this axiom guarantees that all bases have the same cardinality, which is the \emph{rank} of $M$.

We will also make use of two dual operations on matroids: contraction and deletion.
\begin{definition}
    For a matroid $M = (U, \mathcal{B})$ and an element $e\in M$, the matroid \emph{$M$ contract $e$} is $M_e = (U, \mathcal{B}_e)$, where $\mathcal{B}_e = \set{B\in \mathcal{B}: e\in B}$.  Similarly, \emph{$M$ delete $e$} is $M^e = (U, \mathcal{B}^e)$, where $B^e = \set{B\in \mathcal{B}: e\not\in B}$.\footnote{Note that this definition differs slightly from the usual one, in which the element $e$ being contracted or deleted is removed from the ground set.  In our case it will be convenient to leave $e$ in place, so as to more easily identify $\mathcal{B}_e\subseteq \mathcal{B}$ and preserve identities such as $\mathcal{B} = \mathcal{B}_e \cup \mathcal{B}^e$ (with $\mathcal{B}_e\cap \mathcal{B}^e = \emptyset$).}

    A \emph{minor} of $M$ is any matroid which can be obtained from $M$ through a series of contractions and deletions.
\end{definition}

The order in which contractions and deletions are performed does not matter, so we will write $M_I^J$ for the matroid obtained from $M$ by contracting the elements in $I$ and deleting the elements in $J$, and $\mathcal{B}_I^J$ will denote the set of bases of $M_I^J$.

Given a matroid $M$, Feder and Mihail \cite{Feder1992} study the following \emph{basis exchange walk} on the state space $\mathcal{B}$ of bases:
\begin{quote}
    From state $B\in\mathcal{B}$, pick uniformly and independently at random elements $a\in U$ and $b\in B$ and move in the next step to $B' = B\cup\set{a}\setminus\set{b}$ if $B'\in \mathcal{B}$ (else remain at $B$).
\end{quote}
The matroid basis exchange axiom guarantees that this walk is ergodic, and since it is symmetric its stationary distribution is uniform.  It is tentatively conjectured that the basis exchange walk is fast for any matroid (i.e., the mixing time is bounded by some polynomial in $m = \size{U}$), but there is little evidence in favor of this.

On the other hand, \cite{Feder1992} introduced the notion of \emph{balanced} matroids, which capture the notion that for a randomly chosen basis (of the matroid or any of its minors), conditioning on the occurrence of one element (in the basis) makes the occurrence of any other less probable.  They show that for the case of balanced matroids the walk is indeed rapidly mixing, by using decomposition techniques to bound its spectral gap.

\begin{definition}
    A matroid $M = (U,\mathcal{B})$ is \emph{negatively correlated} if for every pair of distinct elements $e,f\in U$
    \begin{align} \label{eqn:negative-correlation}
        \frac{\size{\mathcal{B}_e}}{\size{\mathcal{B}}} \geq \frac{\size{\mathcal{B}_{ef}}}{\size{\mathcal{B}_f}}.
    \end{align}
    The matroid $M$ is \emph{balanced} if $M$ and all of its minors are negatively correlated.
\end{definition}
Negative correlation is equivalent to the very natural condition that for a uniform random basis $B\in\mathcal{B}$, $\Pr[e\in B] \geq \Pr[e\in B | f\in B]$.

Indeed, many common classes of matroids are balanced, including uniform matroids (whose bases are all size-$r$ subsets of the ground set), graphic matroids (with the ground set being the edges of a connected graph and the bases being spanning trees of the graph), matroids of rank $\leq 3$, and regular matroids (matroids which can be represented over every field) \cite{Feder1992,Choe2006}.  We will rely on the following extension of Feder and Mihail's result for balanced matroids, due to Jerrum and Son.

\begin{theorem}[\cite{Jerrum-Son}]\label{thm:balanced-mixing}
    The spectral gap $\lambda$ and log-Sobolev constant $\alpha$ for the basis exchange walk on a balanced matroid $M$ of rank $r$ on a ground set of size $m$ are lower bounded by
    \begin{align}
        \lambda \geq \frac{2}{mr} \qquad\text{and}\qquad \alpha \geq \frac{1}{2mr}.
    \end{align}
\end{theorem}

In particular, via standard mixing time bounds (see \cite{Diaconis1996}), the bound on the log-Sobolev constant implies that the mixing time of the basis exchange walk on a balanced matroid is at most $O(mr\log\log\size{\mathcal{B}})$.

\begin{remark}
Due to these standard bounds and the result for balanced matroids the proof of our main result never explicitly refers to the definitions of $\lambda$ and $\alpha$.  These definitions can, however, be found in Section \ref{sec:lower-bounds}, where they are used to give a lower bound on the mixing time for the Dyck random transposition chain.
\end{remark}

\subsection{The Catalan matroid and other lattice path matroids}
To aid our setting of the Catalan transposition walk in terms of matroids, we refer to an observation of Ardila \cite{Ardila2003} that the set of Dyck paths can be thought of as a matroid.

\begin{definition}
    The \emph{Catalan matroid} of order $n$ is $\mathcal{C}(n) = ([2n], \mathcal{B}(n))$, where the elements of $\mathcal{B}(n)$ are the index-sets of up-steps in Dyck paths of length $2n$.
\end{definition}

\begin{figure}
    \centering
    \begin{tikzpicture}[scale=.4]
        \begin{scope}
        \path[clip] (0,-1) to (0,0) \u \u \u \u \d \u \d \u \u \d \d \d \u \d \d \d to (16, -1);
        \foreach \x in {1,2,3,4,6,8,9,13}{
            \draw[fill=black!5, draw=black!15] (\x-1,-1) rectangle (\x,6);
        }
        \end{scope}
        \draw[thick] (0,0) \u \u \u \u \d \u \d \u \u \d \d \d \u \d \d \d;
        \foreach \x in {1,...,16}{
            \node at (\x-.5, -.5) {\footnotesize$\x$};
        }
        \draw[thin, draw=black!30] (0,0) to (16,0);
    \end{tikzpicture}
    \caption{The Dyck path $(++++-+-++---+---)$ above corresponds to the basis $\set{1,2,3,4,6,8,9,13}$ of the Catalan matroid of order $8$.}
\end{figure}

Ardila notes that this is precisely the transversal matroid for the set system $\mathcal{S} = \set{[1],[3],\dots,[2n-1]}$ (i.e., its bases are the systems of distinct representatives of $\mathcal{S}$).  In short, the representative of the set $[2i-1]$ will be the index of the $i$th up-step in the corresponding Dyck path (although this assignment of distinct representatives need not be unique).  Indeed,

\begin{observation}
    The basis exchange walk on the Catalan matroid is exactly the random transposition walk on Dyck paths.
\end{observation}

In other words, given Theorem \ref{thm:balanced-mixing}, to obtain a mixing time bound for the Dyck transposition walk it suffices to show that the Catalan matroids are balanced.

To this end, it would be convenient if the Catalan matroids belonged to some known class of balanced matroids.  As noted above, the main class of matroids known to be balanced is \emph{regular matroids}, but Ardila notes that $\mathcal{C}(n)$ is not representable over any $\mathbb{F}_q$ for $q\leq n-2$ and thus is not regular.  Transversal matroids also need not be balanced in general: Choe and Wagner \cite{Choe2006} give a transversal matroid of rank 4 which is not balanced.

Thus the bulk of our work here will be to show that the Catalan matroid is balanced, and from this the main mixing result will follow immediately.  For our later discussion it will help to work with a minor-closed class of matroids (which the Catalan matroids certainly are not).  Bonin and de Mier (\cite{Bonin2006}) discuss the following class of \emph{lattice path matroids}, which generalize the Catalan matroids by allowing \emph{any} pair of bounding paths.

\begin{definition}
    For two lattice paths $A\leq B$ from $(0,0)$ to $(m, 2r-m)$, consider the set $L$ of lattice paths $P$ from $(0,0)$ to $(m, 2r-m)$ with $A\leq P \leq B$.  The \emph{lattice path matroid} $\mathcal{L}[A,B]$ (which is of rank $r$ on ground set $[m]$) has as its bases the index sets of up-steps of paths in $L$.
\end{definition}

Although it is not immediately obvious that $\mathcal{L}[A,B]$ is a matroid, \cite{Bonin2006} observes that in fact the lattice path matroid $\L[A,B]$ is the transversal matroid of the set system $\set{[a_1,b_1], [a_2,b_2], \dots, [a_r, b_r]}$, where $a_i$ (resp. $b_i$) is the index of the $i$th up-step in $A$ (resp. $B$).  Indeed, it is shown in that paper that the transversal matroids of a set systems $\set{[a_1,b_1],[a_2,b_2],\dots,[a_r,b_r]}$ with $a_1 \leq a_2 \leq \dots \leq a_r$ and $b_1 \leq b_2 \leq \dots \leq b_r$ are precisely the lattice path matroids. (The set system of this form corresponding to a given lattice path matroid may not be unique; for example, the Catalan matroid is the transversal matroid for both $\set{[1],[3],\dots,[2n-1]}$ and $\set{[1],[2,3],\dots,[n,2n-1]}$.)  The class of basis exchange walks on lattice path matroids also includes the (unconstrained) Bernoulli-Laplace model as the basis exchange walk on a uniform matroid.

In addition to showing this correspondence, Bonin and de Mier show that the class of lattice path matroids is closed under taking minors and duals.  They also define a smaller (in fact, minimal) minor-closed class of matroids containing $\mathcal{C}_n$, which they call \emph{generalized Catalan matroids}, consisting of the lattice path matroids $M(A,B)$ where $A = (+)^r (-)^{m-r}$ is maximal among all paths from $(0,0)$ to $(m, 2r-m)$.  They give several nice properties of generalized Catalan matroids, which we will not go into here as our analysis covers all lattice path matroids.  Indeed, Sohoni \cite{Sohoni1999} has actually already shown the balanced property of generalized Catalan matroids, although he calls them \emph{Schubert matroids} and does not link his result to rapid mixing of the Dyck random transposition chain (although he does mention the Dyck adjacent transposition chain).  Our proof applies to the more general class of lattice path matroids.

Given that the class of lattice path matroids is minor-closed, it suffices for our main result to show that lattice path matroids are negatively correlated, but it is worth noting that the same method can also be used to show negative correlation directly for any minor of a lattice path matroid.

\section{Mixing bound for random transpositions} \label{sec:main-result}

To show negative correlation for lattice path matroids, it will be convenient to use an equivalent formulation of \eqref{eqn:negative-correlation}, which is easily obtained by repeatedly applying the identity $\size{\mathcal{B}} = \size{B_e} + \size{B^e}$,
\begin{align}
    \size{\mathcal{B}_{ef}} \size{\mathcal{B}^{ef}} \leq \size{\mathcal{B}_{e}^{f}} \size{\mathcal{B}_{f}^{e}}\,. \label{neg-cor-equiv}
\end{align}
Now we are ready to prove

\begin{theorem}
    For every pair of lattice paths $A\leq B$ from $(0,0)$ to $(m, 2r-m)$ the lattice path matroid $\L[A,B]$ is negatively correlated.
\end{theorem}

\begin{proof}
Let $\L[A,B] = ([m], \mathcal{B})$ be any lattice path matroid.

To prove inequality \eqref{neg-cor-equiv} for every pair $e < f\in [m]$, we will construct an injective map
\[\varphi_{ef} : \mathcal{B}_{ef} \times \mathcal{B}^{ef} \to \mathcal{B}_{e}^{f} \times \mathcal{B}_{f}^{e}.\]
Note that we can associate members of these sets with lattice paths in $\L[A,B]$ by
\begin{itemize}
\item $\mathcal{B}_{ef}$: paths with up-steps at indices $e,f$,
\item $\mathcal{B}^{ef}$: paths with down-steps at indices $e,f$,
\item $\mathcal{B}_{e}^{f}$: paths with an up-step at index $e$ and a down-step at index $f$, and
\item $\mathcal{B}_{f}^{e}$: paths with a down-step at index $e$ and an up-step at index $f$.
\end{itemize}

Let $P \in \mathcal{B}_{ef}$ and $Q \in \mathcal{B}^{ef}$ be lattice paths and consider the following cases. (The figures shown are for the Catalan matroid.)

\begin{description}
\item[Case 1.] Suppose the paths $P,Q$ intersect (without necessarily crossing) in the region (I) between $e$ and $f$.  Note that this includes all cases where the path $P$ is below $Q$ at $e$ and above $Q$ at $f$ or vice versa.  Take the first such intersection point $x$ in (I) and switch the paths $P, Q$ after $x$ to obtain new paths $P' \in \mathcal{B}_{e}^{f}, Q' \in \mathcal{B}_{f}^{e}$ as shown in Figure \ref{fig:case1}.  Set $\varphi_{ef}(P, Q) = (P', Q').$

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,7);
    \fill[fill=black!15] (5,0) rectangle ++(1,7) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (16,0) rectangle ++(1,7) +(-.5,-1) node {$f$};
    \fill[fill=black!5] (6,0) rectangle (16,7);
    \node at (11,6) {(I)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,7);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \u \d \u \d \d coordinate (x) \u \d \d \u \u \u \d \d \u \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \u \d \u \d \u \u \d \d \d \d \u \u \d \u \d \d;
    \node[red] at (1,2) {$P$};
    \node[blue] at (3,4) {$Q$};
    \path (x) +(0,-0.075) node [label=above:{$x$}, fill, circle, inner sep=1pt] {};
  \end{tikzpicture}\qquad
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,7);
    \fill[fill=black!15] (5,0) rectangle ++(1,7) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (16,0) rectangle ++(1,7) +(-.5,-1) node {$f$};
    \fill[fill=black!5] (6,0) rectangle (16,7);
    \node at (11,6) {(I)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,7);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \u \d \u \d \d coordinate (x) \d \u \u \d \d \d \d \u \u \d \u \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \u \d \u \u \d \d \u \u \u \d \d \u \d \d \d \d;
    \node[red] at (1,2) {$P'$};
    \node[blue] at (3,4) {$Q'$};
    \path (x) +(0.075,-0.075) node [label=above:{$x$}, fill, circle, inner sep=1pt] {};
  \end{tikzpicture}
  \caption{The injection for Case 1.}
\label{fig:case1}
\end{figure}

\item[Case 2.] Suppose the paths do not meet in region (I) and consider the paths $P, Q$ in the region (II) after position $f$. Imagine translating the fragment of path $Q$ after $f$ to $P$ so that their initial points in $f$ coincide. If the imaginary fragment intersects $P$, let $x$ be the first such point of intersection. Construct new paths $P',Q'$ by swapping the segments of $P$ and $Q$ between $f$ and $x$ (including $f$ itself).  See Figure \ref{fig:case2}.  If the new paths $P', Q'$ are both between $A$ and $B$ (so that $P'\in \mathcal{B}_e^f$ and $Q'\in B_f^e$), then we set $\varphi_{ef}(P, Q) = (P', Q')$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,9);
    \fill[fill=black!15] (7,0) rectangle ++(1,9) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (12,0) rectangle ++(1,9) +(-.5,-1) node {$f$};
    \fill[fill=black!5] (13,0) rectangle (22,9);
    \node at (18,8) {(II)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,9);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \d \u \u \d \u \d \u \u coordinate (fQ) \d \d \u \d \u \d \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \d \u \u \d \u \d coordinate (fP) \u \d \u \d \d \u \d \u \d \d;

    \draw[red, thick, dotted] (fQ)+(0,.15)   \u \d \u \d \d coordinate (xQ) \u \d \u \d \d;
    \draw[blue, thick, dotted] (fP)          \d \d \u \d \u coordinate (xP) \d \d;
    \node[red] at (1,2) {$P$};
    \node[blue] at (3,4) {$Q$};
    \node[label=above:{$x$}, fill, circle, inner sep=1pt] at (xQ) {};
    \node[fill, circle, inner sep=1pt] at (xP) {};
  \end{tikzpicture}\qquad
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,9);
    \fill[fill=black!15] (7,0) rectangle ++(1,9) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (12,0) rectangle ++(1,9) +(-.5,-1) node {$f$};
    \fill[fill=black!5] (13,0) rectangle (22,9);
    \node at (18,8) {(II)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,9);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \d \u \u \d \u \d \u \u coordinate (fQ) \u \d \u \d \d \d \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \d \u \u \d \u \d coordinate (fP) \d \d \u \d \u \u \d \u \d \d;

    \draw[blue, thick, dotted] (fQ)             \d \d \u \d \u coordinate (xQ);
    \draw[red, thick, dotted] (fP)              \u \d \u \d \d coordinate (xP);
    \node[red] at (1,2) {$P'$};
    \node[blue] at (3,4) {$Q'$};
    \node[label=above:{$x$}, fill, circle, inner sep=1pt] at (xQ) {};
    \node[fill, circle, inner sep=1pt] at (xP) {};
  \end{tikzpicture}
\caption{The injection for Case 2.}
\label{fig:case2}
\end{figure}

\item[Case 3.] Otherwise, perform the mirror image of the operation from Case 2 in the part (III) up to position $e$, as shown in Figure \ref{fig:case3} and again set $\varphi_{ef}(P, Q) = (Q', P')$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,9);
    \fill[fill=black!15] (7,0) rectangle ++(1,9) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (12,0) rectangle ++(1,9) + (-.5,-1) node {$f$};
    \fill[fill=black!5] (0,0) rectangle ++(7,9);
    \node at (3.5,8) {(III)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,9);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \d \u \u \d coordinate (eQ) \u \d \u \u coordinate (fQ) \d \d \d \u \u \d \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \d \u \u coordinate (eP) \u \d \u \d coordinate (fP) \u \d \u \d \d \u \d \u \d \d;

    \draw[red, thick, dotted] (fQ)           \u \d \u \d \d \u;
    \draw[blue, thick, dotted] (fP)          \d \d \d;

    \draw[red, thick, dotted] (eQ)  to ++(-1,-1) node[draw=none] {}\dl to ++(-1,1) node[draw=none] {} coordinate (xQ);
    \draw[blue, thick, dotted] (eP) to ++(-1,1) node[draw=none] {} to ++(-1,-1) node[draw=none] {}\dl coordinate (xP);
    \node[red] at (1,2) {$P$};
    \node[blue] at (3,4) {$Q$};
    \node[label=above:{$x$}, fill, circle, inner sep=1pt] at (xQ) {};
    \node[fill, circle, inner sep=1pt] at (xP) {};
  \end{tikzpicture}\qquad
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,9);
    \fill[fill=black!15] (7,0) rectangle ++(1,9) +(-.5,-1) node {$e$};
    \fill[fill=black!15] (12,0) rectangle ++(1,9) + (-.5,-1) node {$f$};
    \fill[fill=black!5] (0,0) rectangle ++(7,9);
    \node at (3.5,8) {(III)};

    \begin{scope}
        \path[clip] (0,0) rectangle (22,9);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \d \d \u \u coordinate (eQ) \u \d \u \u coordinate (fQ) \d \d \d \u \u \d \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \u \d coordinate (eP) \u \d \u \d coordinate (fP) \u \d \u \d \d \u \d \u \d \d;

    \draw[blue, thick, dotted] (eQ)  to ++(-1,1) node[draw=none] {} to ++(-1,-1) node[draw=none] {}\dl coordinate (xQ);
    \draw[red, thick, dotted] (eP)   to ++(-1,-1) node[draw=none] {}\dl to ++(-1,1) node[draw=none] {} coordinate (xP);
    \node[red] at (1,2) {$Q'$};
    \node[blue] at (3,4) {$P'$};
    \node[label=above:{$x$}, fill, circle, inner sep=1pt] at (xQ) {};
    \node[fill, circle, inner sep=1pt] at (xP) {};
  \end{tikzpicture}
\caption{The injection for Case 3.  Case 2 fails because the translated segment of $Q$ leaves the valid region before intersecting $Q$.}
\label{fig:case3}
\end{figure}
\end{description}

Case 1 covers all pairs of paths $P,Q$ such that $P$ is above $Q$ at $e$ and below it at $f$ or vice versa.  Case 2 covers all remaining pairs of paths with $P$ above $Q$ at $f$, since in this case the fragment of $Q$ must intersect $P$ and both resulting paths $(P',Q')$ are between $P$ and $Q$.  Finally, Case 3 covers all still remaining pairs of paths with $P$ below $Q$ at $e$.  Hence every pair of paths $P,Q$ is covered by one of these cases.

Finally we must argue that $\varphi_{ef}$ is injective.  Given $P',Q'$ we must be able to determine which case above was applied to produce them, and from this it is simple to recover $P$ and $Q$.  Since we know $e,f$ it is easy to identify the regions (I), (II), (III) in the three cases.  If $P',Q'$ intersect in region (I) then they must have come from Case 1, as neither of the later cases can produce such an intersection.  If they do not, first try to apply the inverse operation for Case 2, which is the same as the forwards operation; if this was not possible for the starting paths $P,Q$ then it is not possible for $P',Q'$ either (as performing the transformation of Case 3 cannot cause the transformation of Case 2 to become valid if it was not already), so if we have a pair of paths $P',Q'$ for which the second move is possible and does result in lattice paths between $A$ and $B$ then we know we must have arrived at it through Case 2.  Otherwise, we were in Case 3.  See Figure \ref{fig:recovery}.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}[scale=.25]
    \path[use as bounding box] (0,0) rectangle (22,8);
    \fill[fill=black!15] (7,0) rectangle ++(1,8)
    \fill[fill=black!15] (12,0) rectangle ++(1,8)

    \begin{scope}
        \path[clip] (0,0) rectangle (22,8);
        \draw[thin, draw=black!40] (0,0.2) \u \u \u \u \u \u \u \u \u \u \u \d \d \d \d \d \d \d \d \d \d \d;
        \draw[thin, draw=black!40] (0,0)   \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d \u \d;
    \end{scope}

    \draw[blue, thick] (0,0.15)
    		\u \u \u \u \d \d \u \u coordinate (eQ) \u \d \u \u coordinate (fQ) \d \d \d \u \u \d \d \d \d \d;
    \draw[red, thick] (0,0)
    		\u \u \d \u \d \u \u \d coordinate (eP) \u \d \u \d coordinate (fP) \u \d \u \d \d \u \d \u \d \d;

    \draw[red, thick, dotted] (fQ)           \u \d \u \d \d \u;
    \draw[blue, thick, dotted] (fP)          \d \d \d;

    \draw[blue, thick, dotted] (eQ)  to ++(-1,1) node[draw=none] {} to ++(-1,-1) node[draw=none] {}\dl coordinate (xQ);
    \draw[red, thick, dotted] (eP)   to ++(-1,-1) node[draw=none] {}\dl to ++(-1,1) node[draw=none] {} coordinate (xP);
    \node[red] at (1,2) {$Q'$};
    \node[blue] at (3,4) {$P'$};
  \end{tikzpicture}
\caption{An example of recovering the case from the result.  Since the paths do not intersect in region (I) we must have come from Case 2 or 3.  Case 2 fails because the translated segment of $P'$ leaves the valid region before rejoining $Q'$, so the paths must have come from Case 3.  Note that the transformation $\phi_{ef}$ is usually not onto.}
\label{fig:recovery}
\end{figure}
\end{proof}

\begin{remark}
    Since Case 1 can be applied whenever $P$ is above $Q$ at $e$ and below at $f$ or vice versa, Case 2 can be applied whenever $P$ is above $Q$ at both $e$ and $f$, and Case 3 can be applied whenever $P$ is below $Q$ at both $e$ and $f$, it is tempting to try to divide the cases more naturally according to the relative heights of $P$ and $Q$ at $e$ and $f$.  However, note that in Cases 2 and 3 (and sometimes Case 1) the resulting paths $P',Q'$ always have $P'$ above $Q'$ at both $e$ and $f$, and this more natural division of cases does not allow us to recover uniquely which transformation was applied.
\end{remark}

\begin{remark}
    While the result of \cite{Bonin2006} that lattice path matroids are minor-closed shows that it suffices for us to prove negative correlation, it is worth noting that the same injection shows negative correlation directly for any minor of a lattice path matroid.  In particular, if we assume that $P$ and $Q$ above are both in the minor $\mathcal{B}_I^J$ (i.e., have up-steps at the indices of $I$ and down-steps at the indices in $J$) then the resulting paths $P'$ and $Q'$ again have this property. Moreover, the injection also implies that lattice path matroids are Rayleigh matroids (a class of matroids studied in \cite{Choe2006}).
\end{remark}

\section{Lower bound} \label{sec:lower-bounds}

To derive a lower bound on the mixing time of the Dyck random transposition chain we will need to refer to the definitions of the spectral gap and log-Sobolev constant.  Recall that for a Markov chain with state space $\Omega$, transition matrix $P$, and stationary distribution $\pi$
\begin{align*}
    \lambda \vcentcolon= \inf_f \frac{\mathcal{E}(f,f)}{\mathop{\mathrm{Var}}_\pi(f)}
    \qquad
    \text{and}
    \qquad
    \alpha \vcentcolon= \inf_f \frac{\mathcal{E}(f,f)}{\mathcal{L}_\pi(f)},
\end{align*}
where
\begin{align*}
    \mathcal{E}(f,f) &= \frac{1}{2} \sum_{x,y} (f(x) - f(y))^2 P(x,y) \pi(x)
    \qquad
    \text{and}
    \qquad\\
    \mathcal{L}_\pi(f) &= \sum_x f(x)^2(\log f(x)^2 - \log \mathop{\mathrm{E}}_\pi(f^2)) \pi(x)
\end{align*}
and the infima are taken over non-constant functions $f:\Omega\to \mathbb{R}$.

Using standard results we can (asymptotically) lower bound the mixing time by the relaxation time $\frac{1}{\lambda}$.
\todo[color=blue!20, linecolor=blue, bordercolor=blue]{Can we give a better bound on $\alpha$ directly by using this test function?  It looks from Mathematica's plots as though $\mathcal{L}_\pi(f) \sim 4/n$, which probably isn't worth showing.}

\todo[color=blue!20, linecolor=blue, bordercolor=blue]{Can we prove an $n^3\log n$ lower bound for Dyck adjacent transpositions, perhaps using a method similar to Wilson's lower bound for unconstrained adjacent transpositions?}

\begin{theorem}
    The spectral gap $\lambda$ and log-Sobolev constant $\alpha$ for the Dyck random transposition chain satisfy
    \[2\alpha \leq \lambda \leq \frac{4}{n}.\]
\end{theorem}

\begin{proof}
It is always the case that $\alpha \leq \lambda/2$ (see, e.g., \cite{Jerrum-Son}), so it suffices to show that $1/\lambda\geq n/4$.

For $x\in \Omega$ a Dyck path of length $2n$, consider the function $f(x)$ giving the number of down-steps of $x$ at even indices.  The number of Dyck paths $x$ of length $2n$ with $f(x) = k$ is precisely the Narayana number $N(n,k) = \frac{1}{n}\binom{n}{k}\binom{n}{k-1}$ (see \cite{Sulanke2002}).
Note that $f$ is $1$-Lipschitz, in that if $x,y$ differ by a single transposition (i.e.\ $P(x,y) > 0$) then $\abs{f(x) - f(y)} \leq 1$.  We can therefore bound $\lambda$ by noting that for a $1$-Lipschitz function we have $\mathcal{E}(f,f) \leq 1/2$, and so it suffices to give a lower bound on $\mathop{\mathrm{Var}}_\pi(f)$,
which is precisely the variance of the Narayana distribution with p.m.f.\ $p(k) = N(n,k)/C_n$.  This distribution is hypergeometric, and its variance is $\frac{(n+1)(n-1)}{4(2n-1)} \geq n/8$ (see, e.g., \cite{Johnson2005}).
\end{proof}

\begin{corollary}
    The (total variation) mixing time of the Dyck random transposition walk is at least $\Omega(n)$.
\end{corollary}

\section{Open questions} \label{sec:open-questions}

Of course, the most immediate open question is whether the $O(n^2\log n)$ mixing time bound for Dyck transpositions is tight.
\todo[color=blue!20, linecolor=blue, bordercolor=blue]{Is there an easy $\Omega(n\log n)$ lower bound, or one we can cite?  We have an $O(1/n)$ bound on spectral gap.}
The general result for lattice path matroids is clearly not tight in every case (as it gives only order $n^2\log n$ mixing time for Bernoulli-Laplace, where we know that the true bound is order $n\log n$), but there are lattice path matroids for which the $O(n^2\log n)$ bound is tight\footnote{One example is the lattice path matroid $\L[A,B]$ where $A = (-+)^n$ and $B=(+-)^n$, which is equivalent to the random walk on the $n$-cube slowed down by a factor of $n$.}.  We believe that the Catalan matroid falls closer to the Bernoulli-Laplace side of the spectrum.
Either way, we hope in the future to work towards a more complete characterization of the mixing rates for various lattice path transposition chains.
\todo[color=blue!20, linecolor=blue, bordercolor=blue]{dy: Can we prove a kind of monotonic result: say, if the lattice region A is contained in the region B, then mixing time for A is at least as for B?}
\todo[color=blue!20, linecolor=blue, bordercolor=blue]{EC: The chain on the region between $+-(+-)^{n-1}$ and $-+(+-)^{n-1}$ mixes in order $n^2$ time, which is faster than the $n^2 \log n$ mixing time for the region between $(+-)^n$ and $(-+)^n$. Might still work for spectral gap?}

It is also not clear what consequences the new bound for Dyck paths might have for mixing on other Catalan structures.  Many well-known Catalan structures suggest natural Markov chains, but it is striking how different they can be---even though there are natural bijections between classes of Catalan structures, they frequently do not preserve natural notions of distance, making standard comparison arguments tricky.  There seems to be no obvious way to leverage even a hypothetical $O(n\log n)$ mixing time result for Dyck transpositions to yield improved bounds for any other Catalan chain. It is worth noting that the classical ($2n+1$-to-one many) Chung-Feller cyclic lemma suggests projecting the chain consisting of all transpositions on the (Bernoulli-Laplace) space of $\{\pm 1\}^{2n+1}$, with $(n+1)$ 1's and $n$ -1's, onto the set of Dyck paths. This projection chain is in fact a Markov chain, as was observed by \cite{CRT10}, and as such inherits a lower bound on the spectral gap of order $1/n$, from that of the Bernoulli-Laplace chain; however, many of the moves in the resulting projection chain are not particularly natural when viewed from the point of Dyck paths, and once again comparisons do not seem to help very much.

An alternate approach would be to come up with a general scheme to bound mixing times which could be modified for any Catalan structure.  For example, it is well-known that the Catalan numbers satisfy the recurrence
\[C_n = \sum_{i=1}^{n} C_{i-1} C_{n-i},\]
and this is evidenced by a recursive structure in almost every combinatorial realization of the Catalan sequence.  It is common to exploit recursive structure in bounding mixing times, and indeed both the triangulation bound of \cite{MRS98} and the balanced matroid bounds of \cite{Feder1992, Jerrum-Son, Jerrum2004} exploit different recursive structures of those objects.  Unlike the recurrence above, the recursive structures used in those papers seem very particular to the Catalan realizations at hand, and do not generalize well to other Catalan chains

Finally, we have mentioned above the conjecture that the basis exchange walk is rapidly mixing for \emph{all} matroids.  As evidence, it would be interesting to show rapid mixing for some class of matroids which is not a subclass of balanced matroids.  One promising candidate might be the class of transversal matroids, for which the basis exchange walk seems closely related to the well-studied problem of walks on bipartite graph matchings.

\section*{Acknowledgements}
The authors thank the Institute for Mathematics and its Applications (IMA) for its excellent research atmosphere, generous support, and hospitality, which were all crucial to the collaboration resulting in the present work. The first two authors also acknowledge additional NSF support by way of the grant DMS-1407657.

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


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Flying A Car Museum, north of 2nd St., west side of Main St.
Beautiful restored vintage cars, gas pumps, pump advertising signs. Only open on Saturday.
Elephant Skull, Albany, TX - < 1 mi.
Giant Texas Longhorn Skull, Albany, TX - < 1 mi.
Restored Sinclair Gas Station, Albany, TX - < 1 mi.
Bank Robbed By Santa Claus, Cisco, TX - 30 mi.


text:
Police Staff
Firearms Police Salary [UK]
Ever wondered how much a police firearms officer makes? Whether you're just curious or are considering a career as an armed officer, this article will take you through everything you need to know, including what they earn, as well as what subscriptions they need to pay for.
Firearms Police Salary: Starting Wage
London Weighting for Firearms Police Officers
Firearms Police Officer Yearly Salary [By Rank]
Firearms Police Salary by Rank
Firearms Police Constable Salary
Firearms Police Sergeant Salary
Firearms Police Inspectors' Salary
Firearms Police Salary vs Take Home Pay
Unsociable Hours
Police Federation
PFOA (Police Firearms Officers Association)
Summary: Firearms Police Salary UK
As we have discussed in our article on police officers' salaries, being a police officer is not going to make you a millionaire on its own. The role is a vocation and is not designed for those that are seeking a fast track to wealth.
The starting wage for a brand new police officer is just over £18,000 per year if you are starting on the PCDA route (earning a degree in policing as you train).
London is an expensive city to live in, so officers wanting to work there, will get a slightly higher salary, known as 'London Weighting'.
Officers working in London (the Metropolitan Police or City of London Police) get additional pay, known as 'London Weighting'.
Due to the cost of living in London, officers here are paid slightly higher than in the rest of the country.
London allowances are; £2,373 per annum pensionable London weighting, and £1,011 per annum non-pensionable London allowance.
There is the potential that as an officer living and working in London that you may qualify for rent or housing allowance.
If you are not entitled to rent or housing allowance, you will receive an additional amount of £3,327, which is to compensate for the increased cost of living. This is the same for all police officers, including firearms officers.
The table below shows the starting wage for police officers and then the hourly rate that police officer gets paid. A full-time police officer works 40 hours per week.
All of the police officers' monthly and hourly pay calculations have been worked out using a 40-hour workweek and are calculated in GBP.
These salaries are the same for firearms trained officers and non firearms trained officers too.
For a quick summary of police officer starting salaries for each police officer rank as of September 2020 check out the below;
➡ Constable Starting Wage: £18,912
➡ Sergeant Starting Wage: £43,965
➡ Inspector Starting Wage: £52,698
*taken from the Police Federations website
Most police forces do not have Inspectors as firearms officers (i.e that actually carry a firearm), however, they are still an active part of the command structure so have been included here.
PCDA 18,912 1,576 10
0 21,402 1,784 11
The first three years for PCDA and the first two years of other entry routes have been struck through. This is because officers in their probation cannot become firearms police officers.
In order to apply to be an armed police officer, you need to be confirmed in post. This means that you need to be signed out of your probation.
1 Removed
From your salary, your pension contributions will automatically be deducted. If you stay with the police pension, 14% will be deducted from your wages and automatically and automatically paid into your pension for you.
For more information on police pensions for officers joining check out this article.
This is no different for firearms officers.
Officers that are undertaking a role that needs them to work night shifts will get additional pay for these hours.
As all forces will have a 24/7 firearms capability, as an armed officer, you will work unsociable hours
Some specialist roles within policing will require officers to be 'on call'.
This means that the officer can be called out from their home address in order to perform specialist duties at short notice.
Some firearms roles may require an officer to be on call.
Officers will get paid an additional wage for this.
With budget constraints, overtime is not as available as it used to be, however, if police officers perform additional duties, they may be eligible to claim overtime for these.
Firearms is a role that is essential for the protection of all, as such, it is an area that over time is more readily available.
Most police officers will be paid members of the Police Federation.
The Police Federation is a staff association that offers support to officers in their work and home life. Membership is less than £25 per month.
Most firearms officers will also subscribe to the PFOA. It is an association that supports those who carry a taser, firearm, or are part of the command structure around the deployment of the same.
At the time of writing. this costs less than £5 per month.
In short, the firearms police salary is the same as that of a non-firearms trained officer. Despite the additional training required in order to undertake the role.
What Should I Study To Become A Police Officer?
Can I Join The Police At 50 [UK]?
What Personal Qualities Do You Need To Become A Special Constable?
Becoming a Special Constable can be a great way of giving back to your local community, as well as gaining new skills and experience. But, what personal qualities do you need to become a special constable? And why are these qualities important? In this article, we will answer these questions.
What's special about the Special Constabulary?
What is the role of Special Constables? Do you have questions about what Special Constables do, what powers they have, or even, how can you become a Special Constable? You're in the right place! We will take you through the role of a Special Constable, the application process and what to expect at training school and beyond!
Can I join the police at 50 is a question we often get. It's not surprising that more people have multiple careers in a lifetime. And maybe becoming a police officer is your second career. So, let's get into what ages you can join the police and other issues you may want o to consider.
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Bonuses are tricky. Here's how to make your bonus work harder for you by reducing the tax impact.
By weighing the tax implications and your long-term goals, you can make your bonus go further.
A tax-advantaged account like a 401(k), IRA, or HSA may make for an effective home for your bonus.
How are you planning to spend your annual bonus? Like with any cash windfall, we all want to use it wisely. But bonuses can be tricky because of taxes.
To use a bonus most tax-efficiently, you'll need to juggle multiple objectives and concerns.
If you're expecting to get more than one bonus per year, it's important to consider all of the possible ways to invest a bonus to maximize its potential value. In this article, we'll review how bonuses are typically taxed, what factors you should be aware of, and how to take advantage of different accounts and investing strategies to make your bonus work harder for you.
Put your bonus to work with Betterment.
How Does a Bonus Get Taxed?
Bonuses are considered "supplemental income" by the IRS, which means they could be withheld differently than your regular salary.
The IRS suggests a flat withholding of 22% from bonuses, and many employers follow that method. (Remember that withholdings are meant to be an estimate of how much you'll owe at the end of the year, not the actual tax itself.) But some employers use the aggregate method, in which your whole bonus is added to your regular paycheck, and the combined amount is withheld at the normal income rate, as though that amount is representative of what you make every paycheck, which could be higher (or lower) than 22%.
Some people believe that bonuses are taxed at a higher rate than ordinary wages, but that's not the case. The aggregate method of withholding can result in bumping you into a higher estimated tax bracket, which creates the illusion that you "keep less of it," but no special tax rates apply just because a payment from your employer is characterized as a bonus. A bonus is like a raise, but when your income goes up, it could do more that just move you to a higher tax bracket—you could potentially lose certain deductions and tax credits.
Bear in mind, while we hope you find this information helpful, you should consult a tax professional to understand your individual circumstances. Betterment is not a tax advisor, so while we like to offer helpful information to get you started, this should not be considered tax advice.
With that said, here are some simple suggestions for how you can use tax-deferred or even taxable accounts to help preserve and grow your windfall.
Before you add your bonus to your 401(k), check with your employer about how bonuses are handled. In some cases, your company may not allow you to make 401(k) contributions using your bonus.
In others, your 401(k) plan may be set up to withhold the same percentage from your bonus as from your paycheck. Thus, if you typically contribute 10% from every paycheck to your 401(k), that same amount could be withheld from your bonus (unless you say otherwise). In the case of a $15,000 bonus, $1,500 would go into your 401(k) for 2019, which may be too little for your aims.
Of course, you can't contribute more than the annual limit, so be sure to check how much you've contributed for the year to date. The contribution limit for your 401(k) for 2019 is $19,000 ($25,000 if you're 50 or older). You can choose any combination of pre-tax or Roth contributions as part of your $19,000/25,000. Not sure which type is good for you? Many participants "split the difference" and contribute 50% pre-tax and 50% Roth. To figure out what kind of contribution might work well for you, use Betterment's traditional vs. Roth 401(k) calculator.
Also, don't assume that a lump-sum deposit is best, especially if your employer matches your 401(k) contributions. A single large deposit might not get the same amount of matching dollars that a comparable amount would if you spread the deposits over time. Betterment's resident CFP® professional Alex Benke notes that it depends on your employer's matching structure. Certain plans offer a "true-up" for matching contributions if you max out early in the year while many plans do not offer that feature.
According to a Forbes report, if you're under 50, earning $240,000 and deposit $18,500 in your 401(k) by April 15, you'd get about $2,100 from your employer's match (assuming a match of 50% up to 6% of contributions). But if you spread out your contributions over the year, you could earn over $5,000 more in employer matching funds. Talk to your employer to find out exactly how they calculate the match.
Now here's the part you may not be aware of: depending on your income and whether you or your spouse is participating in a company retirement plan, you might be able to reduce your taxable income further by contributing to your flexible spending account this year (the maximum is $2,700 for 2019), a health savings account (the maximum for a family is $7,000 for 2019), and a traditional or Roth IRA.
Many people don't realize that you can participate in a company plan and still fund a traditional or Roth IRA. You could contribute to your 401(k) this year, and contribute to a traditional or Roth IRA as well, or a combination of those.
If it's not possible or advantageous to put your money only into tax-deferred accounts, use your windfall to invest by creating "a gift that keeps on giving." You could spend it all, sure, but by investing your windfall in a well-diversified portfolio, you can create an additional source of cash flow that steadily adds to your quality of life, year after year: i.e. a happiness annuity.
Studies show that steady cash flow increases often feel better than a lump sum that's here today, spent on the Canary Islands tomorrow.
Tax deductions can lower the amount of income that you end up paying taxes on. We'll highlight popular above-the-line deductions and below-the-line deductions you may want to consider.


text:
Heidi Zimmermann, avstrijska alpska smučarka, * 1. maj 1946, Zürs, Avstrija.

Nastopila je na Svetovnem prvenstvu 1966, kjer je osvojila srebrno medaljo v veleslalomu in bronasto v kombinaciji. V svetovnem pokalu je tekmovala štiri sezone med letoma 1967 in 1970. V skupnem seštevku svetovnega pokala je bila najvišje na šestnajstem mestu leta 1967. Leta 1969 je postala avstrijska državna prvakinja v veleslalomu.

Zunanje povezave

Avstrijski alpski smučarji

redpajama_set_name: RedPajamaWikipedia


We provide seamless and holistic legal advice in all areas of business law that are relevant to your business or commercial venture. We do not just focus on your proposed or existing venture but on your particular circumstances in order to fully and properly advise you. We care about you and your business and work with you to achieve your business goals.
If you are establishing or purchasing a business we can advise on important matters of business law relating to structuring and tax in addition to critically evaluating what you are actually buying. If you have an existing business we can advise on strategies to maintain the health of the business and advise on any disputes that may arise.
Our philosophy is to create a "win-win" situation for all business partners and to set the terms of agreement between business partners so that they operate fairly and that no partner can take advantage of another, in the event of separation. Our agreements provide practical quick solutions to cause the separation to operate fairly without resource to litigation and without risk of causing the business to go into liquidation.
Talk to one of our expert lawyers today.


Cream puff connoisseurs Croquant Chou Zakuzaku are getting in on the Halloween season fun this year by releasing two tasty treats. The Kuro Zaku and Kuro Zaku Soft will be available from October 1 to October 31.
The Kuro Zaku is an almond crunch croquant filled with a cream that's made with black cocoa powder. Inside that is a purple sweet potato cream that's made with 1:1 custard cream and purple sweet potato paste. The paste is made in Japan and its purple colour gives the dessert that Halloween look.
The Kuro Zaku Soft is an ice cream that's completely black – including the cone. The rich soft serve ice cream melts like fresh cream and includes a white chocolate dip made with bamboo charcoal powder.
Customers who order either of these desserts will take them selves a Halloween face paint seal too.
The shop is designed like a factory so you can eat everything freshly made. Try their delicious products for yourself this Halloween.


Today I will be sharing some simple tips which can be helpful in avoiding or reducing the appearance under-eye circles. Dark circles make us look tired, sick and older. People with dark circles generally feel less confident and conscious of their looks.
Many of us might be very well aware of the tips suggested below so this post can act as alarm for them to get the wheels rolling.
1. Sleep Tight:- Sound and sufficient sleep plays an vital role in good health and well-being . Lack of sleep or errant sleeping schedules can affect your mental as well as physical health. The skin around your eyes looks swollen and shows the effects of fatigue. Adequate sleep helps maintain a healthy balance of the hormones and repair the whole system while you sleep but not getting enough proper sleep can increase the stress levels in the body. To get enough quality sleep at night try to avoid sleeping during day time. Playing a light music at low volume also helps in getting that beauty sleep.
2. Reduce Stress:- Stressing out a lot can adversely affect your skin as well as health. If you are feeling agitated or worried just before going to sleep you might not get a sound sleep. Reduce your stress as it will help you to sleep better. So, release all the tensions and feel relaxed before hitting on to the bed.
3. Drink Plenty of Water:- Dehydration can be the most important reason for dark circles. We lose water from the body through sweating and urinating and it is very important to compensate the loss by drinking more water. It is good to have 8 ounce (3-4 liters) of water every day to keep the skin fresh and moistened. Water also flushes out the toxins from the body thus keeping our skin healthy and reducing the appearance dark circles.
4. Quit Smoking:- You can attain a beautiful, glowing complexion by avoiding smoking. Smoking thins the skin and depletes the collagen in your skin which makes our skin age more rapidly causing wrinkles. So, if you care for your looks I would suggest giving up smoking, alcohol and products high on caffeine. This will help in reducing the appearance of dark under-eye circles and improve your overall health.
5. Wear Sunglasses:– Protect your eyes from the sun by wearing sunglasses. Prolonged exposure to sunlight, especially during summer may adversely affect the delicate under-eye area and can lead to dark circles. Sunglasses also help in preventing wrinkles and fine liners below the eyes caused by harmful sun's UV rays.
6. Avoid rubbing:- The under-eye area is very delicate and we should treat it gently and with care. So, stop rubbing your eyes even if it gets itchy. Keep some eye drops handy to reduce the itchiness or just pat them lightly and resist rubbing them hard.
Sources of Vitamin K:- Spinach, broccoli, cabbage, cucumber, cauliflower, tomatoes and blueberries..
Sources of Vitamin B12:- Fish, meat (especially liver), poultry, eggs, milk, and milk products.
These small changes in the lifestyle may reduce or completely eliminate the appearance of dark-circles. So while you are waiting for these tips to show their effect use an under-eye concealer to hide those dark-circles and stay beautiful.


Adventure junkie Sidharth Malhotra is making the most of his trip to New Zealand and this time the Bollywood actor tried another new thing in the form of kayaking in the Waitemata Harbour.Kayaking within sight of the country's youngest volcanic cone was a wonderful experience for Sidharth."New Zealand has endless opportunities for adventure lovers, and it comes as no surprise that it is one of the best places in the world for the same. It indeed was a once in a lifetime experience for me to learn kayaking at the Waitemata Harbour with well trained guides. A perfect work-out for my arms and shoulders," he said.After that, he also enjoyed a scenic view while on a seaplane and cruise at Auckland."Flying over the beautiful harbour was beautiful. Such a thing can only be experienced in a country like New Zealand. It is a great way to see the city from a different perspective," said Sidharth.He also engaged in archery shooting.