text:
\section{Introduction}
In the Reinforcement Learning (RL) paradigm, an agent interacts with an unknown environment by taking actions and collecting rewards. Throughout this process, the agent strives to maximize its total expected reward, or value $Q$. The optimal value can be recovered with Bellman's equation \cite{dp:1957:bellman:dynamic_programming}, whereby observed rewards are used to update estimates of $Q$ through the unbiased, recursive relation
\begin{align*}
Q(\mathbf{s},\mathbf{a}) = r + \gamma \mathbf{E}[Q(\mathbf{S},\mathbf{A})]	.
\end{align*}
This applies as the agent undergoes a random transition from $(\mathbf{s},\mathbf{a}) \rightarrow (\mathbf{S},\mathbf{A})$.

Bellman's equation motivates many methods for finding the optimal value. Among the most data-efficient are the class of Gaussian Process methods, which replace sample-intensive estimation schemes with a Bayesian non-parametric estimator, based on Gaussian Process regression \cite{rl:deisenroth_rasmussen-pilco,rl:2003:engel_mannor_meir:bayes_meets_bellman_the_gaussian_process_approach_to_temporal_difference_learning,rl:ghavamzadeh_engel_valko-bayesian_policy_gradient_and_actor_critic_algorithm,rl:2017:martin:heteroscedastic,rl:2005:engel_etal:learning_to_control_an_octopus_arm_with_gaussian_process_temporal_difference_methods,rl:2018:martin_etal:spgp_sarsa,rl:2004:rasmussen_kuss:gaussian_processes_in_reinforcement_learning} . These methods have shown to yield state-of-the-art empirical performance in their respective domains, such as model-based and model-free learning.

In this paper, we consider the class of Gaussian Process methods for model-free temporal difference learning \cite{rl:2003:engel_mannor_meir:bayes_meets_bellman_the_gaussian_process_approach_to_temporal_difference_learning,rl:2018:martin_etal:spgp_sarsa}. Specifically, we target the Sparse Pseudo-input Gaussian Process SARSA (\textsc{spgp-sarsa}) method \cite{rl:2018:martin_etal:spgp_sarsa} to improve its online viability with a procedure to perform recursive updates. By extending \textsc{spgp-sarsa} this way, its functionality then covers the full scope of benefits captured by its predecessor, \textsc{gp-sarsa} \cite{rl:2003:engel_mannor_meir:bayes_meets_bellman_the_gaussian_process_approach_to_temporal_difference_learning}; namely, previous results can be reused to compute matrix inverses.

\section{TD Value Estimation as GP Regression}\label{sec:gptd}
\textsc{td} algorithms recover the latent value function with data gathered in the standard \textsc{rl} fashion: at each step, the robot selects an action $\mathbf{a}\in\mathcal{A}$ based on its current state $\mathbf{s}\in\mathcal{S}$, after which it transitions to the next state $\mathbf{s}'$ and collects a reward $R\sim p_r(\cdot|\mathbf{s},\mathbf{a})$. The repeated interaction is described as a Markov Decision Process, $(\mathcal{S},\mathcal{A},p_r,p_s,\gamma)$, associated with the transition distribution $\mathbf{s}'\sim p_s(\cdot|\mathbf{s},\mathbf{a})$, stationary policy $\mathbf{a}\sim\pi(\cdot|\mathbf{s})$, and discount factor $\gamma \in [0,1]$. As the name suggests, \textsc{td} algorithms update a running estimate of the value function to minimize its error difference from the Bellman estimate: $r + \gamma Q(\mathbf{s}',\mathbf{a}') - Q(\mathbf{s}, \mathbf{a})$; $r$ being the observed reward. Once the estimate converges, an agent can select actions from the greedy policy $\pi$, such that $\mathbf{E}_\pi[\mathbf{A}|\mathbf{s}] = \argmax_{\mathbf{a}\in \mathcal{A}} Q(\mathbf{s},\mathbf{a})$.

The Gaussian Process Temporal Difference (\textsc{gptd}) framework improves upon the data efficiency of frequentist \textsc{td} estimation by departing from the contractive nature of Bellman's equation, in favor of a convergence driven by non-parametric Bayesian regression. The data model is based on the random return $Z(\mathbf{x}) = \sum_{t=0}^\infty \gamma^tR(\mathbf{x}_t)$, expressed as a sum of its mean, $Q(\mathbf{x})$, and zero-mean residual, $\Delta Q(\mathbf{x}) = Z(\mathbf{x}) - Q(\mathbf{x})$. Model inputs are state-action vectors $\mathbf{x}\in \mathcal{X}=\mathcal{S}\times\mathcal{A}$, and value differences are used to describe the observation process:
\begin{align}\label{eq:trans_likelihood}
	R(\mathbf{x}) &= Q(\mathbf{x}) -\gamma Q(\mathbf{x}') + [\Delta Q(\mathbf{x})-\gamma\Delta Q(\mathbf{x}')] = Q(\mathbf{x}) -\gamma Q(\mathbf{x}') + \varepsilon(\mathbf{x},\mathbf{x}').
\end{align}
Moving forward, we assume that noise levels, $\varepsilon(\mathbf{x},\mathbf{x}')$, are i.i.d random variables with constant parameters, $\varepsilon \sim \mathcal{N}(0,\sigma^2)$. Under this assumption, transitions exhibit no serial correlation, and the \textsc{spgp-sarsa} model is valid.

Given a time-indexed sequence of transitions $(\mathbf{x}_{t},R(\mathbf{x}_{t}),\mathbf{x}_{t+1})_{t=0}^{N-1}$, the \textsc{gp-sarsa} model stacks variables into vectors to obtain the complete data model: $\mathbf{r} = \mathbf{H}\mathbf{q}(\mathbf{x})+\boldsymbol{\varepsilon}$, where
\begin{align}\label{eq:gptd_model}
	\begin{pmatrix}
		R(\mathbf{x}_0)\\
		R(\mathbf{x}_1)\\
		\vdots\\
		R(\mathbf{x}_{N-1})
	\end{pmatrix} &=
	\begin{pmatrix}
		1 & -\gamma & 0 & \cdots & 0\\
		0 & 1 & -\gamma & \cdots &0 \\
		\vdots & & & & \vdots\\
		0 & 0 &\cdots & 1 & -\gamma
	\end{pmatrix}
	\begin{pmatrix}
		Q(\mathbf{x}_0)\\
		Q(\mathbf{x}_1)\\
		\vdots\\
		Q(\mathbf{x}_N)
	\end{pmatrix} +
	\begin{pmatrix}
		\varepsilon_0\\
		\varepsilon_1\\
		\vdots\\
		\varepsilon_N
	\end{pmatrix},
\end{align}
and $\mathbf{q} \sim \mathcal{N}(\mathbf{0},\mathbf{K}_{qq})$. Notice the commonality Equation \ref{eq:gptd_model} has with a standard \textsc{gp} likelihood model, $\mathbf{y} = \mathbf{f}(\mathbf{x}) + \boldsymbol{\varepsilon}$. Both models assume the outputs, $\mathbf{r} \sim \mathbf{y}$, are noisy observations of a latent function, $\mathbf{q} \sim \mathbf{f}$. What distingushes \textsc{td} estimation is the presence of value correlations, imposed from Bellman's equation and encoded as temporal difference coefficients in $\mathbf{H}$. Used for exact \textsc{gp} regression, Equation \ref{eq:gptd_model} leads to the \textsc{gp-sarsa} algorithm: a non-parametric Bayesian method for recovering latent values \cite{rl:2003:engel_mannor_meir:bayes_meets_bellman_the_gaussian_process_approach_to_temporal_difference_learning}.

As a Bayesian method, \textsc{gp-sarsa} computes a predictive posterior over the latent values by conditioning on observed rewards. The corresponding mean and variance are used for policy evaluation:
\begin{align}
	v(\mathbf{x}_*) &= \mathbf{k}^\top_{r*}(\mathbf{K}_{rr} + \sigma^2\mathbf{I})^{-1}\mathbf{r}, &
	s(\mathbf{x}_*) &= k(\mathbf{x}_*,\mathbf{x}_*) -\mathbf{k}^\top_{r*}(\mathbf{K}_{rr} + \sigma^2\mathbf{I})^{-1}\mathbf{k}_{r*}.
	\label{eq:gptd_predmoms}
\end{align}
Here, $\mathbf{K}_{qq}$ is the covariance matrix with elements  $[\mathbf{K}_{qq}]_{ij} = k(\mathbf{x}_i,\mathbf{x}_j)$, $\mathbf{K}_{rr} = \mathbf{H}\mathbf{K}_{qq}\mathbf{H}^\top$, and $\mathbf{k}_{r*} = \mathbf{H}\mathbf{k}_*$, where $[\mathbf{k}_*]_{i} = k(\mathbf{x}_i,\mathbf{x}_*)$. Subscripts denote dimensionality, e.g. $\mathbf{K}_{qq} \in \mathbb{R}^{|\mathbf{q}|\times |\mathbf{q}|}$.

\section{Sparse Pseudo-input Gaussian Process Temporal Difference Learning} \label{sec:sgptd}
\begin{figure*}
	\centering
	\includegraphics[width=0.3\columnwidth]{gp-sarsa.pdf}
	\includegraphics[width=0.3\columnwidth]{spgp-sarsa-nop.pdf}
	\includegraphics[width=0.3\columnwidth]{spgp-sarsa-opt.pdf}
	\caption{\textbf{Visualizing \textsc{gp}- and \textsc{spgp-sarsa} posteriors:} The exact \textsc{gp-sarsa} posterior (left) is supported with all the training data (dots). The \textsc{spgp-sarsa} posterior (center) uses randomly-initialized pseudo inputs (diamonds). After adjusting the pseudo inputs to maximize likelihood of training data, the posterior (right) is nearly identical to the exact model.}
\end{figure*}
The \textsc{gp-sarsa} method requires an expensive $N\times N$ matrix inversion, costing $\mathcal{O}(N^3)$. To improve computational efficiency, \textsc{spgp-sarsa} algorithm applies the Sparse Pseudo-input Approximation \cite{gp:2006:snelson_ghahramani:sparse_gaussian_processes_using_pseudo_inputs}. Sparsity is induced in the standard data model (Equation \ref{eq:gptd_model}) by expanding the probability space with $M\ll N$ additional pseudo values, $\mathbf{u}$. The corresponding pseudo inputs, $\mathbf{z}\in\mathbf{Z}\subset \mathcal{X}$, act as parameters on the support of the predictive posterior. These extra latent variables obey the same data model as $\mathbf{q}$, but are predetermined, and thus, exhibit no noise. By conditioning $\mathbf{q}$ upon $\mathbf{u}$ and $\mathbf{Z}$, the predictive probability space collapses such that all dense matrix inversions are of rank $M$. This algorithm is called Sparse Pseudo-input Gaussian Process SARSA (\textsc{spgp-sarsa}) \cite{rl:2018:martin_etal:spgp_sarsa}.

The \textsc{spgp-sarsa} predictive posterior is Gaussian, $\mathcal{N}(\tilde{v}(\mathbf{x}),\tilde{p}(\mathbf{x}))$, with parameter functions
\begin{align}\label{eq:pred_moments}
	\tilde{v}(\mathbf{x}_*) &=\mathbf{k}_{u*}^\top\mathbf{M}^{-1}\mathbf{K}_{ur}(\mathbf{Q} + \sigma^2\mathbf{I})^{-1}\mathbf{r} &
	\tilde{p}(\mathbf{x}_*) &= k(\mathbf{x}_*,\mathbf{x}_*) - \mathbf{k}_{u*}^\top(\mathbf{K}_{uu}^{-1} - \mathbf{M}^{-1})\mathbf{k}_{u*}.
\end{align}

\section{Deriving a Recursive Algorithm}
The terms in Equation \ref{eq:pred_moments} that do not depend on the input effectively parameterize the posterior. We denote these parameters as
\begin{align}\label{eq:predictive_mean}
	\boldsymbol{\alpha}_t &= \mathbf{C}_{kk}\mathbf{K}_{kt}\mathbf{H}_t^\top\mathbf{B}_{tt}\mathbf{r}_{t-1}, &
	\mathbf{P}_t &= \mathbf{A}_{kk} - \mathbf{C}_{kk}.
\end{align}

Here we adopt a new notation that allows us to index updates associated with each input variable, $\mathbf{x}$ and $\mathbf{z}$. We drop the $v$ and $u$ designations on matrices in favor of $t$ and $k$, which respectively denote the update index of $\mathbf{x}$ and $\mathbf{z}$. A detailed breakdown of the notation is given below
\begin{itemize}
	\item{$t$: the unique index for the inputs $\mathbf{x}$.}
	\item{$k$: the unique index for the pseudo inputs $\mathbf{z}$}
	\item{$\mathbf{k}_{t-1}(\mathbf{x}_t) = (k(\mathbf{x}_1,\mathbf{x}_t),\cdots,k(\mathbf{x}_{t-1},\mathbf{x}_t))^\top$. Here, the array covers the range $1,\cdots,t-1$, and $\mathbf{x}_t$ is the common argument to all elements.}
	\item{$\mathbf{K}_{k}$: the $k\times k$ covariance matrix of pseudo input evaluations $[\mathbf{K}_k]_{ij}=k(\mathbf{z}_i,\mathbf{z}_j)$}
	\item{$\mathbf{K}_t$: the $t\times t$ covariance matrix of input evaluations $[\mathbf{K}_t]_{ij}=k(\mathbf{x}_i,\mathbf{x}_j)$}
\end{itemize}

This new notation suggests that \textsc{spgp-sarsa} can be used on two timescales. There is the scale, $t$, associated with state transitions, and the scale $k$, associated with adding new pseudo inputs. Although we do not elaborate on when and how to apply multi-timescale updates, we believe this constitutes the subject of interesting future work.

There are four distinct modalities in which \textsc{spgp-sarsa} can be updated:
\begin{enumerate}
	\item{\textit{Offline}: $\mathbf{Z}$ is fixed. $\mathbf{X}$ is fixed.}
	\item{\textit{Offline}: $\mathbf{Z}$ can vary. $\mathbf{X}$ is fixed.}
	\item{\textit{Online}: $\mathbf{Z}$ is fixed. $\mathbf{X}$ can vary.}
	\item{\textit{Online}: $\mathbf{Z}$ can vary. $\mathbf{X}$ can vary.}
\end{enumerate}
Here we consider the third and fourth cases, when both the transition training set and the pseudo set can grow. We decompose the predictive moments into partitioned matrices and apply the partitioned matrix inversion lemma to derive a recursive algorithm for their updates.
\subsection{Partitioned Matrix Inversion Lemma}
Let $\mathbf{K}_t$ be a $t\times t$ symmetric positive definite matrix whose partition is
\begin{align}
	\mathbf{K}_t =
	\begin{pmatrix}
		\mathbf{K}_{t-1} & \mathbf{k}_t\\
		\mathbf{k}_t^\top & k_{tt}
	\end{pmatrix}.
\end{align}
Define $s_t = k_{tt} - \mathbf{k}^\top\mathbf{K}_{t-1}^{-1}\mathbf{k}_t$. Then the inverse is given by
\begin{align}
	\mathbf{K}_t^{-1} =
	\begin{pmatrix}
		\mathbf{K}_{t-1}^{-1} & \mathbf{0}\\
		\mathbf{0} & 0
	\end{pmatrix} +
	\frac{1}{s_t}
	\begin{pmatrix}
		\mathbf{K}_{t-1}^{-1}\mathbf{k}_t\\
		-1
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{k}_t^\top\mathbf{K}_{t-1}^{-1} & -1
	\end{pmatrix}.
\end{align}

\subsection{Partitioning the Fundamental Matrices}
For deterministic transitions, the noise matrix is
\begin{align}
		\mathbf{\Sigma}_t =
		\begin{pmatrix}
			\mathbf{\Sigma}_{t-1} & \mathbf{0}\\
			\mathbf{0} & \sigma^2_{t-1}
		\end{pmatrix}.
\end{align}

The Bellman matrix partition is
\begin{align}
		\mathbf{H}_t &=
	\left(
	\begin{array}{cc}
		\mathbf{H}_{t-1} \ \mathbf{0} \\
		\hline
		 \hbf^\top_{t}
	\end{array}
	\right),
	& \hbf_t &= (0, \cdots, 1 , -\gamma)^\top.
\end{align}
Here, $\mathbf{H}_t \in \mathbb{R}^{t-1\times t}$, with $\hbf_t \in \mathbb{R}^t$.

\begin{align}
		\mathbf{K}_{tt} &=
		\begin{pmatrix}
			\mathbf{K}_{t-1t-1} & \mathbf{k}_{t-1}(\mathbf{x}_t)\\
			\mathbf{k}_{t-1}^\top(\mathbf{x}_t) & k_{tt}
		\end{pmatrix}
\end{align}
Here, we define $\mathbf{k}_{t-1}(\mathbf{x}_t) = (k(\mathbf{x}_1,\mathbf{x}_t) , \cdots , k(\mathbf{x}_{t-1},\mathbf{x}_t))^\top$, $k_{tt}=k(\mathbf{x}_t , \mathbf{x}_t )$.

\begin{align}
	\mathbf{K}_{kk} =
	\begin{pmatrix}
		\mathbf{K}_{k-1k-1} & \mathbf{k}_{k-1}(\mathbf{z}_k)\\
		\mathbf{k}_{k-1}^\top(\mathbf{z}_k) & k_{kk}
	\end{pmatrix}
\end{align}
Here we define $\mathbf{k}_{k-1}(\mathbf{z}_t) = (k(\mathbf{z}_1,\mathbf{z}_t) , \cdots , k(\mathbf{z}_{k-1},\mathbf{z}_t))^\top$, $k_{kk}=k(\mathbf{z}_k , \mathbf{z}_k )$.

\begin{align}
	\mathbf{K}_{tk} &=
	\begin{pmatrix}
		\mathbf{K}_{t-1k-1} & \mathbf{k}_{t-1}(\mathbf{z}_k)\\
		\mathbf{k}_{k-1}^\top(\mathbf{x}_t) & k_{tk}
	\end{pmatrix}
\end{align}
This partition is special, because it can be updated in three ways. We may include a new input $\mathbf{x}_t$, a pseudo input $\mathbf{z}_k$, or both. Here, $\mathbf{x}$ includes the rows with index $t$, and the pseudo inputs, $\mathbf{z}$, include the columns with index $k$. We define $\mathbf{k}_{k-1}(\mathbf{x}_t) = (k(\mathbf{z}_1,\mathbf{x}_t) , \cdots , k(\mathbf{z}_{k-1},\mathbf{x}_t))^\top$, $k_{tk}=k(\mathbf{x}_t , \mathbf{z}_k )$, and $\mathbf{k}_{t-1}(\mathbf{z}_k) = (k(\mathbf{x}_1,\mathbf{z}_k) , \cdots , k(\mathbf{x}_{t-1},\mathbf{z}_k))^\top$. When $\mathbf{x}_t$ is new, we must compute $\mathbf{k}_{k-1}^\top(\mathbf{x}_t)$ and $k_{tk}$. Similarly, when $\mathbf{z}_k$ is new, we must compute $\mathbf{k}_{t-1}(\mathbf{z}_k)$ and $k_{tk}$. When both variables are new, the only element we may reuse is $\mathbf{K}_{t-1k-1}$.

This concludes our analysis of the fundamental matrices. Next we turn our attention to the compound matrices, which are products of those described above.

\subsection{Partitioning Compound Matrices}

\begin{align}
		\mathbf{H}_t\mathbf{K}_{tk}
		&= \Hblk
		\begin{pmatrix}
			\mathbf{K}_{t-1k-1} & \mathbf{k}_{t-1}(\mathbf{z}_k)\\
			\mathbf{k}_{k-1}^\top(\mathbf{x}_t) & k_{tk}
		\end{pmatrix}\nonumber\\
		&=
		\begin{pmatrix}
			\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
			(\mathbf{k}_{k-1}(\mathbf{x}_{t-1}) - \gamma\mathbf{k}_{k-1}(\mathbf{x}_t))^\top & k(\mathbf{x}_{t-1},\mathbf{z}_k) - \gamma k(\mathbf{x}_t,\mathbf{z}_k)
		\end{pmatrix}\nonumber\\
		&=
		\begin{pmatrix}
			\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
			\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) & \Delta k_k(\mathbf{x}_t)
		\end{pmatrix}
\end{align}
Here we define $\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) = \mathbf{k}_{k-1}(\mathbf{x}_{t-1}) - \gamma\mathbf{k}_{k-1}(\mathbf{x}_t)$, and $\Delta k_k(\mathbf{x}_t)= k(\mathbf{z}_k,\mathbf{x}_{t-1}) - \gamma k(\mathbf{z}_k,\mathbf{x}_t)$.

\begin{align}
	\mathbf{H}_t\mathbf{K}_{tt}\mathbf{H}_t^\top &=
	\Hblk \begin{pmatrix}
		\mathbf{K}_{t-1t-1} & \mathbf{k}_{t-1}(\mathbf{x}_t)\\
		 \mathbf{k}_{t-1}^\top(\mathbf{x}_t) & k_{tt}
	\end{pmatrix}
	\Htblk,\nonumber\\
	&=
	\Hblk
	\begin{pmatrix}
		\mathbf{K}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{k}_{t-1}(\mathbf{x}_{t-1})-\gamma \mathbf{k}_{t-1}(\mathbf{x}_{t})\\
		\mathbf{k}^\top_{t-1}(\mathbf{x}_t)\mathbf{H}^\top_{t-1} & k(\mathbf{x}_{t-1},\mathbf{x}_t) - \gamma k(\mathbf{x}_t,\mathbf{x}_t)
	\end{pmatrix},\nonumber\\
	&=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{H}_{t-1}(\mathbf{k}_{t-1}(\mathbf{x}_{t-1})-\gamma \mathbf{k}_{t-1}(\mathbf{x}_{t})) \\
		(\mathbf{k}_{t-1}(\mathbf{x}_{t-1})-\gamma \mathbf{k}_{t-1}(\mathbf{x}_{t}))^\top\mathbf{H}_{t-1}^\top & k_{t-1t-1} -2\gamma k_{t-1t} + \gamma^2 k_{tt}
	\end{pmatrix},\nonumber\\
	&=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{H}_{t-1}\Delta\mathbf{k}_{t-1}\\
		 (\mathbf{H}_{t-1}\Delta\mathbf{k}_{t-1})^\top & \Delta^2 k_{t}
	\end{pmatrix}
\end{align}
Here we define $\Delta\mathbf{k}_{t-1}(\mathbf{x}_t) = \mathbf{k}_{t-1}(\mathbf{x}_{t-1})-\gamma \mathbf{k}_{t-1}(\mathbf{x}_{t})$. We also use $\Delta^2 k_{t}$ to denote the arguments distributed according to a binomial:
\begin{align*}
	\Delta^2 k_{t} &= k\circ [(\mathbf{x}_{t-1},\cdot) - \gamma(\mathbf{x}_{t},\cdot)][(\cdot,\mathbf{x}_{t-1}) - \gamma(\cdot,\mathbf{x}_t)] \\
	&=
	k\circ[(\mathbf{x}_{t-1},\mathbf{x}_{t-1}) - \gamma(\mathbf{x}_{t-1},\mathbf{x}_t) - \gamma(\mathbf{x}_t,\mathbf{x}_{t-1}) +\gamma^2(\mathbf{x}_t,\mathbf{x}_t)],\\
	&=k(\mathbf{x}_{t-1},\mathbf{x}_{t-1}) - \gamma k(\mathbf{x}_{t-1},\mathbf{x}_t) - \gamma k(\mathbf{x}_t,\mathbf{x}_{t-1}) +\gamma^2 k(\mathbf{x}_t,\mathbf{x}_t),\\
	&=k(\mathbf{x}_{t-1},\mathbf{x}_{t-1}) - 2\gamma k(\mathbf{x}_{t-1},\mathbf{x}_t) +\gamma^2 k(\mathbf{x}_t,\mathbf{x}_t).
\end{align*}

\section{Computing Inverse Matrix Partitions}
Three matrices must be inverted. They are:
\begin{itemize}
	\item{$\mathbf{A}_{kk} = \mathbf{K}_{kk}^{-1}$}
	\item{$\mathbf{B}_{tt} = [\text{diag}(\mathbf{H}_t\mathbf{K}_{tt}\mathbf{H}_t^\top - \mathbf{H}_t\mathbf{K}_{tk}\mathbf{K}_{kk}^{-1}\mathbf{K}_{kt}\mathbf{H}_t^\top) + \mathbf{\Sigma}_t]^{-1}$}
	\item{$\mathbf{C}_{kk} = [\mathbf{K}_{kk} + \mathbf{K}_{kt}\mathbf{H}_t^\top\mathbf{B}_{tt}\mathbf{H}_t\mathbf{K}_{tk}]^{-1}$}
\end{itemize}

\subsection{Inverting $\mathbf{K}_{kk}$}
By the matrix inversion lemma, we have
\begin{align}
	\mathbf{A}_{kk} &=
	\begin{pmatrix}
		\mathbf{A}_{k-1k-1} & \mathbf{a}_{k-1}(\mathbf{z}_k)\\
		\mathbf{a}_{k-1}^\top(\mathbf{z}_k) & a_k
	\end{pmatrix} =
	\begin{pmatrix}
		\mathbf{K}_{k-1k-1}^{-1} + s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\mathbf{g}^\top_{k-1}(\mathbf{z}_k) & -s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\\
		-s_k\mathbf{g}_{k-1}^\top(\mathbf{z}_k) & s_k
	\end{pmatrix}.
\end{align}
Here, we have defined:
\begin{align}
	\mathbf{g}_{k-1}(\mathbf{z}_k) &= \mathbf{K}_{k-1k-1}^{-1}\mathbf{k}_{k-1}(\mathbf{z}_k),\\
	1/s_k &= k_{kk} - \mathbf{k}_{k-1}^\top(\mathbf{z}_k)\mathbf{K}_{k-1k-1}^{-1}\mathbf{k}_{k-1}(\mathbf{z}_k).
\end{align}

\subsection{Computing $\mathbf{B}_{tt}$}
To start, we compute the composite matrix $\mathbf{D}_{tt} = \mathbf{H}_t\mathbf{K}_{tk}\mathbf{A}_{kk}\mathbf{K}_{kt}\mathbf{H}_t^\top$
\begin{align*}
	\mathbf{D}_{tt} &=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
		\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) & \Delta k_k(\mathbf{x}_t)
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{A}_{k-1k-1} & \mathbf{a}_{k-1}(\mathbf{z}_k)\\
		\mathbf{a}_{k-1}^\top(\mathbf{z}_k) & a_k
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top & \Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})\\
		(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top & \Delta k_k(\mathbf{x}_t)
	\end{pmatrix},\\
	&=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
		\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) & \Delta k_k(\mathbf{x}_t)
	\end{pmatrix} \\
	&\cdot
	\begin{pmatrix}
		\mathbf{A}_{k-1k-1}\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top + \mathbf{a}_{k-1}(\mathbf{z}_k)(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top & \mathbf{A}_{k-1k-1}\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})+\mathbf{a}_{k-1}(\mathbf{z}_k)\Delta k_k(\mathbf{x}_t)\\
		\mathbf{a}_{k-1}^\top(\mathbf{z}_k)\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top + a_k(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top & \mathbf{a}_{k-1}^\top(\mathbf{z}_k)\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) +  a_k\Delta k_k(\mathbf{x}_t)
	\end{pmatrix},\\
	&=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\Delta\mathbf{D}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{H}_{t-1}\Delta \mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k) \\
		(\mathbf{H}_{t-1}\Delta \mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k))^\top  & \Delta d_t
	\end{pmatrix},
\end{align*}
where we have defined the following elements:
\begin{align*}
	\Delta\mathbf{D}_{t-1t-1} &= \mathbf{K}_{t-1k-1}\mathbf{A}_{k-1k-1}\mathbf{K}_{k-1t-1} + \widetilde{\mathbf{D}}_{t-1t-1},\\
	\widetilde{\mathbf{D}}_{t-1t-1} &= 2\mathbf{K}_{t-1k-1}\mathbf{a}_{k-1}(\mathbf{z}_k)\mathbf{k}_{t-1}^\top(\mathbf{z}_k)+ a_k\mathbf{k}_{t-1}(\mathbf{z}_k)\mathbf{k}_{t-1}^\top(\mathbf{z}_k)\\
	\Delta \mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k) &= \mathbf{K}_{t-1k-1}[\mathbf{A}_{k-1k-1}\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})+\mathbf{a}_{k-1}(\mathbf{z}_k)\Delta k_k(\mathbf{x}_t)] \\
	&+ \mathbf{k}_{t-1}(\mathbf{z}_k)[\mathbf{a}_{k-1}^\top(\mathbf{z}_k)\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) +  a_k\Delta k_k(\mathbf{x}_t)],\\
	\Delta d_{t} &=\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t})[\mathbf{A}_{k-1k-1}\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})+\mathbf{a}_{k-1}(\mathbf{z}_k)\Delta k_k(\mathbf{x}_t)] \\
	&+ \Delta k_k(\mathbf{x}_t)[\mathbf{a}_{k-1}^\top(\mathbf{z}_k)\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) +  a_k\Delta k_k(\mathbf{x}_t)].
\end{align*}

Now we add all the terms and take the diagonal:
\begin{align*}
	\mathbf{B}_{tt}^{-1} &=
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{H}_{t-1}\Delta\mathbf{k}_{t-1}(\mathbf{x}_t)\\
		 (\mathbf{H}_{t-1}\Delta\mathbf{k}_{t-1}(\mathbf{x}_t))^\top & \Delta k_t
	\end{pmatrix}\\
	&
	-
	\begin{pmatrix}
		\mathbf{H}_{t-1}\Delta\mathbf{D}_{t-1t-1}\mathbf{H}_{t-1}^\top & \mathbf{H}_{t-1}\Delta \mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k) \\
		(\mathbf{H}_{t-1}\Delta \mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k))^\top  & \Delta d_t
	\end{pmatrix} +
	\begin{pmatrix}
			\mathbf{\Sigma}_{t-1} & \mathbf{0}\\
			\mathbf{0} & \sigma^2_{t-1}
	\end{pmatrix},\\
	&=
	\begin{pmatrix}
		\mathbf{H}_{t-1}(\mathbf{K}_{t-1t-1}-\Delta{\mathbf{D}}_{t-1t-1})\mathbf{H}_{t-1}^\top + \mathbf{\Sigma}_{t-1}& \mathbf{H}_{t-1}(\Delta\mathbf{k}_{t-1}(\mathbf{x}_t)-\Delta\mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k))\\
		 (\mathbf{H}_{t-1}(\Delta\mathbf{k}_{t-1}(\mathbf{x}_t)-\Delta\mathbf{d}_{t-1}(\mathbf{x}_t,\mathbf{z}_k))^\top & \Delta k_t - \Delta d_t + \sigma^2_{t-1}
	\end{pmatrix}.
\end{align*}
We define $1/b_j = \Delta k_j - \Delta d_j + \sigma^2_{j}$ for $j=1,\cdots , t$. The inverse is simply:
\begin{align}
	\mathbf{B}_{tt} &=
	\begin{pmatrix}
		\mathbf{B}_{t-1t-1} & \mathbf{0}\\
		\mathbf{0} &  b_{t}
	\end{pmatrix}.
\end{align}

\subsection{Computing $\mathbf{C}_{kk}$}
First we compute $\mathbf{F}_{kk} = \mathbf{K}_{kt}\mathbf{H}_t^\top\mathbf{B}_{tt}\mathbf{H}_t\mathbf{K}_{tk}$.
\begin{align}
	\mathbf{F}_{kk} &=
	\begin{pmatrix}
			\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top & \Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})\\
			(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top & \Delta k_k(\mathbf{x}_t)
		\end{pmatrix}
	\begin{pmatrix}
		\mathbf{B}_{t-1t-1} & \mathbf{0}\\
		\mathbf{0} &  b_{t}
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
		\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) & \Delta k_k(\mathbf{x}_t)
	\end{pmatrix},\nonumber \\
	&=
	\begin{pmatrix}
		\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top & \Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})\\
		(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top & \Delta k_k(\mathbf{x}_t)
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{B}_{t-1t-1}\mathbf{H}_{t-1}\mathbf{K}_{t-1k-1} & \mathbf{B}_{t-1t-1}\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k)\\
		b_t\Delta\mathbf{k}_{k-1}^\top(\mathbf{x}_{t}) & b_t\Delta k_k(\mathbf{x}_t)
	\end{pmatrix},\nonumber \\
	&=
	\begin{pmatrix}
		\mathbf{F}_{k-1k-1} &  \mathbf{f}_{k-1}\\
		 \mathbf{f}_{k-1}^\top &  f_{k}
	\end{pmatrix} + b_t
	\begin{pmatrix}
		\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) \\
		\Delta k_k(\mathbf{x}_t)
	\end{pmatrix}
	\begin{pmatrix}
		\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) ,\Delta k_k(\mathbf{x}_t)
	\end{pmatrix},\\
	&=
	\begin{pmatrix}
		\Delta\mathbf{F}_{k-1k-1} &  \Delta\mathbf{f}_{k-1}\\
		 \Delta\mathbf{f}_{k-1}^\top &  \Delta f_{k}
	\end{pmatrix}
\end{align}
Here we define:
\begin{align}
	\mathbf{f}_{k-1} &= \mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top\mathbf{B}_{t-1t-1}\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k) ,\\
	f_k &= \mathbf{k}_{t-1}^\top(\mathbf{z}_k)\mathbf{H}_{t-1}^\top\mathbf{B}_{t-1t-1}\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k).
\end{align}
Adding the terms produces the partitioned $\mathbf{C}_{kk}=\mathbf{K}_{kk} + \mathbf{F}_{kk}$ matrix:
\begin{align}
	\mathbf{C}_{kk}^{-1} &=
	\begin{pmatrix}
		\mathbf{K}_{k-1k-1} + \Delta\mathbf{F}_{k-1k-1} & \mathbf{k}_{k-1}(\mathbf{z}_k) + \Delta\mathbf{f}_{k-1}\\
		(\mathbf{k}_{k-1}(\mathbf{z}_k) + \Delta\mathbf{f}_{k-1})^\top & k_{kk} + \Delta f_{k}
	\end{pmatrix}.
\end{align}
By the matrix inversion lemma, we have:
\begin{align}
	\mathbf{C}_{kk} &=
	\begin{pmatrix}
		\mathbf{C}_{k-1k-1} + w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top & -w_k\tilde{\mathbf{c}}_{k-1}\\
		-w_k\tilde{\mathbf{c}}_{k-1}^\top & w_k
	\end{pmatrix},\\
	\tilde{\mathbf{c}}_{k-1} &= \mathbf{C}_{k-1k-1}(\mathbf{k}_{k-1}(\mathbf{z}_k) + \Delta\mathbf{f}_{k-1}),\\
	1/w_k &= k_{kk} + \Delta f_{k} - \tilde{\mathbf{c}}_{k-1}^\top\mathbf{C}_{k-1k-1}\tilde{\mathbf{c}}_{k-1}.
\end{align}

\subsection{Recursive Parameters}
We may now derive recursions for the parameters $\boldsymbol{\alpha}_t=\mathbf{C}_{kk}\mathbf{K}_{kt}\mathbf{H}_t^\top\mathbf{B}_{tt}\mathbf{r}_{t-1}$ and $\mathbf{P}_{kk}$:
\begin{align}
	\boldsymbol{\alpha}_{t} &=
	\begin{pmatrix}
		\mathbf{C}_{k-1k-1} + w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top & -w_k\tilde{\mathbf{c}}_{k-1}\\
		-w_k\tilde{\mathbf{c}}_{k-1}^\top & w_k
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{K}_{k-1t-1}\mathbf{H}_{t-1}^\top\mathbf{B}_{t-1t-1} & b_t\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t})\\
		(\mathbf{H}_{t-1}\mathbf{k}_{t-1}(\mathbf{z}_k))^\top\mathbf{B}_{t-1t-1} & b_t\Delta k_k(\mathbf{x}_t)
	\end{pmatrix}
	\begin{pmatrix}
		\mathbf{r}_{t-2}\\
		r_{t-1}
	\end{pmatrix},\nonumber\\
	&=
	\begin{pmatrix}
		\boldsymbol{\alpha}_{t-1}\\
		0
	\end{pmatrix} +
	\begin{pmatrix}
		\tilde{\boldsymbol{\alpha}}_{t-1}\\
		\tilde{\alpha}_{t-1}
	\end{pmatrix}.
\end{align}
We have defined:
\begin{align*}
	\tilde{\boldsymbol{\alpha}}_{t-1}&= w_k\tilde{\mathbf{c}}_{k-1}[\tilde{\mathbf{c}}_{k-1}^\top\mathbf{K}_{k-1t-1} - \mathbf{k}_{t-1}^\top(\mathbf{z}_k)]\mathbf{H}_{t-1}^\top\mathbf{B}_{t-1t-1}\mathbf{r}_{t-2}\\
	&+b_t[(\mathbf{C}_{k-1k-1} + w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top)\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) -w_k\tilde{\mathbf{c}}_{k-1}\Delta k_k(\mathbf{x}_t)]r_{t-1},\\
	\tilde{\alpha}_t&=-w_k[\tilde{\mathbf{c}}_{k-1}^\top\mathbf{K}_{k-1t-1} - \mathbf{k}_{t-1}^\top(\mathbf{z}_k)]\mathbf{H}_{t-1}^\top\mathbf{B}_{t-1t-1}\mathbf{r}_{t-2}\\
	& -w_kb_t[\tilde{\mathbf{c}}_{k-1}^\top\Delta\mathbf{k}_{k-1}(\mathbf{x}_{t}) - \Delta k_k(\mathbf{x}_t)]r_{t-1}.
\end{align*}
For the last parameter, we have:
\begin{align}
	\mathbf{P}_{kk} &= \mathbf{A}_{kk} - \mathbf{C}_{kk},\nonumber\\
	&=
	\begin{pmatrix}
		\mathbf{K}_{k-1k-1}^{-1} + s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\mathbf{g}^\top_{k-1}(\mathbf{z}_k) & -s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\\
		-s_k\mathbf{g}_{k-1}^\top(\mathbf{z}_k) & s_k
	\end{pmatrix}
	-\begin{pmatrix}
		\mathbf{C}_{k-1k-1} + w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top & -w_k\tilde{\mathbf{c}}_{k-1}\\
		-w_k\tilde{\mathbf{c}}_{k-1}^\top & w_k
	\end{pmatrix},\nonumber\\
	&=
	\begin{pmatrix}
		\mathbf{P}_{k-1k-1} + s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\mathbf{g}^\top_{k-1}(\mathbf{z}_k) - w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top &
		w_k\tilde{\mathbf{c}}_{k-1} -s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\\
		(w_k\tilde{\mathbf{c}}_{k-1} -s_k\mathbf{g}_{k-1}(\mathbf{z}_k))^\top & s_k - w_k
	\end{pmatrix},\nonumber\\
	&=
	\begin{pmatrix}
		\mathbf{P}_{k-1k-1} & \mathbf{0}\\
		\mathbf{0} & 0
	\end{pmatrix}+
	\begin{pmatrix}
		 \widetilde{\mathbf{P}}_{k-1k-1} & \tilde{\mathbf{p}}_{k-1}
		\\
		\tilde{\mathbf{p}}_{k-1}^\top & \tilde{p}_k
	\end{pmatrix},
\end{align}
where we have defined:
\begin{align*}
	\widetilde{\mathbf{P}}_{k-1k-1} &= s_k\mathbf{g}_{k-1}(\mathbf{z}_k)\mathbf{g}^\top_{k-1}(\mathbf{z}_k) - w_k\tilde{\mathbf{c}}_{k-1}\tilde{\mathbf{c}}_{k-1}^\top,\\
	\tilde{\mathbf{p}}_{k-1} &=w_k\tilde{\mathbf{c}}_{k-1} -s_k\mathbf{g}_{k-1}(\mathbf{z}_k),\\
	\tilde{p}_k &= s_k - w_k.
\end{align*}

On-line updates are possible by unrolling the recursion, starting from basic posterior parameters.

\section{Conclusion}
In this paper we derived formulas for updating the \textsc{spgp-sarsa} algorithm recursively. This allows previous computations to be reused and promotes greater memory efficiency than computing matrix inverses from scratch at each iteration. Promising future work will explore the best practices for adding pseudo inputs and performing experiments to quantify the benefits of the recursive approach.

\section{References}
\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


We love our Windwood couples, and Hannah and Mckinley were certainly no exception. We had the privilege of helping the adorable couple plan for their big day, and the laid back pair was an absolute joy to work with! All the rose colored details of Hannah and Mckinley's special day had the farm looking especially springy! The stunning couple was all smiles for their day!
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text:
Recommended - Blood and Guts in High School by Kathy Acker
09.02.2017 by Ollie //
£8.99 / paperback
Written in the late 1970s, Blood and Guts in High School was met with widespread shock, outrage, and acclaim upon its initial publication in 1984, and remains Acker's seminal work. Via fraught dialogue, hand-written poetry, crude illustrations, and prison journals (with Jean Genet himself even emerging as a central character), the literary pillars of youth, desire, punk, and feminism all collapse in Acker's wake. What emerges is a voice that not only reshaped the avant-garde, but whose impact is still being felt across literature as a whole (as evidenced by Chris Kraus' biography, After Kathy Acker, also published last month.)
Published by Penguin on August 31st 2017
Categories // Recommended Tags // Classics

redpajama_set_name: RedPajamaCommonCrawl


text:
FDA: Newly-approved coronavirus test works in 45 minutes, to be rolled out next week
FOX 5 NY
A patient COVID-19 swab is taken out from a sterile tube as lab technicians carry out a diagnostic test for coronavirus in the microbiology laboratory inside the Specialist Virology Centre at the University Hospital of Wales in Cardiff. (Photo by Ben
The U.S. Food and Drug Administration has announced that it has approved the use of a rapid coronavirus diagnostic test that returns results in approximately 45 minutes.
Cepheid, a California-based molecular diagnostics company behind the test, says that it expects it to begin shipping next week.
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So far in New York, Governor Andrew Cuomo said the state has tested 45,437 people, more than any other state in the nation and more per capita than in China or South Korea. 10,356 people have tested positive for the virus, with 1,603 victims hospitalized.
New Jersey has begun aggressively testing for the virus, with 1,327 positive coronavirus cases found in the state. Governor Phil Murphy said a drive-thru testing site in Bergen County has had to close on Friday and Saturday when it reached capacity after testing a few hundred residents.
NJ's first drive-thru coronavirus testing site temporarily closes due to demand
According to the CDC, as of March 20, 91 public health laboratories in every state in the nation are now offering testing for coronavirus. Up until late February, the CDC was doing the overwhelming majority of testing for the virus, but was largely only capable of testing a few dozen specimens each day.
However, since more and more public health laboratories have joined the fight, since March 1, they have conducted 61,139 tests for coronavirus, compared to just 1,274 from the CDC in that same time period.
So far, according to the CDC, there are 15,219 cases of coronavirus in the United States, causing 201 deaths. The virus has been reported in all 50 states and four territories.
Globally, there are over 287,000 cases of coronavirus confirmed around the globe, with 11,900 deaths. Nearly 89,000 people have recovered.
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Any individual who needs information on a program or service available in Prescott-Russell.
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Adult adoptees, adoptive parents and birth relatives (birth parent, birth grandparent, adult birth sibling) may request a copy of non-identifying information. For adoption disclosure services, the adult adoptee and the birth parent both have to register with the Adoption Disclosure Registry of Ontario.
Can my son and daughter share the same room?
It is preferable that two children of opposite sexes have their own room. That being said, if circumstances are temporary, it may possibly be acceptable. If it is the case, you will need to ensure that there is proper supervision. If we receive a call regarding this matter, we may assess the situation, depending on the circumstances.
Do you offer couple therapy?
If conflicts are family related, such as a difference of opinion in regards to disciplining the children or the family structure, we can offer our services. However, if the conflicts are strictly couple related, we ask you to seek counseling through your employment assistance program, the "Centre de santé communautaire de l'Estrie" or another organization in the Ottawa region offering either public or private services .
The Child and Family Services Act CFSA says that a child under 16 years of age, cannot be left alone without his or her parents making adequate arrangements for his or her supervision and care.
Also, there are other factors that need to be considered when deciding if a child can be left alone for a short period of time such as the child's development level, the child's ability to communicate, the time of day the child is left alone, the duration of the adult's absence, the child's maturity, the child's individual strengths/skills/abilities, the child's access to a responsible adult, the safety of the family environment and the neighborhood.
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Welcome to Playful Meanderings, a new WordPress blog to chat about life, books, travel, recreation and whatever else may come up.
It's so true. I live in this body and I need to take care of it — all of it. My mind needed stimulation outside of the routine. It needed refreshment and the exposure to new ideas, old ideas, points of view and everything that books bring to the reader. Invigorating the sponge upstairs to learn — or re-learn — history, discover the power of deeper thinking, and reshaping my spirit by diving into the written word were among the positive points to taking on this project.
Why 75? The seed got planted by an article a friend sent to me in which the author was expounding on the number of books he had read the previous year. I mistakenly thought my friend had written the article, not just forwarded it to me, and I thought, "He is challenging me! If he can read 75, then I can read 75!" I made my resolution (and I had to come up with a plan to accomplish this ambitious number) and embarked on the project.
Funny thing is, I brought up his momentous reading to this same friend two months back, and he really didn't know what I was talking about: "75 books. No," he replied. "I'm lucky to read 15." Turns out I failed to note the author — something I'm bound to do again in this life. But what the heck, that mistake got me to get on with my journey into books.
Number one: Sounds expensive. Not so much. Turns out I have at least that many that I've never read around my house that I've been meaning to read. Also, I have a library card — fancy that — and the library near my house has many, many more than 75 books that I've never read and been meaning to. Speaking of my library, they have a club, friends of the library, that sells books twice a year- Lots of great books that I've been meaning to read. You get the picture. Then there's this: when I said that I was going to read 75 books — and only to a few people because what if I don't do it? — they gave me some books. Folks, I am swimming in books right now!
Number two: Time. I don't know about you, but I always think of myself as busy. Much too busy to read. Turns out I can bring a book to wherever I am waiting. Yes, I do a fair amount of that, for I am a mother, a worker, a consumer, a patient and a person. I wait for daughters at piano, at guitar, at church, at sporting events. I go to the dentist ( more than I would like) or doctor or places where you wait in line and I get out my book. I just have made it part of the routine to bring the book I'm skipping (or plowing) through currently. Ok, and I've given up most of my television watching. That alone created several hours in the week. I also cheat. What? Yes, a few people have called it that, and judge me if you will, but about 1/3 of the books I have read are during my commute. I am read to through the free audiobooks app. They are (http://librivox.org/ ) all books are in the public domain and read by volunteers into my car stereo. More about all of this in a later post, but I don't call it cheating. I listen, learn and put the time into hearing these "books on tape" just as I would if holding a book in my hand and taking in the words on the printed page. And final note about time. — the more I read, the more I want to read. I've been known to read a little later into the night, forgo the morning paper in favor of my book, shoot out the garden door and read my book in the backyard instead of cleaning my house. Oh the joys of reading!
So where am I now on the delicious journey? Books #42 and #43. What? Yes, because I have the power of the app on audiobooks (which is free, by the way) I am always reading two books at once. No, the two books don't always mesh together, but what the heck, I read two and three books at once when I was in college and I'm an American multitasker at heart.
No, this blog won't be solely dedicated to books and the goal of reading 75 in one year, but the next few posts will certainly be about the topic. I welcome your comments, suggestions and humorous stories about your own goals — be they new years resolutions or some other scheme to change your own life in a positive way.


VCIT Insight is a Level 1 BBBEE company operating within the ICT Sector of South Africa since 2009 across Gauteng, KwaZulu Natal, Western and Eastern Cape. From our humble beginnings as a provider of basic hardware infrastructure and consumables to the South African market, VCIT has evolved into a Value Based Services and Software Provider. In the last 4 years, since 2013, we have been taking on some larger OEM Brands to go to market with as well as investing in a dedicated full time software development Team.
At the core of our Company is a simple set of values based on a belief that our People and non-negotiable Value to our Customers matter as much as Shareholder dividends . The values under pinning our Company focus on helping everyone be the best they can at what they do so that they can amaze and excite our Customers with world class service delivery every day. Our leadership encourage that our People spend time to understand our Customers intimately & authentically and apply disruptive critical systems thinking.
To become a Value Driven Innovation Partner to Our Customers by being deeply committed to solving problems in Their Business.
We will build a highly capable Team of Professionals that are passionately dedicated to Value Engineering and creative problem solving as well as providing world-class software solutions & support. We pride ourselves on doing what we say we're going to do and being easy to do Business with.