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Let $U:=\sum_{a \leq t < b} q_t$ and $V:=\sum_{t \geq b} q_t$, Then

$var( \sum_{t \geq a} q_t) = var(U) + var(V) + cov(U,V) $

where $q_t$ is a bounded deterministic function of a Markov Chain's state at time $t$.

For the above, I am not sure how the variance was split into two such terms and also how the covariance between $U$ and $V$ appears. Can someone help explain?

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1 Answer 1

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Note: $\sum_{t\geq a} q_t=\sum_{a\leq t<b}q_t+\sum_{t\geq b} q_t$

$$\def\E{\mathsf E}\def\var{\mathsf{Var}}\def\cov{\mathsf{Cov}} \begin{align} \var(U+V)&=\E((U+V)^2)-(\E(U+V))^2 \\[1ex]&=\E(U^2)+\E(V^2)+2\E(UV)-\big(\E(U)^2+\E(V)^2+2\E(U)\E(V)\big) \\[1ex]&=\E(U^2)-\E(U)^2~+~\E(V^2)-\E(V)^2~+~2\E(UV)-2\E(U)\E(V) \\[1ex]&=\var(U)+\var(V)+2\cov(U,V) \\[2ex]\var\left(\sum_{t\geq a}q_t\right)&=\var\left(\sum_{a\leq t<b}q_t\right)+\var\left(\sum_{t\geq b}q_t\right)+2~\cov\left(\sum_{a\leq t<b}q_t,\sum_{t\geq b}q_t\right)\end{align}$$

Note factor of $2$ for $cov$. Also, statement is general for two dependent random variables, not particularly for Markov chains.

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