Prove that Cov$(\sum_{k=j}^\infty \rho^k e_{t-k}, \sum_{k=0}^\infty \rho^k e_{t-j-k} ) = \rho^j \frac{\sigma^2}{1-\rho^2}$, where $e_t$ white noise

Question

Let $e_t$ be a white noise, in other words:

• E$e_t = 0$,
• Cov$(e_t, e_{t'})=0$, when $t \not = t'$,
• Var$(e_t) = \sigma^2$ (do not depends on time t)

Let $|\rho| < 1$, $ j>0 $ be constants.

How to prove that

Cov$(\sum_{k=j}^\infty \rho^k e_{t-k}, \sum_{k=0}^\infty \rho^k e_{t-j-k} ) = \rho^j \frac{\sigma^2}{1-\rho^2}$.

When I tried to prove it, I stucked with two infinity sums product. Any help would be truly appreciated.

Answer 1

\begin{align} Cov(\sum_{k=j}^\infty \rho^k e_{t-k}, \sum_{k=0}^\infty \rho^k e_{t-j-k} )&=Cov(\sum_{k=j}^\infty \rho^k e_{t-k}, \sum_{k=j}^\infty \rho^{k-j} e_{t-k} )\\ &=\rho^{-j}Cov(\sum_{k=j}^\infty \rho^k e_{t-k}, \sum_{k=j}^\infty \rho^{k} e_{t-k} )\\ &=\rho^{-j}var(\sum_{k=j}^\infty \rho^k e_{t-k})\\ &=\rho^{-j}\sum_{k=j}^\infty var( \rho^k e_{t-k})\\ &=\rho^{-j}\sum_{k=j}^\infty \rho^{2k} \sigma^2\\ &=\frac{\rho^{j}}{1-\rho^2} \sigma^2\\ \end{align}

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Using $k+j \to k'$ we have \begin{align} \sum_{k=0}^\infty \rho^k e_{t-j-k}&=\sum_{k=0}^\infty \rho^{(k+j)-j}e_{t-(j+k)}\\ &=\sum_{k'=j}^\infty \rho^{k'-j} e_{t-k'}\\ &=\rho^{-j}\sum_{k'=j}^\infty \rho^{k'} e_{t-k'}\\ \end{align}