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I have some trouble solving the following question:

Given is model $X$:

$X_t = (W_t -1)*W_{t-1}$

Where $W_t$ is the stationary process:

$W_t = p*W_{t-1} + Z_t$

Where $Z_t$ is distributed $WN(0,1)$ and $|p|<1$

The question is:

'Show that the expected value of $X$, $E[X] = p/(1-p^2)$. You are allowed to use the formula for the autocovariance function of an autoregressive process.'

I don't see how the autocovariance function relates to the expected value. Help is very much appreciated.

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2 Answers 2

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You have $$X_t=(pW_{t-1}+Z_t-1)W_{t-1}=pW_{t-1}^2+Z_tW_{t-1}-W_{t-1}$$

and taking expectations you get $p$ times the variance of the AR(1) process. The second term is zero since $Z$ and $W$ are independent. The last term is zero since is mean is zero. So then you have $$ E(X_t)=pV(W_{t-1})=p\frac{1}{1-p^2}=\frac{p}{1-p^2} $$

as desired.

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Expand out, so $X_t=W_t*W_{t-1} - W_{t-1}$, then $E(X_t)=E(W_t*W_{t-1}) - E(W_{t-1})$ and notice the first term on the RHS is the autocovariance of W. Now note that $E(W_t*W_{t-1})=E(pW_{t-1}+Z_t)(W_{t-1})=E(pW^2_{t-1}) +0$. To find $E(W^2_{t-1})$ we can then use the formula for the autocovariance functions for an AR process, which is $\gamma(k)=\sigma^2_Z\sum^{\infty}_{i=0} p^i p^{i+k}$. Set $k=0$ and you should be done. Hope this helps, Trurl.

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  • $\begingroup$ I understand the expanding part, but I still can't notice a term that could equal the autocovariance of W. I think I am missing an equation for the autocovariance function. $\endgroup$
    – Bradda
    Commented Oct 4, 2013 at 15:00

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