I have given the AR(1) process as followed:

$$ y_t = phi y_{t-1} + e_t $$

where

$$ e_t sim WN(0,sigma^2)$$

I need to prove that

$$ cov(y_t,y_{t-j}) = phi^j sigma^2[1 + sigma^2 + (sigma^2)^2 + …+ (sigma^2)^{t-1-j}]$$

This is a nonstationary process.

I don't really know where to start at except for the expansion of the AR(1) process. Any help would be appreciated. Thank you guys.

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#### Best Answer

The expression cannot be correct. The correct one is

$${rm Cov}(y_t,y_{t-j}) = phi^j sigma^2[1 + phi^2 + (phi^2)^2 + …+ (phi^2)^{t-1-j}]$$

How can we prove it (show it actually, there is no "proof" involved)? Well,

$$y_t = phi y_{t-1} + e_t \ implies y_t = sum_{k=1}^tphi^{t-k}e_k \ implies y_{t-j} = sum_{k=1}^{t-j}phi^{t-j-k}e_k \$$

So

$${rm Cov}(y_t,y_{t-j}) = Eleft[left(sum_{k=1}^tphi^{t-k}e_kright)left(sum_{k=1}^{t-j}phi^{t-j-k}e_kright)right]$$

Hmmm… the first sum contains errors that lie "in the future" of the second sum… some breaking up of the sums will do it, don't you think? Be careful with the indices, it's all that matters here.

Signal when you solved it.

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