I have a process $Y_t = aX_t + (1-a)X_{t-k}$ where a is between 0 and 1 and $X_t$ is a stationary process with mean $\mu$ and autocovariance function $\gamma_k$

Now I want to find the autocovariance function of $Y_t$ which I started by multiplying both sides by $Y_{t-k}$ and then taking the expectation to get $$E[Y_tY_{t-k}]= aE[X_t X_{t-k}]+E[X^2 _{t-k}]-aE[X^2 _{t-k}]$$

I evaluated the first expectation to just be $\gamma_k$, but would the next two be? Since they're squared, and $E[X^2]=Var[X]$ would they just be $\gamma_0$?


  • 1
    $\begingroup$ $E[X^2] = Var[X]$ if and only if $\mu=0$. $\endgroup$
    – dlnB
    Apr 20 '20 at 21:03

$E[X^2] = Var(X)$ does not hold in general, but instead if and only if $\mu=0$. However, by definition, $E[X_{t-k}^2]=\gamma_0$.

There is a mistake in your algebra of calculating $[Y_tY_{t-k}]$ that gives you

$$[Y_tY_{t-k}]= a[X_t X_{t-k}]+[X^2 _{t-k}]-a[X^2 _{t-k}].$$

It should instead be:

$$[Y_tY_{t-k}]= (aX_t + (1-a)X_{t-k})(aX_{t-k}+(1-a)X_{t-2k}).$$

Expanding and taking expectations gives:

$$E[Y_tY_{t-k}]= a^2E[X_tX_{t-k}]+a(1-a)E[X_{t-k}X_{t-k}]+a(1-a)E[X_{t}X_{t-2k}]+(1-a)^2E[X_{t-k}X_{t-2k}]$$

Replacing expectations with autocovariance function notation gives:


  • $\begingroup$ thank u that's very helpful, do I need a $\mu^2(1-a)$ term on the end because I have $-aE[X^2_{t-k}]= -a(\gamma_0 + \mu^2)$? $\endgroup$
    – clovis
    Apr 20 '20 at 21:18
  • $\begingroup$ Please see the revised version. $\endgroup$
    – dlnB
    Apr 20 '20 at 21:19

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