# Finding the autocovariance function of a process

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 $$\gamma_0$$?

TIA!

• $E[X^2] = Var[X]$ if and only if $\mu=0$. – 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:

$$E[Y_tY_{t-k}]=a^2\gamma_k+a(1-a)\gamma_0+a(1-a)\gamma_{2k}+(1-a)^2\gamma_k.$$

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