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I have the process $y_t=e_t+ae_{t-1}e_{t-2}$ where $e_t$ is iid with mean of $0$ and variance $\sigma^2$

How do I go about mathematically proving that this is a white noise process?

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  • $\begingroup$ I'm not sure how to interpret the expression. Is it $y_t=e_t+ae_{t-1}e_{t-2}$ where $e_t \stackrel{iid}{\sim} N(0,\sigma^2)$? Perhaps you mean the MA(2) process $y_t=e_t+q_1e_{t-1}+q_2e_{t-2}$ ? $\endgroup$ – Peter Leopold Sep 16 '19 at 5:00
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Clearly, the process is zero-mean. Let's check its autocovariance $k\geq 1$: $$\begin{align}E[y_ty_{t-k}]&=E[(e_t+ae_{t-1}e_{t-2})(e_{t-k}+ae_{t-1-k}e_{t-2-k})]\\&=E[e_te_{t-k}]+aE[e_te_{t-1-k}e_{t-2-k}]+aE[e_{t-k}e_{t-1}e_{t-2}]+a^2E[e_{t-1}e_{t-2}e_{t-1-k}e_{t-2-k}]\end{align}$$ Here, $E[e_te_{t-k}]=E[e_t]E[e_{t-k}]=0$, similarly for $E[e_te_{t-1-k}e_{t-2-k}]$. And, for $E[e_{t-k}e_{t-1}e_{t-2}]$, even if $k=1$ or $2$, one of the terms will leave the expectation and make the expression $0$. For the last term, the expectation is non-zero only if $k=0$. So, we have non-zero autocorrelation, $r_y(k)$, only if $k=0$, and the process is zero-mean. Then, this is a white process. Note that you can find the $r_y(0)$ as $1+a^2$ from the above equation.

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