# Proof of white noise process

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?

• 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 $y_t=e_t+q_1e_{t-1}+q_2e_{t-2}$ ? – Peter Leopold Sep 16 '19 at 5:00

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.