# Calculating covariance for a non-strictly-stationary white noise process

I'm reading Hayashi's "Econometrics", and in Chapter 2, page 101 he discusses the following white noise process:

I understand the calculations for the expected mean and variance, but I can't understand why the covariance would be 0 for $i\neq j$.

Since $E(z_i)=0$, the expression for the covariance would be reduced to $E(\cos(iw)\cdotp \cos(jw))$. Would someone mind showing me why this goes to zero? Assuming for the moment that $cos(iw)$ is positive, then $\cos((i+1)w)$ would be negative and so on. But why would there be 0 covariance for any two $i$ and $j$, $i\neq j$? I know this is a simple question, but can't seem to wrap my head around it.

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Use trigonometric identities(scroll down to angle sum and difference identities):

$$\cos(iw)\cdot\cos(jw)=\frac{1}{2}\left(\cos((i+j)w)+\cos((i-j)w)\right)$$

Then substitute this formula to the following equation:

\begin{align} E\cos(iw)\cdot\cos(jw)=\frac{1}{2\pi}\int_0^{2\pi}\cos(iw)\cdot\cos(jw)dw \end{align}

and use the fact that

\begin{align} \int_0^{2\pi}\cos(kw)dw=0 \end{align}

for $k\in\mathbb{N}$.

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 Perfect! Thank you very much. – jefflovejapan Jul 12 '11 at 10:06