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thanks for reading my post. I know its fundamental and rather easy qns but I'm seriously struggling. Please help me, thank you very much!

Let $\boldsymbol{X}$ have a distribution with mean $\mu$ and variance $\sigma^{2}$, and let $\boldsymbol{Z}_{t}$ = $\boldsymbol{X}$ for all $t$ .

(a) Show that {$\boldsymbol{Z}_{t}$} is weakly stationary.

$\mu_{t}$ = $\mu$

$\mu_{t}$ does not depend on $t$

$cov(Z_{t+h},Z_{t})$ = $\sigma^{2}$

$v(t+h,h) = v(h) = v(-h) = \sigma^{2}$

By definition, {$\boldsymbol{Z}_{t}$} is weakly stationary

(b) Find the autocovariance function for {$\boldsymbol{Z}_{t}$}.

$cov(Z_{t+h},Z_{t})$ = $\sigma^{2}$

(c) Suppose $X$ and $\epsilon_{t}$ are IID Normal, show that $y_{t}$ = $X\epsilon_{t}$ is White Noise, but $y_{t}^{2}$$E(y_{t}^{2})$ is not White Noise.

I have no idea if I can just assume for $\epsilon_{t}$ distribution iid normal with mean $0$ and variance $\sigma^{2}$.

Could someone enlighten me with the proof? Also, if my way of answering (a) and (b) is right?

I tried for part(c):

Below is my attempt:

$E(y_{t}) = E(X\epsilon_{t}) = E(X)E(\epsilon_{t}) = \mu * 0 = 0$

$v(0) = cov(y_{t}) = cov(X\epsilon_{t}) = \sigma^{2} $

$v(h) = cov(y_{t},y_{t+h}) = cov(X\epsilon_{t},X\epsilon_{t+h})$ = 0

Thus, $y_{t}$ is a white noise distribution with mean 0 and variance $\sigma_{z}^{2}$.

$y_{t}^{2} = X^{2}\epsilon_{t}^{2} $

$E(y_{t}^{2}) = E(X^{2}\epsilon_{t}^{2}) = E(X^{2})E(\epsilon_{t}^{2})$

$E(y_{t}^{2} - E(y_{t}^{2}) ) = E(y_{t}^{2}) - 0 = 0$

$v(0) = cov(y_{t}^{2} - E(y_{t}^{2})) = cov(y_{t}^{2}) - cov(E(y_{t}^{2})) = cov(X^{2}\epsilon_{t}^{2}) - Cov(0) = \sigma^{4} $

$v(h) = cov(y_{t}^{2} - E(y_{t}^{2}),y_{t+h}^{2} - E(y_{t+h}^{2})) = cov(y_{t}^{2},y_{t+h}^{2}) = cov(X^{2}\epsilon_{t}^{2},X^{2}\epsilon_{t+h}^{2}) = 0$

Since $v(0) = \sigma^{4} \neq \sigma^{2} $

$y_{t}^{2}$$E(y_{t}^{2})$ is not White Noise.

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  • $\begingroup$ You don't need the normality assumption for a & b. $\endgroup$ – Michael R. Chernick Oct 20 '19 at 17:51
  • $\begingroup$ @MichaelChernick Yes, I understand that I do not need normality assumption for (a) and (b). Do you know how to do c? $\endgroup$ – Miss_ LHX Oct 20 '19 at 18:10
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    $\begingroup$ Add the self-study tag & someone may be able to help you with c. $\endgroup$ – Michael R. Chernick Oct 20 '19 at 18:53
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$E(y_{t}^{2} - E(y_{t}^{2}) ) = E(y_{t}^{2}) - 0 = 0$

... You should attempt to explain how you came to conclude that $E(y_{t}^{2}) =0$

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  • $\begingroup$ Honestly I just assumed $E(\epsilon_{t}^{2})= 0$ because I don't know how to prove that $E(y_{t}^{2}) =0$ $\endgroup$ – Miss_ LHX Oct 21 '19 at 1:55
  • $\begingroup$ Note that since $\text{Var}(X)=E(X^2)-E(X)^2$, you should immediately see that $E(X^2)=\sigma^2+\mu^2$. The only way that's zero is if $\mu=0$ and $\sigma^2=0$. $\endgroup$ – Glen_b Oct 21 '19 at 5:49

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