# Time Series assumptions for iid $\epsilon$

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.

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

$$E(y_{t}^{2} - E(y_{t}^{2}) ) = E(y_{t}^{2}) - 0 = 0$$
... You should attempt to explain how you came to conclude that $$E(y_{t}^{2}) =0$$
• Honestly I just assumed $E(\epsilon_{t}^{2})= 0$ because I don't know how to prove that $E(y_{t}^{2}) =0$ – Miss_ LHX Oct 21 '19 at 1:55
• 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 $\sigma^2=0$. – Glen_b Oct 21 '19 at 5:49