# Calculating the Mean and Autocovariance Function of a Piecewise Time Series

I'm taking an introductory class to time series and this was a practice problem assigned for the first week (not homework).

Consider the following time series ($$w_t$$ is iid. Normal($$0, 1$$)):

$$x_t = w_t$$ for $$t = 1, 3, 5, 7,...$$ and

$$x_t = \frac{1}{\sqrt{2}}(w_{t-1}^2 -1)$$ for $$t = 2, 4, 6, 8,...$$

Find the mean function and autocovariance function for $$x_t$$. Are $$x_1$$ and $$x_2$$ identically distributed?

For the mean function, I'm stuck on finding the expectation of $$w_{t-1}^2$$. This is what I've got so far:

$$\mu_{x_t} = E(x_t$$) = $$0$$, if $$t$$ is an odd number (since the expectation of white noise is $$0$$, right?), and $$\mu_{x_t} = E(\frac{1}{\sqrt{2}}(w_{t-1}^2 -1))$$ = $$\frac{1}{\sqrt{2}}E(w_{t-1}^2) - E(\frac{1}{\sqrt{2}})$$, if $$t$$ is an even number. For the even part of the mean function, is the expectation of $$w_{t-1}^2$$ simply $$0$$? I think $$w_{t-1}^2$$ is just white noise, but I don't know if I'm right.

For the autocovariance function, are there 3 possible cases? $$t$$ can be odd or even. In either case, if $$h = 0, 2, 4, 6$$... (i.e. even number), then the autocovariance would simply be the variance. So, for cases where $$h$$ is odd, the autocovariance would be Cov$$(x_t, x_{t + h})$$ = Cov$$(x_{t + h}, x_t)$$. Am I on the right track with this line of thinking? If so, then the covariance would be Cov$$(w_t, \frac{1}{\sqrt{2}}(w_{t-1}^2 -1)$$ $$=$$ Cov$$(w_t, \frac{1}{\sqrt{2}}(w_{t-1}^2))$$ $$-$$ Cov$$(w_t, w_t)$$, right?

Any guidance on this problem and verification (or correction) of my line of thinking would be very helpful. Thank you.

• $w_t$ is iid. Normal(0,1), everything is said. Jan 27, 2018 at 22:36

Clearly for odd $t$:$$E(x_t)=E(w_t)=0$$and for even $t$:$$E(x_t)=\dfrac{1}{\sqrt 2}E(w_{t-1}^2-1)=\dfrac{1}{\sqrt 2}(\sigma_{w_{t-1}}^2-1)=\dfrac{1}{\sqrt 2}(1-1)=0$$Also:$$C(t_1,t_2)=E((x_{t_1}-E(x_{t_1}))(x_{t_2}-E(x_{t_2})))=E(x_{t_1}x_{t_2})$$ according to definition for distinct $t_1$ and $t_2$, $x_{t_1}$ and $x_{t_2}$ are dependent only if $t_1$ is even and $t_2=t_1+1$ or $t_2$ is even and $t_1=t_2+1$. Assume the former case. Then $t_1=t$ is even and $t_2=t+1$ is odd so:$$C(t,t+1)=E(x_tx_{t+1})=E(x_tx_{t+1})=\dfrac{1}{\sqrt 2}E(w_t^3-w_t)=0$$So the covariance function is zero for distinct $t_1$ and $t_2$ and for $t_1=t_2=t$ we have:$$C(t,t)=E(x^2_t)$$for odd $t$:$$C(t,t)=E(w_t^2)=1$$and for even $t$:$$C(t,t)=\dfrac{1}{2}E(w^4_{t-1}-2w^2_{t-1}+1)=\dfrac{1}{2}(E(w^4_{t-1})-1)=\dfrac{1}{2}(3-1)=1$$and we finally obtain:$$C(t_1,t_2)=\delta[t_1-t_2]$$where $\delta[n]=1$ for $n=0$ and zero elsewhere.
For the second question we have $$x_1=w_1$$ and $$x_2=\dfrac{1}{\sqrt 2}(w_1^2-1)$$ so $x_1$ can vary in $(-\infty,\infty)$ but $x_2$ can vary in $[-\dfrac{1}{\sqrt 2},\infty)$ so they can't be identically distributed.