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Let $y_t = \Delta p_t$ denote a time series of asset returns, where $p_t$ are logarithmic prices. $y_t$ is generated by a heteroskedastic MA(1) process
\begin{aligned} y_t &= z_t+\theta z_{t-1}, \\ z_t &= \sqrt{h_t} \epsilon_t, \\ h_t &= \omega+\alpha z_{t}^2+\beta h_{t-1}, \\ \epsilon_t &\sim iid (0,1). \\ \end{aligned} In addition, $\omega>0$, $\alpha>0$, $\beta<1$ and $\alpha+\beta<1$.

Derive expressions for

  • The unconditional mean of the process, $E(y_t)$ ----> For this first point I got $E(y_t) = 0$ since it's the sum of the expected value of the sum of 2 zero-mean variables
  • The unconditional variance $\text{Var}(y_t)$ ----> For this point I am stuck. What I did so far is $$ \text{Var}(y_t) = E(z_t^2+\theta^2 z_{t-1} ^2 ) = E(h_t+\theta^2 h_{t-1}). $$ How can I get the final expression for the unconditional variance?

What is weird for me is that $h_t = \omega+\alpha z_t^2+\beta h_{t-1}$ instead of $h_t = \omega+\alpha z_{t-1}^2+\beta h_{t-1}$. If a consider a typo in $h_t$ then the unconditional variance is $$ \gamma(0) = \frac{\omega(1+\theta^2)}{1-(\alpha+\beta)}. $$ Is my result correct?

Then I am asked to find the autocorrelation function of $y_t$, $\rho(k)$, for $k = 1,2$. I am having trouble to find the autocovariance at lag 1. $$ \gamma(1) = E(z_t+\theta z_{t-1})(z_{t-1}+\theta z_{t-2}) = \theta E(z_{t-1}^2) $$ I would say that $\gamma(1) = \frac{\theta \omega}{1-(\alpha+\beta)}$ but I am completely unsure of this result.

Regarding $\rho(2)$ should be equal to $0$ because $\gamma(2) = 0$.

Can someone tell me how to proceed?

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  • $\begingroup$ Welcome to CV. I initially closed this question because I misread it: it is very difficult to distinguish "$\varepsilon_t$" from "$\epsilon_t$" in some browsers. $\endgroup$
    – whuber ♦
    Commented Aug 23, 2023 at 22:17
  • $\begingroup$ I see , I am changing notation to make the question clearer $\endgroup$
    – V013
    Commented Aug 23, 2023 at 22:19
  • $\begingroup$ I think it should be $z_{t-1}^2$ instead of $z_t^2$ in the conditional variance equation. Then your process is MA(1)-GARCH(1,1). Otherwise you get a circular definition where $z_t$ defines $h_t$ and the other way around. $\endgroup$
    – Richard Hardy
    Commented Aug 24, 2023 at 12:18
  • $\begingroup$ yes, do you think that the equation for the variance I got is correct? $\endgroup$
    – V013
    Commented Aug 24, 2023 at 15:28

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