0
$\begingroup$

Let $y_t = \Delta{p_t}$ denote a time series of asset log-returns, where $p_t$ are logarithmic prices; $y_t$ is generated by the conditionally heteroscedastic MA(1) process

$y_t = \epsilon_t + \theta \epsilon_{t-1}$, where $\epsilon_t = \sqrt{h_t}z_t$ and $z_t\sim i.i.d N (0,1) $ with $|\theta|<1$, $h_t = \omega+\alpha \epsilon_t^2+\beta h_{t-1},\quad \omega>0, \alpha>0 ,\alpha+\beta<1$.

  1. Derive the expressions for the unconditional mean $E(y_t)$, unconditional variance $Var(y_t)$ and autocorrelation function of $y_t, \rho(k),k=1,2,.....$

I computed the unconditional mean being equal to 0 because $E(\sqrt{h_t} z_t)+\theta E(\sqrt{h_{t-1}}z_{t-1})=0 $ for the unconditional variance $Var(y_t)= E(y_t^2)= (h_t z_t^2+\theta^2 h_{t-1} z_t^2)$ = $E(h_t+\theta^2 h_{t-1})$ because $z_t\sim i.i.d N (0,1)$.

From now on I am not sure. Should I substitute $h_t$ to be done? Any help would be really appreciated.

$\endgroup$
2
  • $\begingroup$ I think that should be $\alpha \epsilon^2_{t-1}$ in your definition of $h_t$. $\endgroup$ May 22 at 9:06
  • $\begingroup$ Consider adding a self-study tag. $\endgroup$ May 22 at 11:05

1 Answer 1

0
$\begingroup$

So far so good.

To obtain an expression for the unconditional variance, you need to work out $\mathbb{E}(\epsilon_t^2) =\mathbb{E}(h_t)$. Call this $\sigma^2_t$.

Substituting in $h_t$ as you suggested gives $$ \sigma^2_t = \mathbb{E}(\omega + \alpha \epsilon_{t-1}^2 + \beta h_{t-1}) = \omega + (\alpha + \beta) \sigma^2_{t-1} $$ Now substitute in $\sigma^2_{t-1}$. Can you see what's going to happen?

$\endgroup$
2
  • 1
    $\begingroup$ Introducing $\sigma_t^2$ in addition to $h_t$ seems superfluous and perhaps even confusing, since both refer to the same object. Also, the notation of your expectation operators does not suggest they are conditional, but your second paragraph seems to imply they are. $\endgroup$ May 22 at 11:00
  • $\begingroup$ is the variance equal to $\frac{\omega}{1-\alpha-\beta}+\theta^2 \frac{\omega}{1-\alpha-\beta}$? $\endgroup$
    – user362147
    May 22 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.