Let $y_t = \sqrt{h_t} \epsilon_t$ where $\epsilon_t\overset{ iid}{\sim} N(0,1)$ $h_t = \alpha_0 +\alpha_1 y_{t-1}^2+\beta_1 h_{t-1}$ with $\alpha_0>0, \alpha_1>0, \beta_1<1,\alpha_1+\beta_1<1$

Show that $Cov(y_t^2 y_{t-j}) =E(y_t^2 y_{t-j})= 0$

My guess is that $E(y_t^2) = h_t$ then $E(y_t^2 y_{t-j})= E(h_t*\sqrt{h_{t-j}} \epsilon_{t-j}) $ since $E(\epsilon_{t-j}) = 0$ then all the product inside expectation goes to 0

can someone tell me if this is the correct interpretation please?

  • $\begingroup$ $E(XY)\neq E(X)E(Y)$ but you have used this in your guess. $\endgroup$ Sep 4, 2023 at 14:15
  • $\begingroup$ yeah, I am asked to prove the above result but I can't see another way to prove it $\endgroup$
    – XY0
    Sep 5, 2023 at 7:29
  • 1
    $\begingroup$ I am just pointing out that using a wrong equality is not the way to prove anything. But I suppose we are not disagreeing on that. $\endgroup$ Sep 5, 2023 at 7:41
  • $\begingroup$ Did you mean to write $E(y_t y_{t-j}^2) = 0$? This can be done a bit more easily. $\endgroup$ Oct 12, 2023 at 3:45

1 Answer 1


I don't get zero but generally I would approach this by just writing out the two expressions for $y^{2}_t$ and $y_{t-j}$.

$y_{t} = \sqrt{h_t} \epsilon_t $

Therefore, $y^2_{t} = h_t \epsilon^2_{t} \rightarrow y_{t}^2 = (\alpha_{0} + \alpha_1 y_{(t-1)}^2 + \beta_1 h_{(t-1)}) \times \epsilon^2_{t}$

Also, $y_{t-j} = \sqrt{h_{t-j}} \epsilon_{t-j} = \sqrt{(\alpha_{0} + \alpha_1 y_{(t-j)}^2 + \beta_1 h_{(t-j)})} \times \epsilon_{(t-j)}$

Looking at the two expressions, the only terms that are possibly correlated are

$\sqrt{h_{(t-j)}}$ and $h_{t-1}$.

So, we need to see how $h_{t-1}$ is related to $h_{t-j}$. First write out, the various $h_t$ going back in time.

$h_{t-1} = (\alpha_{0} + \alpha_1 y_{(t-2)}^2 + \beta_1 h_{(t-2}) \times \epsilon_{t-1}$

$h_{t-2} = (\alpha_{0} + \alpha_1 y_{(t-3)}^2 + \beta_1 h_{(t-3}) \times \epsilon_{t-2}$


$h_{t-j-1} = (\alpha_{0} + \alpha_1 y_{(t-j-2)}^2 + \beta_1 h_{(t-j-2}) \times \epsilon_{t-j-1}$

$h_{t-j} = (\alpha_{0} + \alpha_1 y_{(t-j-1)}^2 + \beta_1 h_{(t-j-1}) \times \epsilon_{t-j}$

So, from this recursive relation, one can see that $\sqrt{h_{t-1}}$ is a function of $\sqrt{\beta_{1}^{j+1} h_{t-j}}$.

Therefore, the covariance of $y_{t}$ and $y_{t-j} = \sqrt{(\beta_{1}^{j+1} \beta_{1}^2)}$

Unfortunately, I don't see how the covariance can be zero since $h_{(t-1)}$ is correlated with $h_{(t-j)}$ ? I could have made an algebra mistake somewhere but I don't see how they can't be related ?


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