# covariance between squared returns and past returns

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?

• $E(XY)\neq E(X)E(Y)$ but you have used this in your guess. Sep 4, 2023 at 14:15
• yeah, I am asked to prove the above result but I can't see another way to prove it
– XY0
Sep 5, 2023 at 7:29
• 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. Sep 5, 2023 at 7:41
• Did you mean to write $E(y_t y_{t-j}^2) = 0$? This can be done a bit more easily. Oct 12, 2023 at 3:45

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.

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

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

$$\ldots$$

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

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

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 ?