Why Squared Returns Can Be Uncorrelated with Lagged Returns in GARCH
Summary
The post asks whether a GARCH model implies zero covariance between a squared return and a lagged return. It identifies the conditional variance as a function of past squared returns and variance, then considers how that dependence affects the expectation involving an earlier return.
The questioner's argument assumes that a zero-mean innovation makes the whole expectation vanish. A response questions that conclusion and attempts to trace dependence through the variance recursion. However, the response contains unclear indexing and algebra, and it does not establish a correct result. The exchange is useful chiefly as a prompt to distinguish conditional expectations from unconditional dependence: a centered innovation alone does not justify treating every other factor in a product as independent. Readers should verify the calculation rather than rely on the answer's stated conclusion.
Key ideas
- A GARCH conditional variance depends on earlier squared returns and conditional variances.
- A zero-mean innovation does not by itself make a product expectation zero when its other factors may depend on that innovation.
- The response traces possible dependence through the variance recursion but does not give a reliable derivation.
- Conditional and unconditional expectations must be distinguished when evaluating return moments.
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Full text
# covariance between squared returns and past returns
# 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?
## Answer by mark leeds (score 1)
https://quant.stackexchange.com/a/76578
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 ?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.