1
$\begingroup$

I am studying this paper, and I don't understand the derivation of the covariances at the bottom of page 3090.

Basically I have two shocks: $\varepsilon_{1t}$ has constant volatility $E[\varepsilon_{1t}^2]$ = $\sigma^2_1$ while $\varepsilon_{2t}$ has time varying volatility $E[\varepsilon_{2t}^2]$ = $\sigma^2_{2,t}$. I further assume that:

$E[\varepsilon_{it}|\varepsilon_{jt}]=0$ for $j \neq i$ and every t

$E[\varepsilon_{it}|\varepsilon_{ks}]=0$ fo every $k$ and $s \neq t$

I am struggling to derive these quantities:

$cov(\varepsilon_{1t}^2, \varepsilon_{1t-p}^2)$

$cov(\varepsilon_{1t}^2, \varepsilon_{1t-p} \varepsilon_{2t-p})$

$cov(\varepsilon_{1t}^2,\varepsilon_{2t-p}^2)$

$cov(\varepsilon_{2t}^2,\varepsilon_{1t-p}^2)$

$cov(\varepsilon_{1t}\varepsilon_{2t},\varepsilon_{1t-p}^2)$

$cov(\varepsilon_{1t}\varepsilon_{2t}, \varepsilon_{1t-p}\varepsilon_{2t-p} )$

From the paper it seems that all these quantities are equal to 0, to get the two equations at the bottom. I don't understand why. For the first quantity I have $cov(\varepsilon_{1t}^2, \varepsilon_{1t-p}^2)= E[\varepsilon_{1t}^2\varepsilon_{1t-p}^2] - (\sigma^2_1)^2$ but it is not clear to me how to show that $E[\varepsilon_{1t}^2\varepsilon_{1t-p}^2] = (\sigma^2_1)^2 $ to obtain 0.

Similarly for the other quantities, why $E[\varepsilon_{1t}^2\varepsilon_{1t-p} \varepsilon_{2t-p}]$ would be equal to 0?

$\endgroup$
1
  • 1
    $\begingroup$ All results can be obtained by applying the Law of Iterated Expectations a sufficient number of times $\endgroup$ Commented Feb 5, 2022 at 1:54

1 Answer 1

1
$\begingroup$

A useful implication of conditional mean independence is: $E[\varepsilon_{it}|\varepsilon_{ks}]=0 \implies E[\varepsilon_{it}\varepsilon_{ks}]=0,$ and more generally, $E[\varepsilon_{it}|\varepsilon_{ks}]=0 \implies E[\varepsilon_{it}m(\varepsilon_{ks})]=0,$ for any function $m$.

This can be applied to your case: if $$E[\varepsilon_{1t}^2] =E[\varepsilon_{1t}^2|\varepsilon_{1s}] = \sigma_1^2,$$ then (for $m$ defined as the square): $$ E[(\varepsilon_{1t}^2-\sigma_1^2)\varepsilon_{1s}^2] = 0 \Leftrightarrow E[\varepsilon_{1t}^2\varepsilon_{1s}^2] = \sigma_1^2\sigma_1^2,$$

and so, $cov(\varepsilon_{1t}^2,\varepsilon_{1s}^2)=0.$

Further interesting results for the covariance of squared and product of random variables are included in this paper:
Bohrnstedt, G. W. and A. S. Goldberger, 1969, On the Exact Covariance of Products of Random Variables, Journal of the American Statistical Association, 64, 1439-1442.

Hope that it helps...

$\endgroup$
3
  • $\begingroup$ Do you refer to equation (15) of the paper? In my framework I only have conditional mean independence of the shocks. I don't understand how this might be exploited in that equation. $\endgroup$
    – Giorgetto
    Commented Jan 31, 2022 at 14:29
  • $\begingroup$ Conditional mean independence is quite a strong requirement as $E[\varepsilon_{it}|\varepsilon_{ks}]=0 \implies E[\varepsilon_{it}\varepsilon_{ks}]=0,$ and more generally, $E[\varepsilon_{it}m(\varepsilon_{ks})]=0,$ for any function $m$. $\endgroup$
    – Bertrand
    Commented Jan 31, 2022 at 16:37
  • $\begingroup$ @Giorgetto: I updated my former post, and give more details on how conditional mean independence helps to find some results. $\endgroup$
    – Bertrand
    Commented Feb 2, 2022 at 7:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.