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:

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

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




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

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?

  • 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


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...

  • $\begingroup$ 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

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