# Derivation of autocovariances Lewis (2021) RES

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?

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

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...

• 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. Jan 31, 2022 at 14:29
• 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 $m$. Jan 31, 2022 at 16:37
• @Giorgetto: I updated my former post, and give more details on how conditional mean independence helps to find some results. Feb 2, 2022 at 7:33