<p>A useful implication of conditional mean independence is: <span class="math-container">$E[\varepsilon_{it}|\varepsilon_{ks}]=0 \implies E[\varepsilon_{it}\varepsilon_{ks}]=0,$</span> and more generally, <span class="math-container">$E[\varepsilon_{it}|\varepsilon_{ks}]=0 \implies E[\varepsilon_{it}m(\varepsilon_{ks})]=0,$</span> for any function <span class="math-container">$m$</span>.</p>
<p>This can be applied to your case: if
<span class="math-container">$$E[\varepsilon_{1t}^2] =E[\varepsilon_{1t}^2|\varepsilon_{1s}] = \sigma_1^2,$$</span>
then (for <span class="math-container">$m$</span> defined as the square):
<span class="math-container">$$ E[(\varepsilon_{1t}^2-\sigma_1^2)\varepsilon_{1s}^2] = 0 
\Leftrightarrow E[\varepsilon_{1t}^2\varepsilon_{1s}^2] = \sigma_1^2\sigma_1^2,$$</span></p>
<p>and so, <span class="math-container">$cov(\varepsilon_{1t}^2,\varepsilon_{1s}^2)=0.$</span></p>
<p>Further interesting results for the covariance of squared and product of random variables are included in this paper:<br />
Bohrnstedt, G. W. and A. S. Goldberger, 1969,
On the Exact Covariance of Products of Random Variables, <em>Journal of the American Statistical Association</em>, 64, 1439-1442.</p>
<p>Hope that it helps...</p>

CC BY-SA 4.0
2022-01-31T14:29:39.833Z
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.
Giorgetto
CC BY-SA 4.0
2022-01-31T16:37:59.680Z
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 $m$.
Bertrand
CC BY-SA 4.0
2022-02-02T07:33:43.327Z
@Giorgetto: I updated my former post, and give more details on how conditional mean independence helps to find some results.
Bertrand
CC BY-SA 4.0
2022-01-31T12:55:43.963Z
2022-02-01T21:19:12.120Z
2022-02-01T21:19:12.120Z
Bertrand
<p>I am studying <a href="https://www.restud.com/paper/identifying-shocks-via-time-varying-volatility/" rel="nofollow noreferrer">this paper</a>, and I don't understand the derivation  of the covariances at the bottom of page 3090.</p>
<p>Basically I have two shocks: <span class="math-container">$\varepsilon_{1t}$</span> has constant volatility <span class="math-container">$E[\varepsilon_{1t}^2]$</span> = <span class="math-container">$\sigma^2_1$</span> while <span class="math-container">$\varepsilon_{2t}$</span> has time varying volatility <span class="math-container">$E[\varepsilon_{2t}^2]$</span> = <span class="math-container">$\sigma^2_{2,t}$</span>. I further assume that:</p>
<p><span class="math-container">$E[\varepsilon_{it}|\varepsilon_{jt}]=0$</span> for <span class="math-container">$j \neq i$</span> and every t</p>
<p><span class="math-container">$E[\varepsilon_{it}|\varepsilon_{ks}]=0$</span> fo every <span class="math-container">$k$</span> and <span class="math-container">$s \neq t$</span></p>
<p>I am struggling to derive these quantities:</p>
<p><span class="math-container">$cov(\varepsilon_{1t}^2, \varepsilon_{1t-p}^2)$</span></p>
<p><span class="math-container">$cov(\varepsilon_{1t}^2, \varepsilon_{1t-p} \varepsilon_{2t-p})$</span></p>
<p><span class="math-container">$cov(\varepsilon_{1t}^2,\varepsilon_{2t-p}^2)$</span></p>
<p><span class="math-container">$cov(\varepsilon_{2t}^2,\varepsilon_{1t-p}^2)$</span></p>
<p><span class="math-container">$cov(\varepsilon_{1t}\varepsilon_{2t},\varepsilon_{1t-p}^2)$</span></p>
<p><span class="math-container">$cov(\varepsilon_{1t}\varepsilon_{2t},  \varepsilon_{1t-p}\varepsilon_{2t-p} )$</span></p>
<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 <span class="math-container">$cov(\varepsilon_{1t}^2, \varepsilon_{1t-p}^2)= E[\varepsilon_{1t}^2\varepsilon_{1t-p}^2] - (\sigma^2_1)^2$</span>
but it is not clear to me how to show that <span class="math-container">$E[\varepsilon_{1t}^2\varepsilon_{1t-p}^2] = (\sigma^2_1)^2 $</span>
to obtain 0.</p>
<p>Similarly for the other quantities, why
<span class="math-container">$E[\varepsilon_{1t}^2\varepsilon_{1t-p} \varepsilon_{2t-p}]$</span>
would be equal to 0?<a href="https://www.restud.com/paper/identifying-shocks-via-time-varying-volatility/" rel="nofollow noreferrer"></a></p>

CC BY-SA 4.0
2022-02-05T01:54:47.033Z
All results can be obtained by applying the Law of Iterated Expectations a sufficient number of times
Grada Gukovic
<econometrics><time-series><stochastic-processes><variance>
Derivation of autocovariances Lewis (2021) RES
Giorgetto
# Derivation of autocovariances Lewis (2021) RES

(Asked by: username_0 on 2022-01-31)

I am studying [this paper](https://www.restud.com/paper/identifying-shocks-via-time-varying-volatility/), 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?

**Tags:** econometrics, time-series, stochastic-processes, variance

> **Comments:**
> **username_1**: All results can be obtained by applying the Law of Iterated Expectations a sufficient number of times  

---

## 1 answers

### Answer by username_2 on 2022-01-31. Score: 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...

> **Comments:**
> **username_0**: 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.  
> **username_2**: 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 $m$.  
> **username_2**: @username_0: I updated my former post, and give more details on how conditional mean independence helps to find some results.  

---

