Given the following stochastic process: $$X_t = \epsilon_t\epsilon_{t-1}~~~~,~\epsilon_t \sim N(0, \sigma^2)$$

Which is the variance of the process $X_t^2$?

Firstly I defined the process $X_t$ as:

$$W_t = (\epsilon_t\epsilon_{t-1})^2= \epsilon_t^2\epsilon_{t-1}^2$$

Then its expected value as:

$$E[W_t]= E[\epsilon_t^2\epsilon_{t-1}^2]=E[\epsilon_t^2]E[\epsilon_{t-1}^2]= \sigma^4$$

And the variance as:

$$\text{Var}[W_t]=E[(\epsilon_t^2\epsilon_{t-1}^2-E[X_t])^2]$$ $$=E[\epsilon_t^4\epsilon_{t-1}^4]-2E[\epsilon_t^2\epsilon_{t-1}^2]\sigma^4+\sigma^8 = \sigma^8 - 2\sigma^8+\sigma^8=0$$

I am not able to understand where is my error.

  • 1
    $\begingroup$ Are $\epsilon_t$ and $\epsilon_{t-1}$ independent? Because you are using the property of independence in your derivations. Also, are you interested in the variance of $X_t$ or $W_t$? You start with the former but end with the latter. $\endgroup$ Feb 15, 2017 at 17:11
  • 2
    $\begingroup$ is $\mathbb{E}\epsilon^4 = \sigma^4$? Think carefully... $\endgroup$
    – jbowman
    Feb 15, 2017 at 17:30
  • $\begingroup$ Yeah here is the problem..... $E\sigma^4$ is the kurtosis $\endgroup$
    – Archimede
    Feb 15, 2017 at 17:36
  • $\begingroup$ $E[\epsilon^4]=3\sigma^4$ $\endgroup$
    – Julius
    Feb 15, 2017 at 23:23
  • $\begingroup$ Hint: first solve the general problem of computing the variance of $XY$ in terms of moments of $X$ and $Y$ where $X$ and $Y$ are independent random variables. Apply this to the case where $X=\epsilon_t^2$ and $Y=\epsilon_{t-1}^2$. That reduces the problem to finding the second and fourth moments of a zero-mean Normal variable. $\endgroup$
    – whuber
    Feb 15, 2017 at 23:29

2 Answers 2


The mistake of the OP lies in the last line of the post: in reality

$$E[\epsilon_t^4\epsilon_{t-1}^4] \neq \sigma^8$$

because, under assumed independence,

$$E[\epsilon_t^4\epsilon_{t-1}^4] = E[\epsilon_t^4]\cdot E[\epsilon_{t-1}^4]$$


$$E[\epsilon_t^4] = E[(\epsilon_t^2)^2] > [E(\epsilon_t^2)]^2 = (\sigma^2)^2 = \sigma^4$$

due to Jensen's Inequality.


\begin{eqnarray*} Var(X_t^2) &:=& Var(W_t) \\ &=& E[W_t^2] - E[W_t]^2 \\ &=& E[X_t^4] - E[X_t^2]^2 \\ &=& E[(\varepsilon_t\varepsilon_{t-1})^4] - E[(\varepsilon_t\varepsilon_{t-1})^2]^2 \end{eqnarray*}

Recall that the product of two standard normal random variables (that is, $N(0,1)$) is distributed according to a $\chi^2_1$ distribution. Then,

\begin{eqnarray*} \varepsilon_t\varepsilon_t &=& \sigma^2\frac{\varepsilon_t}{\sigma}\frac{\varepsilon_t}{\sigma} \\ &:=& \sigma^2Y \\ \Rightarrow Y &\sim& \chi^2_1 \\ \Rightarrow \sigma^2Y &\sim& \Gamma\left(\frac{1}{2},2\sigma^2\right) \\ \Rightarrow \varepsilon_t\varepsilon_t &\sim& \Gamma\left(\frac{1}{2},2\sigma^2\right) \end{eqnarray*}

The second implication in the above work is taken from this link. Note that the exact same holds for $\varepsilon_{t-1}\varepsilon_{t-1}$ because $\varepsilon_t$ and $\varepsilon_{t-1}$ follow the same distribution. Thus we can write $X^4$ as $(\varepsilon_t\varepsilon_t)^2(\varepsilon_{t-1}\varepsilon_{t-1})^2$, where $X$ follows a $\Gamma\left(\frac{1}{2},2\sigma^2\right)$ distribution.

For any random variable $X$, $E[X^n] = M^{(n)}_X(0)$, where $M^{(n)}_X(0)$ is the $n$th derivative of the moment generating function (mgf) of $X$, evaluated at 0. (This is Theorem 2.3.7 in Casella and Berger's Statistical Inference.) The moment generating function for a $\Gamma(\alpha,\beta)$ variable is $M_X(t)=\left(\frac{1}{1-\beta t}\right)^\alpha$. Then...

\begin{eqnarray*} E[X_t^2] &=& M^{(2)}_X(0) \\ &=& \frac{d^2}{dt^2}\left(\frac{1}{1-\beta t}\right)^\alpha\biggr\rvert_{t=0} \\ &=& \frac{d^2}{dt^2}(1-\beta t)^{-\alpha}\biggr\rvert_{t=0} \\ &=& \frac{d}{dt}(-\alpha\cdot(1-\beta t)^{-\alpha-1}\cdot-\beta)\biggr\rvert_{t=0} \\ &=& \frac{d}{dt}(\alpha\beta(1-\beta t)^{-\alpha-1})\biggr\rvert_{t=0} \\ &=& \left((-\alpha-1)\cdot\alpha\beta(1-\beta t)^{-\alpha-2}\cdot-\beta\right)\biggr\rvert_{t=0} \\ &=& \left(\beta(\alpha+1)\alpha\beta(1-\beta t)^{-\alpha-2}\right)\biggr\rvert_{t=0} \\ &=& \left(\frac{\beta(\alpha+1)\alpha\beta}{(1-\beta t)^{\alpha+2}}\right)\biggr\rvert_{t=0} \\ &=& \left(\beta^2\alpha(\alpha+1)\right) \end{eqnarray*}

You can find $E[X^4]$ in the same manner or by using formulas available online. Be careful, however, as some Web sites will have different parameterizations of the Gamma distribution.

\begin{eqnarray*} E[X^4] &=& M^{(4)}_X(0) \\ &=& \beta^4\alpha(\alpha+1)(\alpha+2)(\alpha+3) \end{eqnarray*}

From the top, we had: \begin{eqnarray*} Var(X_t^2) &=& E[X_t^4] - E[X_t^2]^2 \\ &=& \beta^4\alpha(\alpha+1)(\alpha+2)(\alpha+3) - \left(\beta^2\alpha(\alpha+1)\right)^2 \\ &=& \beta^4\alpha(\alpha+1)(\alpha+2)(\alpha+3) - \beta^4\alpha^2(\alpha+1)^2 \\ Var(X_t^2) &=& \beta^4\alpha(\alpha+1)\left[(\alpha+2)(\alpha+3)-\alpha(\alpha+1)\right] \end{eqnarray*}

  • 1
    $\begingroup$ The product of two independent standard Normals cannot possibly have a $\chi^2$ distribution, because half the time that product is negative! $\endgroup$
    – whuber
    Feb 15, 2017 at 22:21
  • 1
    $\begingroup$ You're right. Let me reconsider and edit my post. $\endgroup$
    – Matt Brems
    Feb 15, 2017 at 22:27

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