2
$\begingroup$

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.

$\endgroup$
5
  • 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

1
$\begingroup$

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,

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

and

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

due to Jensen's Inequality.

$\endgroup$
1
$\begingroup$

\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*}

$\endgroup$
2
  • 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

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.