Prove that the Variance of a normal distribution is (sigma)^2 (using its moment generating function).
What I did so far:
$Var(X) = E(X^2) – (E(X))^2$
$$E(X^2) = Mx'(0) = \frac r{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2)$$
$E(X) = Mx''(0) = \frac {r^2}{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2)$
So $Var(X) = \frac r{\sqrt{2\pi}*\sigma} * exp(-[(x-\mu)/\sigma]^2/2) -(r^4)/[(2\pi)*(\sigma^2)] * exp(-[(x-\mu)/\sigma]^2)$
However, I dont see how I can proceed from here. How can I take these terms together? Could anyone please help me out?
Best Answer
It is not clear from the wording of the question whether you have been given the mgf or not. If you have not, then we need to calculate it. A sketch of how to do this is given in the remark at the end. But for now let us assume we know the mgf.
Suppose we know that $M_X(t)$, the mgf of the random variable $X$, is given by $M_X(t)=\exp(a t+\frac{1}{2}b t^2)$. We want to show that $a$ is the mean of $X$ and $b$ is the variance of $X$.
Calculate $M_X'(t)$. We get $$M_X'(t)=(a+bt)\exp(a t+\frac{1}{2}b t^2).$$ So $M_X'(0)=a$. That means that $a$ is the mean $\mu$ of $X$, that is, $a=E(X)$.
To find $E(X^2)$, we find the second derivative of $M_X(t)$ at $t=0$. We get $$M_X''(t)=(a+bt)^2 \exp(a t+\frac{1}{2}b t^2)+b\exp(a t+\frac{1}{2}b t^2).$$ Set $t=0$. We get $M_X''(0)=a^2+b$. So $E(X^2)=a^2+b$.
Finally, the variance of $X$ is $E(X^2)-(E(X))^2$. (Please note: it is not $E(X^2)-E(X)$.) We have $$E(X^2)-(E(X))^2=(a^2+b)-a^2=b.$$
Remark: We sketch how to find the mgf of the general normal. Recall that $M_X(t)=E(e^{tX})$. Thus $$M_X(t)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} \exp(tx)\exp(-(x-\mu)^2/(2\sigma^2))\,dx.$$ The product of the exponentials is $\exp\left(-\frac{x^2-2x(\mu+t\sigma^2)+\mu^2}{2\sigma^2}\right)$. Now make the change of variable $\sigma u=x-(\mu +t\sigma^2)$. After some mildly tedious algebra, and using the fact that $\int_{-\infty}^\infty e^{-w^2/2}\,dw=\sqrt{2\pi}$, we get the mgf.