About gaussian distribution - calculation of variance (change of variables step)

Question

According to these slides, in slide 3:

Why we do not change the x of exp with (x-mu)^2 ? I did not understand this part.

Answer 1

We have:- $$\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}(x-\mu)^{2}\exp(-\frac{(x-\mu)^{2}}{2\sigma^{2}})\,dx$$

We are then substituting $\frac{x-\mu}{\sqrt{2}\sigma}=y$

So $dy=\frac{dx}{\sqrt{2}\sigma}$

Using this we get:-

$$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}2\sigma^{2}y^{2}\exp(-y^{2})\,dy$$

Now use the "convenient trick" to get :-

$$\frac{2\sigma^{2}}{\sqrt{\pi}}\cdot\frac{1}{2}\cdot\sqrt{\pi}=\sigma^{2}$$

Hence you have the result.

You can also proceed like this:-

Assume $\mu=0$.

Then $$\int_{-\infty}^{\infty}x^{2}\frac{1}{\sigma\sqrt{2\pi}}\exp(-\frac{x^{2}}{2\sigma^{2}})\,dx=\frac{1}{\sqrt{2\pi}\sigma}\frac{1}{2}\frac{\sqrt{\pi}}{\sqrt{\frac{1}{2^{3}\sigma^{6}}}}=\sigma^{2}$$(By using the "convenient trick").

Now if $X\sim N(\mu,\sigma^{2})$ then $X-\mu\sim N(0,\sigma^{2})$.

So $Var(X-\mu)=\sigma^{2}$.

So $$E((X-\mu)^{2})=\sigma^{2} $$.

$\implies E(X^{2}+\mu^{2}-2X\mu)=\sigma^{2}$

$\implies E(X^{2})+\mu^{2}-2\mu E(X)=\sigma^{2}$

So $E(X^{2})-\mu^{2}=\sigma^{2}\implies E(X^{2})-(E(X))^{2}=Var(X)=\sigma^{2}$ .

There are other ways to calculate the integral $\int_{-\infty}^{\infty}x^{2}e^{-x^{2}}\,dx$.

You can also use Gamma Function.