Method of moment generating function normal distribution

Quote:

Originally Posted by lllll

Suppose that $Y$ is normally distributed with mean 0, and unknown variance $\sigma^2$ . Using the method of moment generating functions, show that $\frac{Y^2}{\sigma^2}$ has a $\chi^2$ distribution with 1 df.

I know that the mgf of a normal distribution is $\exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right)$ , so if I was to replace the variable with what was given in the question then I would get:

$\exp\left(0 t +\frac{(Y^2/\sigma^2)^2 \cdot \sigma^2}{2} \right) = \exp\left(\frac{Y^4}{2\sigma^2} \right)$ which looks nothing like the mgf of a $\chi^2$ distribution.

Mr F says: You need to stop thinking like this. You cannot just substitute like it's a change of variable. This is a grave mistake that is a common to a lot of your solutions.

Mr F quarantines the above. ---------------------------------------------------------------------------------------------

the other way I was thinking of doing solving this was using the definition, so I would get:

$m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y-\mu)^2\right] \cdot \exp\left(\frac{ty^2}{\sigma^2}\right) \ dy$ Mr F says: Good move!

$m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 \cdot \left(\frac{ty^2}{\sigma^2}\right)\right] \ dy$ Mr F says: NO! When you multiply exponentials with the same base, you ADD the powers (this is a basic index law). You do not multiply the powers.

$m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{ty^4}{2\sigma^4}\right)\right] \ dy$

at which point I don't know how to proceed

Quote:

Originally Posted by lllll and edited in red by Mr F

[snip]
$m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 {\color{red}+} \left(\frac{ty^2}{\sigma^2}\right)\right] \ dy$
[snip]

Then:

$m_{Y^2/\sigma^2} (t) = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ - y^2 \left(\frac{1}{2\sigma^2} - \frac{t}{\sigma^2}\right)\right] \, dy$ $= \frac{1}{\sigma\sqrt{2\pi}} \int^{+\infty}_{-\infty} \exp \left[ - y^2 \left(\frac{1 - 2t}{2\sigma^2}\right)\right] \, dy$

Now make the substitution $x = \frac{y}{\sigma} \sqrt{\frac{1 - 2t}{2}} \Rightarrow dy = \sigma \sqrt{\frac{2}{1-2t}} \, dx$ :

$= \frac{1}{\sqrt{\pi}} \, \frac{1}{\sqrt{1 - 2t}} \, \int^{+\infty}_{-\infty} \exp \left[ - x^2 \right] \, dx$

$= \frac{1}{\sqrt{\pi}} \, \frac{1}{\sqrt{1 - 2t}} \, \sqrt{\pi}$

(using the well known result)

$= \frac{1}{\sqrt{1 - 2t}}$ .