Moment generating functions of normal distribution

Question

If I have a normal distribution $X$ with mean 0 and variance $\sigma^2$ for $\sigma>0$, how would I find the moment generating function of $Y=X^2$?

I can find the moment generating function of a normal distribution. But I'm not sure how that changes if I'm squaring the distribution.

Thank you in advance!

The PDF of $Y$ is $$ f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{\large -\frac{y}{2\sigma^2}} $$ and the MGF of $Y$ is $$M_Y(t)=\dfrac{1}{\sqrt{1-2\sigma^2t}}.$$ — Tunk-Fey, May 29 '14 at 09:36

Bombyx mori · Accepted Answer · 2014-05-29T08:36:27.360

2

The square of a standard normal distribution is a chi-squared distribution, and its moment generating function can be looked up from here:

http://en.wikipedia.org/wiki/Chi-squared_distribution

I think you can proceed formally and compute it by $E(e^{tX^{2}})$ by definition as well. This might be easier in practice.

To make matter simple I will do this for the standard normal, we have $$ f_{X}(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}} $$ and we want to integrate $$ \frac{1}{\sqrt{2\pi}}e^{(t-\frac{1}{2})x^{2}} $$ We can use a scale transformation $u=\sqrt{1-2t}x$. Then we change the integral to be $$ \int^{\infty}_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^{2}}{2}}du*\frac{1}{\sqrt{(1-2t)}}=\frac{1}{\sqrt{(1-2t)}} $$ as desired. I think the general case is similar.

edited May 29 '14 at 08:36

answered May 29 '14 at 08:03

Bombyx mori

19,638
6
52
112

I am trying to compute it by $E(e^{tX^{2}})$, but I'm not sure how I would go about doing that. The squared part is what is confusing me. – Q1O3 May 29 '14 at 08:07
You just formally integrate. – Bombyx mori May 29 '14 at 08:07
There is a post from the other direction, see here: http://math.stackexchange.com/questions/71516/moment-generating-function-of-the-chi-square-distribution?rq=1 but the derivation look difficult. – Bombyx mori May 29 '14 at 08:08
Sorry I'm still confused. I'm just not sure how to set out the integral. – Q1O3 May 29 '14 at 08:13
Updated. I will leave the general case to you. – Bombyx mori May 29 '14 at 08:18
Note that: $$ \dfrac{X^2}{\sigma^2}\sim\chi^2_1\quad,\text{ if }\mu=0. $$ Not $$ X^2\sim\chi^2_1. $$ – Tunk-Fey May 29 '14 at 08:21
Yes, this is a good point. I do not know what the shifted version would be, in fact. – Bombyx mori May 29 '14 at 08:31
That makes sense, thanks – Q1O3 May 29 '14 at 09:08

Moment generating functions of normal distribution

1 Answers1

Linked