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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!

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  • $\begingroup$ 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 $$M_Y(t)=\dfrac{1}{\sqrt{1-2\sigma^2t}}.$$ $\endgroup$ – Tunk-Fey May 29 '14 at 9:36
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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.

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  • $\begingroup$ 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. $\endgroup$ – Q1O3 May 29 '14 at 8:07
  • $\begingroup$ You just formally integrate. $\endgroup$ – Bombyx mori May 29 '14 at 8:07
  • $\begingroup$ There is a post from the other direction, see here: math.stackexchange.com/questions/71516/… but the derivation look difficult. $\endgroup$ – Bombyx mori May 29 '14 at 8:08
  • $\begingroup$ Sorry I'm still confused. I'm just not sure how to set out the integral. $\endgroup$ – Q1O3 May 29 '14 at 8:13
  • $\begingroup$ Updated. I will leave the general case to you. $\endgroup$ – Bombyx mori May 29 '14 at 8:18

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