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If X~(0,1), integrate to find the moment generating function of a random variable $X^2$ and identify the distribution of $X^2$ using the moment generating function.

E[$e^{tx^2}]=\int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}\over{\sqrt{2\pi}}}$

which reduces to

=$1\over{\sqrt{2\pi}}$ $\int_{-\infty}^{\infty}{e^{tx^2}e^{-x^2}}$ =$1\over{\sqrt{2\pi}}$ $\int_{-\infty}^{\infty}{e^{x^2(t-\frac12)}}$

and thus I am stuck. I'm sure there must be some trick to this (like completely the square for the mgf of a standard normal variable X) but I can't figure out what it might be.

Any hints?

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1 Answer 1

up vote 2 down vote accepted

The mgf will only be defined when $t\lt 1/2$.

We need $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-x^2(1/2-t)}\, dx.$$ The exponent can be written as $-(x^2/2)(1-2t)$.

Make the change of variable $u=x\sqrt{1-2t}$. Then $dx =\frac{1}{\sqrt{1-2t}}\,du$. Note that as $x$ travels from $-\infty$ to $\infty$, so does $u$. You should end up with something like $$\frac{1}{\sqrt{1-2t}}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-u^2/2}\,du.$$ Now we recognize that the integral part is $1$.

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