Suppose that $Z\sim N(0,1)$ and let $V=Z^2$. Prove that $V\sim \chi^2(1)$.

I want to use the method of moment generating functions, because I already understand the proof using the method of distribution functions. I will show my work, and then where I got stuck.

Since $Z\sim N(0,1)$, then $\mu=0$ and $\sigma^2=1$, and we have

$$M_V(t) =E[e^{tV}]=E[e^{tZ^2}]=\int_{-\infty}^\infty e^{tz^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}dz=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz.$$

At this point, I'm out of ideas. I want to eventually get something that looks like $\frac{1}{(1-2t)^{\frac{1}{2}}}$. Could I get a hint please?

  • 1
    $\begingroup$ You should not be confusing the random variable denoted by capital $Z$ with the variable with respect to which you're integrating, denoted by lower-case $z$. I cleaned that up. $\endgroup$ Sep 10 '14 at 2:07


The general PDF for a normal distribution is given by:

$$ f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

You should attempt to solve the integral by fitting a normal distribution and cancelling it out by realising that it integrates to 1. Currently:

$$ \mu=0 $$ $$ \frac{1}{2\sigma^2}=\frac{1}{2}-t $$

So, solve for $\sigma$ and multiply accordingly to make the integral the pdf of a normal distribution (integrates to 1) whatever is left over should give you the result you're looking for.

Hope this helps


If $a^2=1-2t$, then $t-\frac12= -\frac12\left(1-2t\right)=-\frac{a^2}{2}$, so \begin{align} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})} \, dz & = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-z^2 a^2/2} \, dz \\[10pt] & = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-w^2/2} \left(\frac{dw}a\right) \\[10pt] & = \frac1a\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-w^2/2} \, dw \\[10pt] & = \frac1a = \frac{1}{\sqrt{1-2t}}. \end{align}


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