moment generating function technique

Question

If $X$ was a random variable with a distribution $\mathrm{Normal} ( 0, 1 )$, using moment generating function technique we have to show that $Y= X^2$ has the Chi-square distribution with $1$ degree of freedom. My problem is that I am used to solving such questions when $Y$ is a linear transformation of $X$! This is my first time to face such a question.

Answer 1

First, for the calculation we will need the well-known result that $$ \int_{-\infty}^\infty e^{-ax^2} \; dx = \sqrt{\pi/a} $$ when $a>0$. Also, in general we have that if $X$ is a random variable with density function $f(x)$ then, for any function $h$ of $X$ we have $$ \DeclareMathOperator{\E}{E} \E \left( h(X) \right) = \int_{-\infty}^\infty h(x) f(x) \, dx. $$ Calculating the moment-generating function of $X^2$ when $X$ have the standard normal distribution: $$ \begin{align} \phi_{X^2}(t) &= \E \left( e^{tX^2} \right) \\ &= \int_{-\infty}^{\infty} e^{tx^2} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \; dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-(1/2-t)x^2} \; dx \\ &= (1-2t)^{-1/2}. \end{align} $$ when $t<1/2$. That you can compare with the moment-generating function of the chi-squared distribution (with one degree of freedom). Unicity of moment-generating functions yield the result.