The question: It's given that $Z = N(0,1)$ (the standard normal random variable), $Y= Z^2$. Calcualte the probability density function of Y.
I tried the following: ($F_Y$ is the distribution function of Y)
$F_Y(t) = P(Y\le t) = P(Z^2 \le t) = P(-\sqrt{t} \le Z \le \sqrt{t}) = \int_{-\sqrt{t}}^{\sqrt{t}} f_Z(x)dx = \int_{-\sqrt{t}}^{\sqrt{t}} (1/\sqrt{2\pi})e^{-x^2/2}dx$
But now I'm stuck since this integral doesn't have an antiderivative. What can I do to proceed?
Thanks
By symmetry, the fundamental theorem of calculus and the chain rule, you get for all $t>0$
$$\frac{d}{dt}\int_{-\sqrt{t}}^{\sqrt{t}} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx=2\frac{d}{dt}\int_0^{\sqrt{t}} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx= \frac{2}{\sqrt{2\pi}} e^{-t/2} \cdot \frac{1}{2\sqrt{t}}.$$
Since $\Gamma(1/2)=\sqrt{\pi}$, this is equal to
$$ \frac{(1/2)^{(1/2)}}{\Gamma(1/2)}e^{-(1/2)t}t^{(1/2)-1}.$$
Do you know this expression?