Calculating integral with standard normal distribution. 

I have a problem to solving this,
Because I think that for solving this problem, I need to calculate cdf of standard normal distribution and plug Y value and calculate.
However, at the bottom I found that Integral from zero to infinity of 1 goes to infinity and I cannot derive the answer. Can you tell me what's the problem and what can I do?
I appreciate for your help in advance:)
 A: You have managed to state a closed form for $F_x(x)$.  
This is not in fact possible for a normal distribution, so you have an error in your integral about a third of the way down: the integral of $\displaystyle e^{-\frac12 t^2}$ is not  $\displaystyle -\frac{1}{t} e^{-\frac12 t^2}$

There is an easier solution, but it uses a shortcut which your teacher might not accept here:


*

*For a non-negative random variable, you have $E[Y] =  \displaystyle\int_{t=0}^\infty \Pr(Y \gt t) dt$.

*$X^2$ is indeed non-negative, so you want $E[X^2]$.

*For $X$ with a standard $N(0,1)$ Gaussian distribution, $E[X^2]=1$.     
A: Let $X\sim N(0,1)$ with pdf $f_X(x)$ and $Y=X^2$. The pdf $f_Y(y)$ of $Y$ is
$$
f_Y(y)=f_X(\sqrt y)\left| \frac{1}{2}y^{-\frac{1}{2}}\right|+f_X(-\sqrt y)\left| -\frac{1}{2}y^{-\frac{1}{2}}\right|=\frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}y^{-\frac{1}{2}}\quad \text{for }y>0
$$
Recalling that $\Gamma(1/2)=\sqrt \pi$, we may write
$$
f_Y(y)=\frac{1}{2^{1/2}\Gamma(1/2)}e^{-\frac{y}{2}}y^{-\frac{1}{2}}\quad \text{for }y>0
$$
and then integrating and using the Gamma function you will find the right result 1.
Incidentally, if $X$ is normal $N(0,1)$ then $X^2$ is chi squared with one degree of freedom $\chi^2_1$.
