Calculating integral with standard normal distribution.

Question

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:)

Answer 1

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 $\displaystyle -\frac{1}{t} e^{-\frac12 t^2}$

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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, $E[X^2]=1$.

Answer 2

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 $\chi^2_1$.