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## Homework Statement

The question ask to find [itex]E[X]; E[X^2]; Var(X)[/itex] for the standard normal distribution [itex]f(x)=1/\sqrt{2\pi}e^{-x^2/2}[/itex]

## Homework Equations

I found

[itex]

\begin{align}

E[X]&=\int_{-\infty}^\infty \! x*1/\sqrt{2\pi}e^{-x^2/2} \, \mathrm{d} x\\

&=1/\sqrt{2\pi} ( \int_{-\infty}^0 \! x*e^{-x^2/2} \, \mathrm{d} x + \int_{0}^{\infty} \! x*e^{-x^2/2} \, \mathrm{d} x )\\

&= 1/\sqrt{2\pi} ( (-1-0) + (0+1) )\\

&= 0

\end{align}

[/itex]

I found [itex]E[X^2]=1[/itex]

and [itex]Var[X]=1[/itex]. So far no problem.

## The Attempt at a Solution

Now, I am asked to find the pdf, [itex]E[Y], Var[Y][/itex] of [itex]Y=|X|[/itex]

Since the inverse of [itex]Y=|X|[/itex] is [itex]X=|Y|[/itex] and the derivative of the inverse is 1. The transformation seems fairly straight forward.

[itex]f(y)=1/\sqrt{2\pi}e^{-|y|^2/2}[/itex]

I am just not sure about the domain of y, isn't it exactly the same as X because x is squared? Once I get that I should be able to use the same integration technique as above to find [itex]E[Y]; Var[Y][/itex] Wouldn't these values be exactly the same as [itex]E[X]; Var[X][/itex]? Something seems wrong here.

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