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If X is a standard normal random variable, with mean of 0 and standard deviation of 1, how would you go about calculating the density function of Y=√|X|. I figure that we will be able to break it up into two parts, one where the density function equals zero for y<0, but am a little lost finding the equation for the other part!

Cheers

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Hint: I would probably start by trying to write $F_Y$ in terms of $F_X$. So for $y>0$ you have $$F_Y(y)=P(Y<y)$$ $$=P(|X|<y^2)$$ $$=P(-y^2<X<y^2)$$ $$=F_X(y^2)-F_X(-y^2)$$ Now you can differentiate using the chain rule and write $f_Y$ in terms of $f_X$.

Addendum: Note that for this particular CDF, symmetry implies that $F_X(-u)=1-F_X(u)$. This allows a further simplification $$F_Y(y)=2F_X(y^2)-1$$

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Since the density of X is given by the symmetric $\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}dx$, the density of Z=|X| is just twice that for X when Z >= 0, i.e., $\sqrt\frac{2}{\pi}e^{-z^{2}/2}dz$.
Now $Y=\sqrt{Z}$ is 1-1 on the non-negative reals, so $dz = 2ydy$ for $y>0$ and we get the density function for Y by substitution: $\sqrt\frac{2}{\pi} 2y e^{-y^{4}/2}dy$ for $y>0$ (and 0 elsewhere).

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