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How would I go about finding this density function? Thanks

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  • $\begingroup$ (1) Do you know how to find the partial distribution function for $Z$? (2) Do you know how to find the partial distribution function for $Y$, if you have $Y = f(Z)$ and you know the PDF of $Z$? $\endgroup$ Jul 8 '16 at 11:45
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    $\begingroup$ See here. $Z^2$ is distributed as Chi-Square. So possible duplicate $\endgroup$
    – Shailesh
    Jul 8 '16 at 12:07
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The cumulative distribution function for $Y$ is: $$ F_Y(y)=\operatorname{Prob}(Y<y)=\operatorname{Prob}(-\sqrt{y}<z<\sqrt{y})=F_Z(\sqrt{y})-F_Z(-\sqrt{y})=2F_Z(\sqrt{y})-1 $$ Now to get the density differentiate with respect to $y$, and of course to differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ with respect to $x$ is $f_Z(x)$ the density of $Z$, which you know as $Z\sim N(0,1)$.

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    $\begingroup$ In fact, what you end up with is a chi-squared distribution $\endgroup$ Jul 8 '16 at 11:58
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    $\begingroup$ You may know that and i may know that (with 1 DF) but it does not answer the question as asked which was how you go about finding the density function. $\endgroup$ Jul 8 '16 at 12:03
  • $\begingroup$ @ConradTurner Thanks for the answer! I understand the answer but just had one more query. I've got the formula for the normal probability density function but to differentiate the cumulative distribution function for Y stated above, I would need to find cumulative distribution function for Z up to $\sqrt{y}$, but how could I do that? $\endgroup$
    – silverjoe
    Jul 8 '16 at 13:31
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    $\begingroup$ @ConradTurner I thought it was useful to have some kind of relevant link $\endgroup$ Jul 8 '16 at 14:14
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    $\begingroup$ @SamuelOpiyo To differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ with respect to $x$ is $f_Z(x)$ the density of $Z$. $\endgroup$ Jul 8 '16 at 15:59

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