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Given: Uniform distribution. $Y$ is a continuous random variable over $[0, 1]$, and $W=Y^2$. I want to find the pdf for $W$.

Here's what I have: $F_W (w) = P(W\leq w)= P(Y^2\leq w)=P(Y\leq w^{1/2})=F_Y(w^{1/2})$. Differentiating and using the chain rule gives the cdf $f_W(w)=f_Y(w^{1/2})/[2w^{1/2}]$.

How do I simplify this to $1/[2(w^{1/2})]$? In other words, why is $f_Y(w^{1/2})=1$ for any $w \in [0, 1] $?

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up vote 1 down vote accepted

In what you wrote, there is no reason to think $f_Y(w^{1/2})= 1$. That would be the case if it were given that $Y$ is uniformly distributed in the interval $[0,1]$. If not, then it might be true only of special values of $w$, or there might be no such values of $w$, and all of that would depend on what the distribution of $Y$ is.

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Sorry, I edited the problem. I got it now, thank you! – cap Apr 15 '12 at 23:14

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