# CDF method for uniform distribution

I was asked to find the density distribution for $U=Y^2$ where U~Uniform(0,1) and Y is a density distribution with $3/2(y^2)$ for $-1 \leq y \leq 1$ and 0 elsewhere.

I solved for

$$P(U \leq u) = P(Y^2 \leq u) = P(Y \leq \sqrt u ) = P(- \sqrt u \leq Y \leq \sqrt u)$$

Then I took the integral to find the CDF:

$$P(- \sqrt u \leq Y \leq \sqrt u) = \int_{0}^{\sqrt u} 3/2 y^2 dy = 1/2 \sqrt u$$

I also know that I should take the derivative of the CDF to find the PDF of the function, which would be $\frac{1}{4 \sqrt u}$

My only problem is now defining the range in which the density function $\frac{1}{4 \sqrt u}$ exists. Since Y lives in $-1 \leq y \leq 1$ and $\sqrt u$ can never be negative, does that mean that the density function exists for just $u \leq 1$? Would greatly appreciate someone who could point me in the right direction..

Note; it appears that you have misinterpreted the density of $Y$, since $\int_{-1}^1 \frac 3{2y^2}\mathrm d y$ does not converge, while $\int_{-1}^1 \frac {3y^2}{2}\mathrm d y = 1$ as required.
Also if $U=Y^2$ then $U$ cannot be uniformly distributed.   However, for the given support of $Y$, then $U$ is indeed distributed over $[0;1]$.
Other than that, and the bound on the integral, you almost had it. What you are after is: $$f_U(u) = \left(\frac{\mathrm d~~}{\mathrm d u} \int_{-\surd u}^{+\surd u} \frac{ 3 y^2}2 \mathrm dy\right) ~\mathbf 1_{u\in[0;1]}$$
Now use the following fact to resolve the above without even integrating.$$\dfrac{\mathrm d~~}{\mathrm d u}\int_{h(u)}^{g(u)} k(y)~\mathrm d y = g'(u)~k(g(u))-h'(u)~k(h(u))$$
• Thank you for the reply! I apologize for being unclear. The density Y was indeed $\frac {3/2} \cdot y^2$ in my textbook. Also, thanks for the wonderful explanation! Commented Sep 16, 2016 at 3:36