Finding probability density function from an uniform variable

Question

I need help with this problem: Let $X \sim U(0, 1)$. Find the probability density function of $Y = X^2$ (it does also have for $W = \sqrt{X}$ and for $V = 1-X$, but I think that the idea must be the same).

So, I think that Y is a chi-square (but don't know how to prove) and for the others I have no idea at all, as the only thing given is an uniform variable.

Thanks in advance.

Answer 1

Actually, $X^2$ would be $\chi_1^2$ if we had $X\sim N(0,\,1)$, but that's not what we have here. The strategy you want is to find the CDF, then differentiate. For any $y$ in $Y$'s support $[0,\,1]$,$$P(Y\le y)=P(X\le\sqrt{y})=\sqrt{y}\implies f_Y(y)=\frac{1}{2\sqrt{y}}.$$You should be able to find the PDFs of $W,\,V$ the same way, but bear in mind $V$ is a decreasing function of $X$, so$$P(V\le v)=P(X\ge1-v)=1-P(X\le1-v).$$You can verify this subtlety ensures $f_V(v)\ge0$.