Find the density of Y = min (sqrt(X), 2-sqrt(X)) where X is a continuous random variable

Question

Find the density of Y = min (sqrt[X], 2-sqrt[X]) where X is a continuous random variable with density f(x) = x^3/64, 0<=x<=4, f(x)=0 elsewhere. I just want someone to point me in the right direction of how to find the minimum mentioned above, so I can find the density of Y.

Answer 1

It holds for the density $f_Y$ of Y

$$\begin{align*}f_Y(y) &= \left(P(Y \le y)\right)' \\ &= \left(1 - P(Y > y)\right)' \\ &= -\left(P(\sqrt{X} > y, 2-\sqrt{X} > y)\right)' \\ &= -\left(P(y \le \sqrt{X} \le 2 - y)\right)' \\ &= -\left(F_X((2-y)^2) - F_X(y ^2)\right)'\end{align*}$$

Calc the derivative using $F'_X = f$ and plug in you can do by yourself…