Suppose that we're given four independent random variables $X_1,X_2,X_3$ and $X_4$ and their probability density function is given by:
$f(x)= 3(1-x)^2 $ for $0<x<1$ and otherwise $f(x)=0$.
If $Y$ is the minimum of $X_1,X_2,X_3,X_4$, how can I find the P.D.F and C.D.F of $Y$?
Well, I guess because all random variables are independent we have:
$f(x_1,x_2,x_3,x_4)=81(1-x_1)^2(1-x_2)^2(1-x_3)^2(1-x_4)^2$
I first wanted to say that since $f(x_1,x_2,x_3,x_4)$ is symmetric with respect to $x_1,x_2,x_3,x_4$ I can assume without loss of generality that $0<x_1\leq x_2 \leq x_3 \leq x_4<1$ but then I think it won't help me since it ruins the independence of the random variables.
I don't know what I should do to find the P.D.F of $Y=\min\{X_1,X_2,X_3,X_4\}$. I'm not looking for a full solution at this step, only a hint would be enough. I know how to solve this problem for two variables by considering the cases $X_1 \leq X_2$ and $X_1 > X_2$ separately, but I have no ideas how I can solve this for four variables.