PDF and CDF of $\min\{X_1,X_2,X_3,X_4\}$ for $(X_k)$ i.i.d. with PDF $f(x)=3(1-x)^2$ on $(0,1)$ 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.
 A: Hint: express the event that $ Y\ge y $ in terms of the $ X_i $
A: The magical Dirac delta function is the "correct" way of handling such problems;-). Define $Y:= X_1 \wedge X_2 \wedge X_3 \wedge X_4$ and here you go:
\begin{eqnarray}
\rho_Y(x) &:=& \int\limits_{[0,1]^4} \delta \left( x - x_1 \wedge x_2 \wedge x_3 \wedge x_4  \right) \prod\limits_{j=1}^4 \rho_X(x_j) dx_j \\
&=& 4! \int\limits_{0 < x_1 < x_2 < x_3 < x_4 <1} \delta \left( x - x_1 \right) \rho_X(x_1) dx_1 \cdot \prod\limits_{j=1}^4 \rho_X(x_j) dx_j \\
&=& 4! \int\limits_0^1 \delta\left(x- x_1 \right) \rho_X(x_1) \cdot \underbrace{\left[
\frac{1}{6} - \frac{1}{2} F_X(x_1) + \frac{1}{2} F^2(x_1) - \frac{1}{6} F^3(x_1)
\right]}_{\int\limits_{x_1 < x_2 < x_3 < x_4 < 1} \prod\limits_{j=2}^4 \rho_X(x_j) dx_j} d x_1 \\
&=& 4 (1- F_X(x) )^3 \cdot \rho_X(x)
\end{eqnarray}
Now the only thing we need is to prove the result used in the third line from the top. We use $F_X(x)^{'} = \rho_X(x)$ and we  have:
\begin{eqnarray}
&&\int\limits_{x_1 < x_2 < x_3 < x_4 < 1} \prod\limits_{j=2}^4 \rho_X(x_j) dx_j= \\
&& \int\limits_{x_1 < x_2 < x_3 < 1} \rho_X(x_2) \rho_X(x_3) \left[1 - F_X(x_3) \right] dx_2 dx_3=\\
&& \int\limits_{x_1 < x_2 < 1} \rho_X(x_2) \cdot \left[ \frac{1}{2} - F_X(x_2) + \frac{1}{2} F_X^2(x_2)\right] dx_2= \\
&& \frac{1}{6} - \frac{1}{2} F_X(x_1) + \frac{1}{2} F_X^2(x_1) - \frac{1}{6} F_X^3(x_1)
\end{eqnarray}
as it should be.
A: Hint 1: $F_Y(y)=\mathbb{P}\{Y\leq y\}=\mathbb{P}\{min\{X_1,X_2,X_3,X_4\}\leq y\}$. Now think about when the event can happen. Clearly it happens if all four RVs are $\leq$ y. It also happens when three are $\leq y$ and one is not. Etc.
Hint 2: Events described above are mutually exclusive. Your$X_i$ are iid sampled from $F_X$
Hint 3: When trying to make sense of Hint 1 think of it in terms of "$n$ choose $k$"
Hint 4: This is straightforward but once you find $F_Y$ you just differentiate and use chain rule to find $f_Y$. 
I hope this helps and at the same time doesn't give it all away.
A: The CDF of Y can be written as $$\begin{split}
F_Y(y)=&\text{Pr}\left[\min_{k=1,\ldots,4}X_k\leq y\right]\\
=&1-\text{Pr}\left[\min_{k=1,\ldots,4}X_k\geq y\right]\\
=&1-\prod_{k=1}^4\text{Pr}\left[X_k\geq y\right]\\
=&1-\left(1-\text{Pr}\left[X_k\leq y\right]\right)^4\\
=&1-\left(1-F_X(y)\right)^4
\end{split}$$
All you need to find then is $F_X(y)$, and then find the PDF as $\frac{dF_Y(y)}{dy}$
