TITLE: Finding the density for $\min\{X, Y\}$
QUESTION [3 upvotes]: Problem: Let $X$ and $Y$ be independent and suppose that each has a $\text{Uniform}(0,1)$ distribution. Let $Z = \min\{X, Y\}$. Find the density $f_Z(z)$ for $Z$. Hint: It might be easier to first find $\mathbb{P}(Z > z)$.
Attempted Solution:
Given that $X, Y \sim \text{Uniform}(0,1)$, how do we not just have the following?
$$
f_Z(t) = f_X(t) = f_Y(t) = \begin{cases} 1 & \text{if } 0 \le t \le 1 \\
                                          0 & \text{otherwise}
                            \end{cases}
$$
Of course I'm highly suspicious of this answer because it's not making use of the fact that $X$ and $Y$ are independent, nor is it making use of the provided hint.

REPLY [0 votes]: $ F_Z(z) = \Pr(Z\leq z) = \Pr( \{(x,y) \in [0,1]\times[[0,1] ; min(x,y) \leq z\})$
$$ = \int\int_{A_z} f(x,y) dy dx $$
For each $z$ value, $ A_z = \{(x,y) \in [0,1]\times[[0,1] ; min(x,y) \leq z\} $ is the collection of $(x,y)$ where the minimum of the 2 is less than $z$. 
There are two cases that can exist; either $x<y$ or $y<x$. Since $X$ and $Y$ are continuous, we can ignore the case where $x=y$. We can write $A_z$ as the following union.
$$ A_z = \{(x,y); x\leq z, x<y \} \cup \{(x,y); y \leq z , y<x \} $$
These two sets are disjoint and the probability of disjoint sets is the sum of the probabilities.
$$ Pr(A_z) = Pr( \{(x,y); x\leq z, x<y \}) + Pr(\{(x,y); y \leq z , y<x \}) $$
Because $X$ and $Y$ are independent;
$$ f(x,y) = f(x)f(y) = 1*1 = 1 $$
Thus, we have that,
$$ F_Z(z) = \int\int_{A_{z_1}} 1 dy dx + \int\int_{A_{z_2}} 1 dxdy $$
Before we start integrating, we need to figure out the bounds for the integrals. For the integral on the left: 

If $x<y$, then the range of $y$ has to be between $x$ and $1$.
Since we are concerned about the $x$ values for which $x<z$, and because $z$ ranges from $0$ to $1$, then $x$ has to range between $0$ and $z$. 

It's the same thought process for the second integral. Therefore, we have that:
$$ F_Z(z) = \int_0^{z}\int_x^{1} 1 dydx +  \int_0^{z}\int_y^{1} 1 dxdy  $$
$$ = 2z - z^2 $$
Finally
$$ f_Z(z) = F'_Z(z) = 2(1-z) $$
if $0 \leq z \leq 1 $.
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