Finding the density for $\min\{X, Y\}$ 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.
 A: Note that
$$
\begin{split}
F_Z(z) &= \mathbb{P}[Z \le z] \\
       &= 1 - \mathbb{P}[Z > z] \\
       &= 1 - \mathbb{P}[\min\{X,Y\} > z] \\
       &= 1 - \mathbb{P}[X > z, Y > z] \quad \text{now apply independence}\\
       &= 1 - (1-F_X(z))(1- F_Y(z)) \\
       &= F_X(z) + F_Y(z) - F_X(z)F_Y(z).
\end{split}
$$
Can you finish it?
A: Consider $\mathbb{P}(Z > z) = \mathbb{P}(\min\{X, Y\} > z)$.
If $\min\{X, Y\} > z$, it follows that $X > z$ and $Y > z$. 
[This hopefully isn't too difficult to see! If this doesn't make sense to you, grab two numbers. Choose the smallest one. Find a number that this number is greater than (say $k$). Then the other number should be greater than $k$ as well! 
E.g., suppose I have two numbers: $2$ and $4$. I grab the number $2$ since it is the smallest. $2$ is greater than $1$, for example. $4$ should be greater than $1$ too.]
Hence, 
$$\mathbb{P}(\min\{X, Y\} > z) = \mathbb{P}(X > z \text{ and } Y > z) = \mathbb{P}(X > z)\mathbb{P}(Y > z)$$
by independence. 
Now $X$ and $Y$ are identically distributed, so $$\mathbb{P}(X > z) = \mathbb{P}(Y > z) = \int_{z}^{1}1\text{ d}x = 1-z\text{, } z \in [0, 1]\text{.}$$
This gives
$$\mathbb{P}(Z > z) = (1-z)^2\text{, } z \in [0, 1]\text{.}$$
The CDF is then
$$\mathbb{P}(Z \leq z) = 1 - \mathbb{P}(Z > z) = 1-(1-z)^2\text{, } z \in [0, 1]$$
with value $0$ if $z < 0$ and $1$ if $z > 1$.
This has derivative
$$f_{Z}(z) = -2(1-z)(-1) = 2(1-z)\text{, } z \in [0, 1]$$
and $0$ elsewhere.
A: $$F_X(x) = \mathbb{P}[X \le x]$$
$$ = \int_{-\infty}^{x} 1_{[0,1]}(t) dt$$
If $x \in [0,1]$, then we have
$$F_X(x) = \int_{-\infty}^{0} 1_{[0,1]}(t) dt + \int_{0}^{x} 1_{[0,1]}(t) dt$$
$$ = \int_{-\infty}^{0} 0 dt + \int_{0}^{x} 1 dt$$
$$ = 0 + (x - 0) = x$$
If $x > 1$, then we have
$$F_X(x) = \int_{-\infty}^{0} 1_{[0,1]}(t) dt + \int_{0}^{x} 1_{[0,1]}(t) dt$$
$$ = \int_{-\infty}^{0} 0 dt + \int_{0}^{1} 1_{[0,1]}(t) dt + \int_{1}^{x} 1_{[0,1]}(t) dt$$
$$ = 0 + \int_{0}^{1} 1 dt + \int_{1}^{x} 0 dt$$
$$ = 1 (1-0) = 1$$
If $x < 0$, then we have
$$F_X(x) = \int_{-\infty}^{x} 1_{[0,1]}(t) dt$$
$$ = \int_{-\infty}^{x} 0 dt$$
$$ = 0$$
Hence,
$$\mathbb{P}[X \le x] = \max\{\min\{x,1\},0\}$$

$$F_Z(z) = \mathbb{P}[Z \le z]$$
$$= 1 - \mathbb{P}[Z > z]$$
$$=1 - \mathbb{P}[\min\{X,Y\} > z]$$
$$=1 - \mathbb{P}[X > z, Y > z]$$
$$=1 - \mathbb{P}[X > z] \mathbb{P}[Y > z] \ \text{by independence}$$
$$=1 - (1 - \mathbb{P}[X \le z]) (1 - \mathbb{P}[Y \le z])$$
$$=1 - (1 - \mathbb{P}[X \le z] - \mathbb{P}[Y \le z] + \mathbb{P}[X \le z]\mathbb{P}[Y \le z])$$
$$= \mathbb{P}[X \le z] + \mathbb{P}[Y \le z] - \mathbb{P}[X \le z]\mathbb{P}[Y \le z]$$
$$= \max\{\min\{z,1\},0\} + \max\{\min\{z,1\},0\} - \max\{\min\{z,1\},0\}\max\{\min\{z,1\},0\}$$
$$= 2(\max\{\min\{z,1\},0\}) - (\max\{\min\{z,1\},0\})^2$$
For $z \in [0,1]$, we have
$$F_Z(z) = 2(\max\{\min\{z,1\},0\}) - (\max\{\min\{z,1\},0\})^2$$
$$ = 2(z) - (z)^2$$
$$\to f_Z(z) = 2 - 2z$$
For $z \notin [0,1]$, we have
$$F_Z(z) = 2(\max\{\min\{z,1\},0\}) - (\max\{\min\{z,1\},0\})^2$$
$$ = 2(0) - (0)^2$$
$$ = 0$$
$$\to f_Z(z) = 0$$
A: $ 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 $.
A: For sake of completion I will share this approach:
Let be $Z:=\min\{X,Y\}$. Then,
\begin{align*}
&P(Z\leq z)=P(\{X\leq z\}\cup \{Y\leq z\})\\
&=P(\{X\leq z\})+P(\{Y\leq z\})-P(\{X\leq z\}\cap \{Y\leq z\}), \text{ by inclusion/exlcusion principle}\\
&=P(\{X\leq z\})+P(\{Y\leq z\})-P(\{X\leq z\}\cdot P( \{Y\leq z\}), \text{ by independence}\\
&=F_X(z)+F_Y(z)-F_X(z)F_Y(z).
\end{align*}
If $f_X,f_Y$ are the density and $F_X(z),F_Y(z)$ the associated distributions functions, then taking the derivate yields:
$$
f_X(z)+f_Y(z)-f_X(z)F_Y(z)-F_X(z)f_Y(z).
$$
