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.
Best Answer
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.