Finding the density for $\min\{X, Y\}$

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

$$F_X(x) = \mathbb{P}[X \le x]$$

$$ = \int_{-\infty}^{x} 1_{[0,1]}(t) dt$$

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

$$F_Z(z) = 2(\max\{\min\{z,1\},0\}) - (\max\{\min\{z,1\},0\})^2$$

$$ = 2(0) - (0)^2$$

$$ = 0$$

$$\to f_Z(z) = 0$$

Answer 4

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:

$$ f_X(z)+f_Y(z)-f_X(z)F_Y(z)-F_X(z)f_Y(z). $$