# 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.

• Because for instance the probability that $\min(X,Y)\leq \frac{1}{2}$ is the probability that both $X$ and $Y$ are at most $\frac{1}{2}$, which is $\frac{1}{2}\cdot \frac{1}{2}= \frac{1}{4}$ by independence; not $\frac{1}{2}$ as your answer would give. (Also, I suggest you use the search features of the website, this question or variants have been asked numerous times.) – Clement C. Jun 7 '16 at 16:40

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.

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?

• Suggestion: You may want to add a note where you use the independence. – Clement C. Jun 7 '16 at 16:42
• @ClementC. done... – gt6989b Jun 7 '16 at 16:46
• How did you get to the last line? – BCLC Jun 16 '16 at 19:38
• @ClementC. What about the last line? – BCLC Jun 16 '16 at 19:38
• By independence, and there is a typo in the current last line: $$\mathbb{P}\{X>z, Y>z\} = \mathbb{P}\{X>z\}\mathbb{P}\{ Y>z\} = (1-\mathbb{P}\{X\leq z\})(1-\mathbb{P}\{ Y\leq z\}) = (1-F_X(z))(1-F_Y(z)).$$ – Clement C. Jun 16 '16 at 19:42

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

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

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

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

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

$$= 0$$

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

$F_Z(z) = \Pr(Z\leq z) = \Pr( \{(x,y) \in [0,1]\times[[0,1] ; min(x,y) \leq z\})$ $$= \int\int_{A_z} f(x,y) dy dx$$ 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 $0 \leq z \leq 1$.