# How to find joint distribution of $\min(U_0,U_1)$ and $\min(U_1,U_2)$ where $(U_0,U_1,U_2)$ are i.i.d Uniform?

I have this homework question where there are 3 random variables $$(U_0,U_1,U_2)$$ which are independent and uniform in the interval $$[-1,1]$$.

I have two other random variables $$(X,Y)$$ defined as follows: $$X=\min(U_0,U_1)\quad,\quad Y=\min(U_1,U_2)$$

I am asked to find joint pdf of $$X$$ and $$Y$$.

I have some rough ideas but any hint to suggest how to solve this is very appreciated. Thanks a lot in advance.

For $$X$$, I calculated $$P(X=x)$$ as follows: $$P(X=x)=P(U_0>x|U_1=x) \,\,\, \cup \,\,\, P(U_1>x|U_0=x)$$ and since these belong to different probability spaces and $$U_0$$ and $$U_1$$ are independent: $$P(X=x)=P(U_0>x)+P(U_1>x)$$

Is it going ok so far?

• As a self-study question, it is supposed to come with further details on what you tried and where you get stuck. – Xi'an Nov 18 '18 at 16:50
• "Find the joint pdf of $X$ and $Y$" is an invitation to compute $\Pr(X\le x\text{ and }Y\le y)$ for arbitrary numbers $x$ and $y.$ – whuber Nov 18 '18 at 17:58
• Your approach is unlikely to be successful, because uniform distributions are continuous. This implies both $X$ and $Y$ have continuous distributions, whence it is the case that $P(X=x)$ is zero no matter what the value of $x$ may be. – whuber Nov 18 '18 at 19:38

## 1 Answer

Here is a hint for a possible algebraic approach:

The idea is to use the fact that for any $$k$$,$$U_0,U_1,U_2>k \iff \min(U_0,U_1,U_2)>k$$

Note that, along the lines of $$P(A\cap B^c)=P(A)-P(A\cap B)$$, we can write

$$P(Y>y,X\le x)=P(Y>y)-P(Y>y,X>x)\tag{1}$$

Find the set of admissible values of $$(x,y)$$, i.e. the support of $$(X,Y)$$.

Then again using the previous idea, we have for all $$(x,y)$$,

\begin{align} P(X\le x,Y\le y)&=P(X\le x)-P(X\le x,Y>y) \\&=1-P(X>x)-P(Y>y)+P(X>x,Y>y)\qquad,\text{ using }(1) \\&=1-P(U_0,U_1>x)-P(U_1,U_2>y)+P(U_0,U_1>x\,,\,U_1,U_2>y) \\&=1-P(U_0>x,U_1>x)-P(U_1>y,U_2>y) \\&\quad +P(U_0>x,U_1>\max(x,y),U_2>y) \end{align}

Simplifying the above would give you the joint distribution function of $$(X,Y)$$.

The joint density is then $$f_{X,Y}(x,y)=\frac{\partial^2 }{\partial x\partial y}P(X\le x,Y\le y)\quad,\text{ for all }(x,y)$$