Distribution of Max of 2 Uniforms with different support

Question

I've got 2 independent draws from these two distributions :$X\sim U(0,1)$ and $Y\sim U(0,2)$.

I want to find $E(\max(X_,Y))$.

I know that for two (0,1) independent Uniforms:

$P(\max(X,Y)<z)=P(X<z)P(Y<z)=z^2$

However now when the support changes I'm struggling to figure it out. I thought maybe splitting it over the 2 supports because when z leaves X's support the CDf is 1. I tried to do a conditional thing like: $P(\max(X,Y)<z |z<1)$ but then I'd have to find P(z<1).

Context: Mechanism design problem with 2 bidders in a first price sealed bid auction so I wanted to calculate expected revenue.

Answer 1

Let $Z=\max(X,Y)$. It's always true by independence that
$$ F_Z(z) = P(\max(X,Y) \le z) = P(X \le z)P(Y \le z). $$

If $0 \le z \le 1$, then $P(\max(X,Y) \le z) = z^2/2$. If $1 < z \le 2$ then $P(\max(X,Y) \le z) = z/2$.

You get the density by taking the derivative. The density is $f_Z(z) = z 1(0 \le z \le 1) + 1(1 < z \le 2)/2$. Then

$$ E[\max(X,Y)] = \int_0^1 z^2dz + \int_1^2 z/2 dz = \frac{1}{3} + 1 - \frac{1}{4} $$

We can check this by taking the simulating data and taking the sample average with some R code. It should be close to our answer by the law of large numbers.

num <- 1000
x <- runif(num,0,1)
y <- runif(num,0,2)
df <- cbind(x,y)
mean(apply(df, 1, max))  # [1] 1.08731