Let $(Ω, F, P)$ be a probability space, and $X, Y : Ω → R$ be random variables.
let $ω → M(ω) := max\{X(ω), Y (ω)\}$.
I guess it is easy to show that $M(\omega)$ is a random variables.
$M(ω) := max\{X(ω), Y (ω)\}=\{X\le x\}\cap\{Y\le y\}$
Now assume $X, Y$ are independent and uniform random variables, for
all Borel subsets B of the unit square $[0, 1]^2$
, $P((X, Y ) ∈ B) = λ(B)$, where $λ$ is the
continuous uniform distribution on $[0, 1]^2$
. Determine the cumulative distribution function
as associated to the probability measure $B → P(M ∈ B), B ∈ \mathcal{B}(R)$.
I didn't quite get the question. Could some give me some hints?
I thought the result would be CDF $P(M\le m)=\lambda ([-\infty,m]\times[-\infty,m])$.
However, it seems way too easy. I notice that in the question it emphase that $\lambda$ is a distribution over $[0,1]^2$ instead of $R^2$. Why is that?
Thanks in advance
Best Answer
There is no such thing as uniform distribution on the whole of $\mathbb R^{2}$. It is given that $P(((X,Y ) \in B) $ is the 'area' or the two dimensional Lebesgue measure of $B$. The correct answer is $P(M \leq m)=\lambda ([0,m]\times [0,m])=m^{2}$.