two independent random variables and their conditional probability

Question

I am facing trouble trying to verify a seemingly simple statement:

Let $X, U$ be independent variables. Then, $P(X\geq U \mid U) = P(X \geq U)$.

Is this statement true? Thanks in advance.

Answer 1

The proposed statement is not true.

Suppose $\Pr(X=0) = 1/2 = \Pr(X=1)$ and $\Pr(U=0) = 1/2 = \Pr(U=1)$ and $X,U$ are independent.

Then $\Pr(X\ge U \mid U=0)=1$ and $\Pr(X\ge U \mid U=1) = 1/2.$

Thus $\Pr(X\ge U\mid U) = \begin{cases} 1 & \text{if } U=0, \\ 1/2 & \text{if } U =1. \end{cases}$

This conditional probability $\Pr(X\ge U\mid U)$ is a random variable that is completely determined by the value of $U,$ and is not constant, but equal to $1$ or to $1/2,$ each with probability $1/2.$

On the other hand, $\Pr(X\ge U) = 3/4$ since the event $X\ge U$ occurs in any of three of the four equally probable outcomes, which are $(X,U) = (0,0),\,(1,0),\, (0,1),\, (1,1).$

So the result is false.

One the other hand, it is true that $\operatorname E(\Pr(X\ge U\mid U)) = \Pr(X\ge U).$ And if $A$ is any event at all, then $\operatorname E(\Pr(A\mid U)) = \Pr(A),$ i.e. the prior expected value of the posterior probability is equal to the prior probability. That is the law of total probability.