why CCDF of x is equal to expectd value of probability of x given y

291 Views Asked by At

I found in an article a fact that is commonly used in scientific papers which mentions that:

$\mathbb{P}(f(r)>T) = \mathbb{E}[\mathbb{P}(f(r)>T\,|\,r)]$

($\mathbb{E}$ is with respect to $r$)

That means the CCDF of $f(r)$ is equal to the expectation ,with respect to $r$, of the probability of $f(r)$ greater than $T$ given $r$.

I tried to prove this fact but without any success. Can someone help me to prove that? Is it true?

Thanks.

1

There are 1 best solutions below

1
On

By relying on the tower property of conditional expectation, and the fact that the expectation $\mathbb{E}[\mathbb{1}_{\{A\}}]$ of the indicator function is just the probability $\mathbb{P}(A)$ of the event on which the indicator function is defined, you have the following chain of equalities: $$ \mathbb{P}[f(r)>T]= \mathbb{E}\left[\mathbb{1}_{\{f(r)>T\}}\right]=\mathbb{E}\left[\mathbb{E}[\mathbb{1}_{\{f(r)>T\}}|\,r]\right]= \mathbb{E} \left[ \mathbb{P}(f(r)>T\,|\,r) \right].$$