Compute probability using indicator function

Question

I am proofing the Tail Probability Expectation Formula, I have some idea but I don't know it is correct or not. $$ E[X] = \sum_{n=1}^{\infty}P\{X \geq n\}$$ $$$$ First I create an indicator function. $$ 1_{\{X>n\}}= \begin{cases} 1 & ,x>n \\ 0 & ,otherwise \end{cases} $$

Then,

$$E[X] = E[E[X|1_{\{X>n\}}]]$$ $$ E[X] = E[P\{X>n\}]$$ Since X is non-negative, we have $n>0$ (I am wondering it is correct or not). We have... $$E[X] = \sum_{n=0}^{\infty}P\{X>n\}$$

Then increment the index by 1 since $n>0$. $$ E[X] = \sum_{n=1}^{\infty}P\{X \geq n\}$$ Hence as desired.

Answer 1

Hmm, I don't see how you go from your first $E(X)$ to your second... $P(X > n) = E(1_{X > n})$.

Here's a hint for how to do it. Recall that $$ E(X) = \sum_{n=1}^\infty n P(X = n). $$ So you have one lot of $P(X = 1)$, two of $P(X = 2)$, three of $P(X = 3)$, etc. Imagine these 'boxes'. Each number $n$ represents $P(X = n)$

123456789...
 23456789...
  3456789...
   456789...
    56789...
 ... ... ...

There is one 1 because you count $P(X = 1)$ once; there are two 2s because you count $P(X = 2)$ twice. If you add up all of these, you get $E(X)$. The formula $$ E(X) = \sum_{n=1}^\infty n P(X = n) $$ corresponds to adding up column by column: first $n = 1$ does the one term in te first column, then $n = 2$ does the two terms in the second column, etc.

Instead, try to add up all the elements a row at a time. Your first term is the entire first row: $$ P(X = 1) + P(X = 2) + P(X = 3) + \cdots = P(X \ge 1). $$ Your second term is the second row: $$ P(X = 2) + P(X = 3) + P(X = 4) + \cdots = P(X \ge 2). $$ Can you show rigorously that the $n$-th term is $P(X \ge n)$? If so, once you sum up all the rows, you can the desired formula

Answer 2

The equation $E[X]=E[E[X\mid 1_{X\ge n}]]$ does not imply $E[X]=E[P\{X>n\}]$. However, here is a method possibly more in line with your original idea.

We start with the following deterministic equality, valid for any nonnegative integer $x$: $$ x=\sum_{n=1}^{\infty} {\bf 1}_{x\ge n} $$ In words, a nonengative integer is equal to the number of positive integers which are less than or equal to it.

Now, replace $x$ with the random variable $X$ to get $X=\sum_{n=1}^\infty 1_{X\ge n}$, and take the expected value of each side. Done!