A more advanced proof for showing $\mathbb{E}[X]=\int_{0}^{\infty}\mathbb{P}(X\geq t)dt$

Question

I want to show that for a non-negative random variable $X: \Omega \to [0,+\infty)$ in a probability space $(\Omega, \mathcal{F}, \mathbb{P})$:

$\mathbb{E}[X]=\int_{0}^{\infty}\mathbb{P}(X\geq t)dt$

I know that I can replace $\mathbb{P}(X\geq t)$ with $\mathbb{P}(X > t)$ and then I'd have:

$\int_{0}^{\infty} (1-F_{X}(x))dx = \int_{0}^{\infty}\int_{x}^{\infty} f_{X}(y)dydx = \int_{0}^{\infty}\int_{0}^{y}dx\: f_{X}(y)dy = \int_{0}^{\infty}yf_{X}(y)dy = \mathbb{E}[X]$

And I know that both sides of the equation can be $\infty$. I established the relation for X being an indicator function, and now I want to establish it for X as a simple non-negative function and then for non-negative measurable functions.

I'd appreciate any help.

If you have established the result for indicator functions then you can readily have it for simple functions since they are just sum of finitely many indicator functions. Now use the fact that simple functions are dense in $L^1$ and conclude the result by DCT. — mudok, Nov 01 '18 at 16:47
You cant use density function since it may not exist all the time. Write it as $\int_0^\infty\int_{X>t}dpdt$. Here $p$ is the probability measure. — mudok, Nov 01 '18 at 16:52
You may see Wiki's page for expectation -> alternative formula for expectation. You will find a proof there. — GNUSupporter 8964民主女神地下教會, Nov 01 '18 at 16:55
This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. — Lee David Chung Lin, Nov 13 '18 at 12:42

score 1 · Answer 1 · answered Nov 01 '18 at 16:57

Given the comment of @GnuSupporter, I will solve it for a continuous random variable $X$.

Certainly, if $X$ is a non-negative function the statement follows from linearity of expectation and from the fact that for disjoint $A_1 \dots A_n$ you have that $$\mathbb{P}(\sum_{i=1}^nc_i1_{A_i}\geq t)=\sum_{i=1}^n\mathbb{P}(c_i1_{A_i}\geq t)$$

For the rest, you have that there is an increasing sequence of simple positive random variables $X_n$ s.t. $\lim_{n \rightarrow \infty}X_n=X$ by the Structure Theorem.

We now use MCT repeatedly:

$$\mathbb{E}[X]=\mathbb{E}[\lim_{n \rightarrow \infty}X_n]=\lim_{n \rightarrow \infty}\mathbb{E}[X]=\lim_{n \rightarrow \infty }\int_0^{\infty}\mathbb{P}(X_n\geq t)dt=\int_0^{\infty}\lim_{n\rightarrow \infty}\mathbb{P}(X_n\geq t)dt$$

We've used the MCT to get the first and fourth equalities, noting that for $X_n\geq X_{n-1}$ we have $$\mathbb{P}(X_n \geq t) \geq \mathbb{P}(X_{n-1}\geq t)$$

Finally, a standard result in probability gives that for $(A_n)_{n \geq 1}$ increasing events we get $$\lim_{n \rightarrow \infty}\mathbb{P}(A_n)=\mathbb{P}(\lim_{n \rightarrow \infty}A_n)$$

Take $A_n=\{X_n\geq t\}$ to obtain the result

A more advanced proof for showing $\mathbb{E}[X]=\int_{0}^{\infty}\mathbb{P}(X\geq t)dt$

1 Answers1