show that if $X\ge 0$ , $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.

Question

if $X$ is a random variable and also let $X\ge 0$ , I want to show $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.

Answer 1

Show $X \le \sum_{n=0}^\infty I_{X > n}$. Note the right hand side is nothing more than $\text{ceil}(X)$, where "ceil" or "ceiling" means round up to the next integer.

Answer 2

Suppose the random variable $X$ takes values in $\left\{ 0,1,2,\ldots\right\} $ then:

$$E\left[X\right]=\sum_{k=1}^{\infty}kP\left\{ X=k\right\} $$

$$=\sum_{k=1}^{\infty}\sum_{n=1}^{k}P\left\{ X=k\right\} $$

$$=\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}P\left\{ X=k\right\} $$

$$=\sum_{n=1}^{\infty}P\left\{ X\geq n\right\}$$

$$=\sum_{n=0}^{\infty}P\left\{ X>n\right\} $$

So you even have $=$ instead of $\leq$.