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

Define the sets $A_n=\{x\in \mathbb{R}:x>n\}$, for $n=0,1,2\dots$.

For any fixed $\omega$, let $n_0$ be the smallest integer such that $X(\omega)\leq n_0$. Since $X(\omega)\geq 0$, we have $$ X(\omega)\leq n_0 = \sum_{n=0}^{n_0} I_{A_n} (X(\omega)) = \sum_{n=0}^\infty I_{A_n} (X(\omega)) \, , $$ yielding $$ \mathrm{E}[X]\leq \sum_{n=0}^\infty \mathrm{E}[I_{A_n} (X)]=\sum_{n=0}^\infty P(X>n) \, . $$

Answer 2

You have

$$EX=\int_0^{\infty}xdF(x)$$

Notice that $dF(x)=-d(1-F(x))$ and that $P(X>t)=1-F(t)$ and use integration by parts.

Now show that for monotone decreasing positive function

$$\sum_{n=0}^\infty f(n)\ge\int_0^{\infty} f(t) dt$$

Combine these two results and you get your desired result. Hint for the second, recall Riemman sums.

Answer 3

You don't specify anything about $X$. Is it the general case/what is its support?

If it is for discrete r.v.s, can you say something about the relationship between

$\sum_{n = 0}^{\infty} n P(X = n)$ and $\sum_{n = 0}^{\infty} P(X > n)$?

Consider:

\begin{eqnarray} &0 P(0)& +\, 1& P(1)& +\, 2 &P(2)& +\, 3 &P(3)& +& ...&\\ & &\\ &= & [ &P(1)& +&P(2)& +&P(3)& +& ...&]\\ & & [ & & +&P(2)& +&P(3)& + &...&]\\ & & [ & & & & +&P(3)& + &...&]\\ & & [& & & & & & &...&] \end{eqnarray}

Can you see a way to do it now? (Though I believe this approach establishes a stronger result than you have)

To consider it more generally than the discrete case, see here, and then adapt the above idea.

That is, can you see how establish a relationship between $∑^∞_{n=0}P(X>n)$ and a similar-looking integral that would the give the required inequality?