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

If $X$ is a random variable and also let $X\ge 0$.

I want to show $E(X)\le \sum_{n=0}^{\infty}P(X>n)$.

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 $$\mathrm{E}[X]\leq \sum_{n=0}^\infty \mathrm{E}[I_{A_n} (X)]=\sum_{n=0}^\infty P(X>n) \, .$$

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.

• Unless I've missed something, I believe $dF(x)=-d(1-F(x))$. However, that's what you need for the integration by parts approach. – Glen_b -Reinstate Monica Nov 28 '13 at 9:04

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?

• in this question do not mention $X$ is discrete or not – peter Nov 28 '13 at 7:49
• I believe you can use something very like the above approach to establish the inequality in the general case. See the link I gave (that's why I said 'more generally') and think about how to construct something similar to the above for the general case. Can you establish a relationship between $\sum_{n = 0}^{\infty} P(X > n)$ and a similar-looking integral? – Glen_b -Reinstate Monica Nov 28 '13 at 8:54
• It's not necessary to specify the support for this problem. The non-strict inequality allows for any general RV to be computed in such a fashion if the prover is comfortable with using the general measure-theoretic integral. – AdamO Nov 28 '13 at 17:12
• @AdamO Yes, I understood that this question can be done generally, but being a set problem there are two issues one must keep in mind: (i) posters of such problems are working at a variety of levels, and (ii) unfortunately often, important details in the question are left out when it comes to posting time. It's often necessary to make sure one isn't solving a much harder problem than was being asked (for risk of the solution being beyond them). There were aspects of the way the question was originally worded that made me think the real question could be more basic than the general case...(ctd) – Glen_b -Reinstate Monica Nov 28 '13 at 21:49
• (ctd)... but even if I had not understood that it could be done in the general case (though I mention the general case several times above), mpiktas' solution would have immediately convinced me otherwise. – Glen_b -Reinstate Monica Nov 28 '13 at 21:50