# Expected value inequality for non-negative random variable

Let $$X$$ be a non-negative random variable. Using the fact the next inequality holds: $$n\ \mathbb{I}_{n\leq X < n+1} \leq X\ \mathbb{I}_{n\leq X < n+1} \leq (n+1)\ \mathbb{I}_{n\leq X < n+1}.$$

I need to prove the following inequality:

$$\sum_{n=1}^{\infty} \mathbb{P}\left(X \geq n\right) \leq \mathbb{E}(X) \leq 1 + \sum_{n=1}^{\infty} \mathbb{P}\left(X\geq n\right).$$

My attempt:

Let $$n \in \mathbb{N}$$, then, using the first inequality and applying $$\mathbb{E}$$, then: $$\mathbb{E}[n\ \mathbb{I}_{n\leq X < n+1}] \leq \mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \mathbb{E}[(n+1)\ \mathbb{I}_{n\leq X < n+1}].$$

Then, summing over $$n$$ then $$\sum_{n=0}^{\infty}\mathbb{E}[n\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}\mathbb{E}[(n+1)\ \mathbb{I}_{n\leq X < n+1}]$$

Using expected value properties, it holds $$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}(n+1)\ \mathbb{P}[n\leq X < n+1]$$

Then $$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] \leq \sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] \leq \sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] + \sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1].$$

Here's the deal:

• For $$\sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}]$$ (my) intuition says that because we're summing over all possible values of $$n$$, and the indicator function is considering the set of $$\{n\leq X < n+1\}$$, then we're considering all the half-open intervals $$[n,n+1)$$, since they're disjoint, the union of all of them gives us $$[0,\infty)$$ and so, since $$X$$ is a non-negative r.v., then $$\sum_{n=0}^{\infty}\mathbb{E}[X\ \mathbb{I}_{n\leq X < n+1}] = \mathbb{E}[X]$$.

• For $$\sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1]$$, again, since we're considering all the posible values of $$n$$ and we're looking for $$n\leq X < n+1$$, then essentially we're looking for the probability of $$X \in [0,\infty)$$, since $$X$$ non-negative, then $$\sum_{n=0}^{\infty} \mathbb{P}[n\leq X < n+1] = 1$$.

• Up to this point, if my intuition in the last two arguments is correct, all that remains is to prove $$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] = \sum_{n=1}^{\infty}\mathbb{P}[X\geq n]$$ but here's where I'm having trouble. I know that, if we're talking about discrete r.v. then: $$\sum_{n=1}^{\infty}\mathbb{P}[X\geq n] = \mathbb{P}[X=1] + 2\mathbb{P}[X=2] + \ldots = \sum_{n=1}^{\infty}n\ \mathbb{P}[X=n]$$ and in this case $$\{n\leq X < n+1\}$$ reduces to $$\{X=n\}$$, and so, $$\sum_{n=0}^{\infty}n\ \mathbb{P}[n\leq X < n+1] = \sum_{n=1}^{\infty}n\ \mathbb{P}[X= n] = \sum_{n=1}^{\infty}\mathbb{P}[X\geq n]$$ and the proof will be done. But the problem is that, the only hypothesis I have for $$X$$ is that it is non-negative, so I would make a mistake considering $$X$$ discrete r.v.

I'm lost by this point to prove it in general, and I don't know if I'm missing any other property of expected value that could be useful. Any help would be appreciated.

Let $$b_n=P(X \geq n)$$. Then $$\sum\limits_{n=0}^{\infty} nP(n \leq X .