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 <n+1)=\sum\limits_{n=0}^{\infty} n (b_n-b_{n+1})=1(b_1-b_2)+2(b_2-b_3)+...=b_1+b_2+b_3+...= \sum\limits_{n=1}^{\infty} P(X \geq n)$.


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