0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.