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Let $X$ be a random variable with positive integer range and finite mean. To show that $$E[X]= \sum_{k =0}^{\infty} P(X>k).$$

Proof: I showed using induction that $$\sum_{k=0}^n P(X>k) = \sum_{t=1}^n (t \cdot P(X=t)) + (n+1)P(X>n)$$ If we apply $n \to \infty$, we would get the required result if if we show that as $n \to \infty$, $(n+1)P(X>n)$ tends to $0$. I'm having difficulty proving that.

I know that $n \cdot P(X=n)$ tends to $0$ as $n \to \infty$ because $X$ has a finite mean.

(PS: I do know this question is duplicate but in the previous question I didn't find any proof resembling this. Also to show that $(n+1)P(X>n)$ tends to 0 can be a separate question by itself.)

Lee David Chung Lin
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user385655
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  • For future questions please check the TeX-tutorial. – saz Dec 04 '16 at 07:34
  • You are asking why $(n+1)P(X=n)\to0$ when $X$ is nonnegative and integer valued and $E(X)$ is finite. This is not at all what the title of the question says, so you should modify it and you should make clear from the start of the question that this is what you are asking, and not a proof of the formula for $E(X)$. Re this question itself, note that $(n+1)P(X=n)\leqslant2nP(X=n)$ and that you say you know that $nP(X=n)\to0$, thus... – Did Dec 04 '16 at 10:43
  • Even if you hurriedly accepted an answer after my comment was posted, please address the points in it. – Did Dec 04 '16 at 10:48
  • @Did Am I missing something? The OP knows that $n P(X=n) \to 0$, but (s)he has to prove that $n P(X>n) \to 0$... why do you think that the estimate $(n+1) P(X = n) \leq 2n P(X=n)$ is helpful to establish this convergence....? (However, I totally agree with the first part of your comment.) – saz Dec 04 '16 at 12:45
  • @saz Ah, you are right, naturally. – Did Dec 04 '16 at 13:58
  • Even though this post is slightly different, I’d like to link it to the current choice of mother post. Also see the meta post for (abstract) duplicates. – Lee David Chung Lin Nov 13 '18 at 13:30

2 Answers2

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Note that

$$n \mathbb{P}(X>n) = n \int_{\{X>n\}} \, d\mathbb{P} \leq n \int_{\{X>n\}} \frac{X}{n} \, d\mathbb{P}.$$

Since $X$ is intgrable, we can apply the dominated convergence theorem to conlude

$$\lim_{n \to \infty} (n \mathbb{P}(X>n)) \leq \lim_{n \to \infty} \int_{\{X>n\}} X \, d\mathbb{P}=0.$$

Finally, using that

$$\lim_{n \to \infty} \frac{n+1}{n} = 1$$

we get

$$\lim_{n \to \infty} (n+1) \mathbb{P}(X>n)=0.$$

Remark: The above proof does not only work for integer-valued integrable random variables, but for any non-negative integrable random variable. For the particular case of integer-valued random variables, the proof can be reformulated as follows:

Note that

$$n \mathbb{P}(X>n) = n \sum_{k=n+1}^{\infty} \mathbb{P}(X=k) \leq n \sum_{k=n+1}^{\infty} \frac{k}{n} \mathbb{P}(X=k).$$

Since the expectation $\mathbb{E}(X) = \sum_{k=0}^{\infty} k \mathbb{P}(X=k)$ is finite, we get

$$\lim_{n \to \infty} n \mathbb{P}(X>n) \leq \lim_{n \to \infty} \sum_{k=n+1}^{\infty} k \mathbb{P}(X=k)=0.$$

saz
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\begin{align} E[X] &= \sum_{k=1}^\infty k\cdot P(X=k)\\ &= \quad P(X=1)\\ &\qquad + P(X=2) + P(X=2)\\ &\qquad + P(X=3) + P(X=3) + P(X=3)\\ &\qquad + P(X=4) + P(X=4) + P(X=4) + P(X=4)\\ &\qquad\vdots\\ &= P(X > 0) + P(X >1) + P(X>2) + P(X >3) + \cdots & \text{add each column}\\ &= \sum_{k=1}^\infty P(X \ge k). \end{align}

angryavian
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