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Let $X\ge 0$ be a random variable such that takes values in $\{1,2,...\}$. Show that: $$\Bbb E(X)=\sum_{n=1}^\infty \Bbb P(X\ge n)$$

So since, $$\Bbb E(X)=\sum_{n=1}^\infty x\ \Bbb P(X=x)$$ and we can rewrite $\Bbb P(X=x)= \Bbb P(X\ge x\ \cap\ X<x)$, but got stuck here since I don't know how to connect this with $\Bbb P(X\ge n)$. So I try to go backwards from $\Bbb P(X\ge n)$ but could get much:

$$\Bbb E(X)=\sum_{n=1}^\infty \Bbb P(X\ge n)=\sum_{n=1}^\infty (1-\Bbb P(X< n))$$

Any ideas how to proceed from this two ideas or any other approach would be appreciated.

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I think in the second equality you mean $$E[X] = \sum_{n=0}^\infty n P(X=n)$$We want to show this equals $$\sum_{n=0}^\infty P(X\ge n)$$

Let's fix a particular $n$, call it $n_0$. In the first sum, this will contribute with $n_0P(X=n_0)$. In the second, how many times will it be counted? Well since $$P(X \ge n) = P(X = n) + P(X = n+1 ) + \dots$$, we have that

$P(X \ge 1)$ will contain $P(X = n_0)$. But so will $P(X \ge 2)$ and so on, until $P(X \ge n_0)$. So how many terms? from $1$ to $n_0$ there are exactly $n_0$ terms.

So how many times has $P(X = n_0)$ been counted? Exactly $n_0$ times, so the contribute in the second sum is $n_0P(X = n_0)$, just like in the first sum.

Since this holds for any $n_0$, we have equality

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