in some lecture notes, I stumbled upon the following formula : that if $X$ is a non-negative real-valued random variable then :

$\mathbb{E}[X] = \sum_{n \geq0} \mathbb{P}(X \geq n)$

I tried looking for proofs on MSE and elsewhere but only found for general cases which involve measure theory.

for this specific discrete case I tried the following :

$$\begin{align} & \sum_{n \geq0} \mathbb{P}(X \geq n) = \sum_{n \geq0}\sum_{k \geq n}\mathbb{P}(X =k )= \sum_{k \geq 0}\mathbb{P}(X =k ) + \sum_{k \geq 1}\mathbb{P}(X =k ) + \cdots \\ & \end{align}$$

if you expand each series term by term and align each same term one under the other, you can notice that there's :

one $\,\mathbb{P}(X =0)$ term, two $\,\mathbb{P}(X =1)$ terms, three $\,\mathbb{P}(X =2)$ terms and so on, in general we have $n+1$ $\,\mathbb{P}(X =n)$ terms

problem is If I apply same reasoning to the usual definition of the expectation then I find only $n$ $\,\mathbb{P}(X =n)$ terms

is the formula in the lecture notes wrong or is my reasoning wrong ? also I'd like a more formal proof where I can see the importance of the condition of positiveness.

thanks !


1 Answer 1


The formula is wrong. Th sum on the right has to start with $n=1$. Note that If $X=0$ then LHS $=0$ and RHS $=1$. You have got exactly $\sum nP\{X=n\}$ because there are $n$ terms each equal to $P\{X=n\}$ for $n=0,1,2...$. This proves the modified formula.

  • $\begingroup$ thing is put this way there's no $P\{X=0\}$ because $n = 0$ while there's one in the 2nd definition I'm trying to prove $\endgroup$ Nov 11, 2018 at 20:31
  • 1
    $\begingroup$ @rapidracim Sorry for noticing that the sum started with $n=0$. Please see my revised answer. $\endgroup$ Nov 11, 2018 at 23:19

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