This is from the book Fundamentals of Probability with Stochastic Processes by Saeed Ghahramani, pages 249-250 which asserts, for any random variable $X$ that is non-negative, expectation of $X$ is

$$ E(X)= \int_{0}^{\infty} [1-F(t)]dt= \int_{0}^{\infty} P(X>t) dt $$

Where $F(t)$ is a cumulative distribution function. Somehow this is equal to $\int_{0}^{\infty}x. P(X=x)dx$ and I don't see it.Though the proof is provided in the book, I find it lacking and wasn't completely satisfied with it.

I would like to see a proof of the discrete analog of the above expression

$$ E(X) = \sum_{x=0}^{\infty}x. P(X=x)= \sum_{x=0}^{\infty} [1- F(x)] = \sum_{x=0}^{\infty} \sum_{y=0}^{x} P(Y=y) $$

This is a very basic probability question.


1 Answer 1


The discrete case, assume that $X \ge 0$ takes non-negative integer values. Then we can write the expectation as $$ \DeclareMathOperator{\E}{\mathbb{E}} \DeclareMathOperator{\P}{\mathbb{P}} \E X = \sum_{k=0}^\infty k \P(X=k) $$ Now, we will first write this as a double sum, and then change the order of summation. Observe that $k = \sum_{j=0}^{k-1} 1$ (the case $k=0$ gives an upper limit that is smaller than the lower limit, we take that as the empty sum, which is zero). This gives $$ \E X = \sum_{k=0}^\infty \sum_{j=0}^{k-1} 1 \cdot \P(X=k) $$ Now, in this double sum we sum first on $j$, which clearly goes to $\infty$. Observe that in the inner summation the indices satisfy the inequality $$ 0 \le j \le k-1 $$ Solving that for $k$ gives $ k \ge j+1$, which then gives the limits of summation in the new inner sum: $$ \E X = \sum_{j=0}^\infty \sum_{k=j+1}^\infty \P(X=k) = \sum_{j=0}^\infty \P(X > j) $$ which is the result. The continuous case is similar.

  • 1
    $\begingroup$ +1 This technique is called summation by parts. $\endgroup$
    – whuber
    Commented Sep 26, 2017 at 20:12

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