# Intuition for expectation of discrete random variable that takes positive integers

If $$X$$ is a discrete random variable that takes values on the positive integers, it is true that

$$E(X) = \sum_{k=1}^{\infty} P(X \ge k)\;.$$

I know how to prove this (by expressing the summand as a sum of all probabilities going from $$n=k$$ to $$n=\infty$$ and then playing around with the double sums) but I'm having trouble developing an intuition for why the above result holds.

I am thinking of a number line where the equation says that if I'm stepping through $$k$$ from $$1$$ to $$\infty$$ and take the sum of probabilities of everything to the right of $$k$$ at each step, that gives the expectation of $$X$$, which does not make sense to me.

This can be flipped to be $$1 - \sum_{k=1}^{\infty} P(x \le k)$$, whose interpretation would then be stepping though $$k$$ but summing the probabilities of everything to the left of $$k$$ at each step, but this does not help me understand either.

The result is from exercise 4.7.8 in Rice's statistics book.

One way to see this is as follows:

For a discrete random variable $$X$$, we obviously have that

$$E[X] = \sum_{k=1}^\infty kP(X=k)$$

so our goal is to see how $$\sum_{k=1}^{\infty} P(X\geq k)$$ relates to that. Since $$X$$ is a discrete rv, we also have that

$$P(X>n) = \sum_{k=n}^{\infty} P(X=k)$$

which in words is saying that $$P(X>n)$$ is the same as just summing every individual $$P(X=k)$$ from $$n$$ onwards. Then if we were to sum all of the $$P(X>n)$$ from $$n=1$$ to $$\infty$$, we have

$$\sum_{n=1}^\infty P(X>n) = \sum_{n=1}^\infty \sum_{k=n}^{\infty} P(X=k)$$

Things get confusing with this double sum, so let's take a moment to think about it intuitively. When $$n=1$$, we have one $$P(X=k)$$ for every single $$k=1,\dots,\infty$$. When $$n=2$$, we have one $$P(X=k)$$ for every single $$k=2,\dots,\infty$$. So every $$P(X=k)$$ gets another count, except the first one! And it's clear to see that after $$n=1$$, we will never count another $$P(X=1)$$ since the sums are $$k=n,\dots,\infty$$ for any $$n$$. So we only get one $$P(X=1)$$. To make things easier on our brain, let's just store them in an infinite dimensional matrix $$A$$ where the $$(i,k)$$-th element is whether $$P(X=k)$$ gets a count for when $$n=i$$ ($$1$$ if gets a count, $$0$$ otherwise). So when $$n=1$$ (the first row of the matrix), every $$P(X=j)$$ gets a count. So the first row is just an infinite vector of $$1$$s. For $$n=2$$, every $$P(X=j)$$ for $$j\geq 2$$ gets a count, so the first entry is $$0$$, and the rest are all $$1$$. For $$n=3$$, every $$P(X=j)$$ for $$j \geq 3$$ gets a count, so the first two entries is $$0$$, and the rest are all $$1$$.

The pattern is easy to see: for $$n=i$$, every $$P(X=j)$$ for $$j \geq i$$ will get a count. So let's do this till infinity, and let's count how many times we count each $$P(X=j)$$. For $$j=1$$, we only count it once. For $$j=2$$, we count it twice. For $$j=i$$, we count it $$i$$ times. So it must be that

$$\sum_{n=1}^\infty P(X>n) = \sum_{n=1}^\infty \sum_{k=n}^{\infty} P(X=k) = 1\times P(X=1) + 2\times P(X=2) + 3\times P(X=3) + \dots$$

But wait! That looks awfully like that first equation we wrote! And indeed, it is equal to $$\sum_{k=1}^\infty P(X=k)$$, which is equal to $$E[X]$$.

The intuition is that by counting sums of $$P(X=k)$$ in the way described, we are counting each one $$k$$ times, which is the natural way to describe the 'average' value of a discrete random variable, since this is another way to write that we are weighting each value $$k$$ by $$P(X=k)$$.