# Proof of $\mathbb{E}(Y) = \sum_{t = 0}^{\infty} \mathbb{P}(Y > t)$ for the infinite case?

Let $$Y$$ be a non-negative integer random variable, then show that:

$$\mathbb{E}(Y) = \sum_{t = 0}^{\infty} \mathbb{P}(Y > t)$$

Attempt:

We can clearly see that $$Y = \sum_{t = 0}^{\infty} \mathbf{1}_{Y > t}$$.

And by the linearity of expectation for countable sums (see here: Expected value of infinite sum) we can see that if $$\sum_{t = 0}^{\infty} \mathbb{P}(Y > t) < \infty$$ , then the above result is shown. But what happens when $$\sum_{t = 0}^{\infty} \mathbb{P}(Y > t) = \infty$$, can we somehow show that $$\mathbb{E}(Y) = \infty$$ in this case, or does this not hold?

• Use Tonelli's Theorem. Commented Feb 16, 2023 at 7:41
• Alternatively, for arbitrary $C>0$ you can choose $N$ such that $\sum_{t=0}^N P(Y>t)>C$ and hence $\mathbb E(Y\wedge (N+1))>C$. Commented Feb 16, 2023 at 7:47
• @geetha290krm Could you state Tonelli's theorem as an answer. The version mentioned in Wikipedia allows for interchanging integrals in the double integral of a measurable function or interchanging summations in the double summation of non negative series. I can't seem to relate this to the expectation of a countable sum of rvs.
– Anon
Commented Feb 16, 2023 at 8:01
• Any infinite sum is integration w.r.t counting measure and expectation is also an integral. Commented Feb 16, 2023 at 8:04

$$\mathbb{E}[Y]=\mathbb{E}\left[\sum_{j=0}^\infty\mathbf{1}_{Y> j}\right]=\lim_{n\to\infty}\mathbb{E}\left[\sum_{j=0}^n\mathbf{1}_{Y> j}\right]=\lim_{n\to\infty}\sum_{j=0}^n\mathbb{E}[\mathbf{1}_{Y>j}]=\sum_{j=0}^\infty\mathbb{P}(Y>j).$$