Part of a solution to an exercise in the book Stochastic Processes From Application to Theory (exercise 87) is the following summations:

$$\sum_{i\ge1}P\left(I_i \ge i \right) = \sum_{j \ge i\ge1}P\left(I_i = j \right) = \sum_{j \ge 1} \sum_{1 \le i \le j} (1)P(I_1 = j) = E(I_1)$$

My question is how to make sense of the various summations?

Firstly, $E(I_1) = (0)P(I_1 = 0) + (1)P(I_1 = 1) + (2)P(I_1 = 2)+...$

The first summation make sense: $$\sum_{i\ge1}P\left(I_i \ge i \right)=[P(1)+P(2)+...] +[P(2)+P(3)+…]=(1)P(1)+(2)P(2)...$$

However I am really lost on how to obtain the same result from the second and third summation? Can anyone help explain?

Thank you.

Edit: $I_n$ is a sequence of independent random variables.

  • $\begingroup$ Sorry yes they are. To be honest I was that obsessed with just tyring to figure out how to interpret the numerous summations, I didn't even consider the finer points of the question. $\endgroup$ – Tejay Lovelock Mar 6 at 7:01

Assuming the $I_i$'s are identically distributed,\begin{align*}\sum_{i\ge1}\mathbb P\left(I_i \ge i \right) &= \sum_{i\ge1}\sum_{j \ge i}\mathbb P\left(I_i = j \right)\\&= \sum_{\underbrace{j \ge i\ge1}_{\text{summation}\\\text{order does}\\\text{not matter}}}\mathbb P\left(I_i = j \right)\\ &= \sum_{j \ge 1} \underbrace{\sum_{1 \le i \le j} \mathbb P \underbrace{(I_i = j)}_{\text{independent}\\\text{of index }i}}_{\text{$j$ terms}}\\&= \sum_{j \ge 1} j\, \mathbb P(I_1 = j)\\ &= \mathbb E(I_1)\end{align*}


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