How is $\sum_{k \geq 0} \sum_{i \geq k+1} \mathbb{P}(X=i) = \sum_{i \geq 1} \sum_{k=0}^{i-1}\mathbb{P}(X=i) = \sum_{i \geq1} i \mathbb{P}(X=i) $?

Question

EDIT:

Here's a way to remember without thinking hard about it that I came with from the channel of blackpenredpen:

$\sum_{k \geq 0} \sum_{i \geq k+1} \mathbb{P}(X=i) $

We have $1\leq \underline{k+1 \leq i \leq \infty}$ , in the interior sum, $i$ goes from $k+1$ to $\infty$

In which case the exterior sum takes $k$ from $0$ to $\infty$

Switching the interior sum to $k$ and the exterior to $i$ ,

We have $\underline{1\leq k+1 \leq i} \leq \infty$ , in the interior sum, $k$ goes from $0$ to $i-1$

In which case the exterior sum takes $i$ from $1$ to $\infty$

That is :

$\underline{0\leq k \leq i-1} \leq \infty$

$= \sum_{i \geq 1} \sum_{k=0}^{i-1}\mathbb{P}(X=i)$

By summing the terms $i$ times (from $0$ to $i-1$, that is $i$ times) we get:

$= \sum_{i \geq1} i \mathbb{P}(X=i) $

Answer 1

I find it easier to avoid a changing domain of summation. Your first sum is equivalent to $$\sum_k \sum_i P(X=i)1\{i\geq k+1\}$$ While your second sum is $$\sum_i \sum_k P(X=i)1\{k\leq i-1\}$$ Note the functions are now the same except expressed differently for emphasis and the only difference is the exchange in sums which is legal for non negative terms.