Is there a way to visualise $ \sum_{k \geq 0} \mathbb{P}(X>k) = \mathbb{E} [X] $?

Question

Let X be a random variable taking values in N

$ \sum_{k \geq 0} \mathbb{P} (X>k) \dots (0)$

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

$= \sum_{i \geq 1} \sum_{k=0}^{i-1}\mathbb{P}(X=i) \dots (2) $

$= \sum_{i \geq1} i \mathbb{P}(X=i) \dots (3) $

$= \mathbb{E} [X] $

I am having a problem with the double sum, I don't see how to think of coming up with this, or how do we go from (0) to (2) , the way I reason by default is like :

$\mathbb{E} [X] = \sum_{k \in N} k \mathbb{P}(X=k) = 1 P(1) + 2 P(2) + ... $ then I block

Answer 1

I'm not sure if this is "visual" in the sense that you want, but I think the following is a nice illustration. As you note, we have $\mathbb{E}[X]=\sum_{n=1}^{\infty}n\mathbb{P}(X=n)$. Think of this as summing all the terms in the following picture: $$ \begin{matrix} \mathbb{P}(X=1)&&\\ \mathbb{P}(X=2)&\mathbb{P}(X=2)&\\ \mathbb{P}(X=3)&\mathbb{P}(X=3)&\mathbb{P}(X=3)\\ \vdots&&&\ddots \end{matrix} $$ Now, if you sum these terms row by row, you get $\sum_{n=1}^{\infty}n\mathbb{P}(X=n)$. If you instead sum column by column, you get $\sum_{n=1}^{\infty}\mathbb{P}(X\ge n)$.

You can picture any exchange of double sums in a similar manner. The mathematical justification for this and much more general types of exchanges is given by Fubini's and/or Tonelli's theorem.

Answer 2

0->1

What is P(X>K)? For example P(X>3)=P(4)+P(5)+P(6)...........

or P(X>k)=$\sum_{r=k+1}^{\infty} P(r) $.Now can you viisualize how we moved from 0->1

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1->2

How many times do we count P(3), for P(X>0),P(X>1) and P(X>2).(3 times)

How many times do we count P(k),for P(X>0),P(X>1),P(X>2)..........P(X>(k-2)) and P(X>(K-1)) ,hence k times.

Now can you visualize 1->2

Answer 3

$$\sum_{k\ge 0}\mathbb{P}(X>k)$$ $$=\sum_{k>0}\mathbb{P}(X\ge k)$$ $$=\mathbb{P}(X\ge 1)+\mathbb{P}(X\ge 2)+\mathbb{P}(X\ge 3)+\dots$$ $$=\mathbb{P}(X=1)+2\mathbb{P}(X\ge 2)+\mathbb{P}(X\ge 3)+\dots$$ $$=\mathbb{P}(X=1)+2\mathbb{P}(X=2)+3\mathbb{P}(X\ge 3)+\dots$$ $$\vdots$$ $$=\mathbb{P}(X=1)+2\mathbb{P}(X=2)+3\mathbb{P}(X=3)+\dots+k\mathbb{P}(x=k)+\dots$$

Edit:

$$\mathbb{P}(X\ge 1)=\mathbb{P}(X=1)+\mathbb{P}(X\ge 2)$$