Prove $\mathsf E(N)=\sum_{i=1}^\infty \mathsf P(N\geqslant i)$

Question

We want to prove that $$\mathsf E(N)=\sum_{i=1}^\infty \mathsf P(N\geqslant i)$$

We are given a hint that $$\sum_{i=1}^\infty\mathsf P(N\geqslant i)= \sum_{i=1}^\infty\sum_{k=i}^\infty \mathsf P(N=k)$$

and then instructed to switch the order of the sums.

I can't figure out how this works out. Please help.

Answer 1

Switching the sums means moving from $i\in\{1\ldots\infty\}, k\in\{i\ldots\infty\}$ to the equivalent domain of $k\in\{1\ldots\infty\}, i\in\{1\ldots k\}$. Because both are equivalent to $1\leqslant i\leqslant k < \infty$ for integers $i,k$.

$$\begin{align} \sum_{i=1}^\infty\sum_{k=i}^\infty f(i,k) &= \sum_{k=1}^\infty \sum_{i=1}^{k} f(i,k) \end{align}$$

Now instead of $f(i,k)$ you have $\mathsf P(N=k)$, an expression only in $k$. Can you take it from here?

Answer 2

An alternative (more transparent) way to prove it is the following $$\begin{align*}P(N\ge1)&=P(N=1)+P(N=2)+P(N=3)+P(N=4)+\ldots\\P(N\ge2)&=\phantom{P(N=1)+!!}P(N=2)+P(N=3)+P(N=4)+\ldots\\P(N\ge3)&=\phantom{P(N=1)+!!P(N=2)+!!}P(N=3)+P(N=4)+\ldots\\P(N\ge4)&=\phantom{P(N=1)+!!P(N=2)+!!P(N=3)}+P(N=4)+\ldots\\\ldots&\end{align*}$$ Now if you sum up both sides you have that $$\begin{align*} \sum_{i=1}^{\infty}P(N\ge i)&=1\cdot P(N=1)+2\cdot P(N=2)+3\cdot P(N=3)+4\cdot P(N=4)+\ldots\\&=\sum_{i=1}^{\infty}i\cdot P(N=i)=\mathsf E[N]\end{align*}$$

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The above statement is correct, assuming that your random variable takes values in $\{1,2,3,\ldots\}$. However it is missing from the text of the exercise.