Proof of $\sum_{k=0}^n k \text{Pr}(X=k) = \sum^{n-1}_{k=0} \text{Pr}(X>k) -n \text{Pr}(X>n)$

Question

$X$ is a random variable defined in $\mathbb N$. How can I prove that for all $n\in \mathbb N$?

• $ \text E(X) =\sum_{k=0}^n k \text{Pr}(X=k) = \sum^{n-1}_{k=0} \text{Pr}(X>k) -n \text{Pr}(X>n)$
• $\text E(X) =\sum_{k=0}^n k \text{Pr}(X=k)=\sum_{k\ge 0} \text{Pr}(X>k) $

Answer 1

For part $a)$, use Thomas' hint. You get $$ \sum_{i=0}^{n}k(P(X>i-1)-P(X>i)). $$

This develops as $P(X>0)-P(X>1)+2P(X>1)-2P(X>2)+3P(X>2)-3P(X>3)+\cdots nP(X>n-1)-nP(X>n)$

for part $b)$:

In general, you have

$\mathbb{E}(X)=\sum\limits_{i=1}^\infty P(X\geq i).$

You can show this as follow: $$ \sum\limits_{i=1}^\infty P(X\geq i) = \sum\limits_{i=1}^\infty \sum\limits_{j=i}^\infty P(X = j) $$

Switch the order of summation gives \begin{align} \sum\limits_{i=1}^\infty P(X\geq i)&=\sum\limits_{j=1}^\infty \sum\limits_{i=1}^j P(X = j)\\ &=\sum\limits_{j=1}^\infty j\, P(X = j)\\ &=\mathbb{E}(X) \end{align}

$$\sum\limits_{i=0}^{\infty}iP(X=i)=\sum\limits_{i=1}^\infty P(X\geq i)=\sum\limits_{i=0}^{\infty} P(X> i)$$

Answer 2

Hint: $Pr(X=k) = Pr(X>k-1)-Pr(X>k)$