Proof of expectation of discrete random variable

149 Views Asked by At

In the textbook "Computing and Probability" by Mitzenmacher and Upfal, they give the following proof that the expectation of a discrete random variable equals $\sum_{x=1}^\infty P(X\geq x)$:

\begin{align*} \sum_{x=1}^\infty P(X\geq x) &=\sum_{x=1}^\infty \sum_{n=x}^\infty P(X=n)\\ &=\sum_{n=1}^\infty \sum_{x=1}^n P(X=n)\\ &=\sum_{n=1}^\infty nP(X=n)=EX. \end{align*}

How do we go from the first line to the second line (the interchange of summations)?

1

There are 1 best solutions below

1
On BEST ANSWER

$$ \begin{bmatrix} P(X=1)&P(X=2)& P(X=3)\cdots\\ & P(X=2)& P(X=3)\cdots\\ & & P(X=3)\cdots\\ & & \ddots \end{bmatrix} $$

Instead of summing the rows and adding up the row totals, sum up the columns and add up the column totals. We can safely interchange order of summation since we are dealing with a series with non-negative terms.