Thread: Proof involving expectation of a random variable.

1. Proof involving expectation of a random variable.

Hi all,

I had this problem as part of an example in class. I really didn't understand what was happening so I wondered if anybody could enlighten me as to what's going on here. Any help is greatly appreciated.

E(N) = \sum_{k=1}^{\infty}kP(N=k)= \sum_{k=1}^{\infty}\sum_i^kP(N=k)

= \sum_{i=1}^{\infty}\sum_k^{\infty}P(N=k)

= \sum_{i=1}^{\infty}P(N\ge i)

tex isn't working for me lately, here

3. Originally Posted by matheagle

E(N) = \sum_{k=1}^{\infty}kP(N=k)= \sum_{k=1}^{\infty}\sum_i^kP(N=k)

= \sum_{i=1}^{\infty}\sum_k^{\infty}P(N=k)

= \sum_{i=1}^{\infty}P(N\ge i)

tex isn't working for me lately, here
I'm trying to see what's going on in your statement, but it's tricky without the latex.

E(N) is = the sum of [ k*P(N=k)], i think i understand, but what exactly is going on in the double summation in the first line?

4. replace k with the sum from 1 to k, 1+1+...+1=k

5. Awesome, thanks!

Any ideas about the second part involving the sum [i*P(N>=i)]?

6. same idea, but I tried to avoid a triple sum

\sum_{i=1}^{\infty} i P(N\ge i)

=\sum_{i=1}^{\infty} i \sum_{k=i}^{\infty}P(N=k)

=\sum_{k=1}^{\infty}\sum_{i=1}^k i P(N=k)

now use \sum_{i=1}^k i =k(k+1)/2

=\sum_{k=1}^{\infty}{k(k+1)\over 2} P(N=k)

=E(N(N+1))/2