mooshupork34
Junior Member
- Joined
- Oct 29, 2006
- Messages
- 72
The following was confusing me so any help would be greatly appreciated!
The expectation of a Poisson distribution E(X) is equal to the sum of k going from 1 to infinity of k * (w^k)/k! * e^-w, which equals we^-w * the sum of k going from 1 to infinity of w^(k-1) over (k-1)!, which equals w.
Show that the variance of X is also equal to w.
The expectation of a Poisson distribution E(X) is equal to the sum of k going from 1 to infinity of k * (w^k)/k! * e^-w, which equals we^-w * the sum of k going from 1 to infinity of w^(k-1) over (k-1)!, which equals w.
Show that the variance of X is also equal to w.