| Line 3: | Line 3: | ||
Now take the derivative with respect to r and you get: | Now take the derivative with respect to r and you get: | ||
| − | <math>\sum_{n=0}^\infty nr^ | + | <math>\sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2</math> |
You can use this equation to simplify your expected value. | You can use this equation to simplify your expected value. | ||
Revision as of 12:25, 18 September 2008
The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated for of the commonly used geometric series. $ \sum_{n=0}^\infty r^n = 1/(1-r) $
Now take the derivative with respect to r and you get:
$ \sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2 $
You can use this equation to simplify your expected value.
