m (2a Ken Pesyna moved to 4.2a Ken Pesyna: Improperly named originally) |
|||
| (4 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| − | The E[x] equation can be simplified (rid the summation term) by using a differentiated | + | The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated form of the commonly used geometric series equation: <math>\sum_{n=0}^\infty r^n = 1/(1-r)</math> |
| + | |||
| + | Now take the derivative with respect to r and you get: | ||
| + | |||
| + | <math>\sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2</math> | ||
| + | |||
| + | You can use this equation to simplify your expected value. | ||
Latest revision as of 06:43, 15 October 2008
The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated form of the commonly used geometric series equation: $ \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.
