Editing Geometric distribution (section)

==== Proof of expected value ====
Consider the expected value <math>\mathrm{E}(X)</math> of ''X'' as above, i.e. the average number of trials until a success. 
The first trial either succeeds with probability <math>p</math>, or fails with probability <math>1-p</math>. 
If it fails, the '''remaining''' mean number of trials until a success is identical to the original mean - 
this follows from the fact that all trials are independent.

From this we get the formula:

: <math>\operatorname \mathrm{E}(X) =  p + (1-p)(1 + \mathrm{E}[X]) ,</math>

which, when solved for <math> \mathrm{E}(X) </math>, gives:

: <math>\operatorname E(X) = \frac{1}{p}.</math>

The expected number of '''failures''' <math>Y</math> can be found from the [[linearity of expectation]], <math>\mathrm{E}(Y) = \mathrm{E}(X-1) = \mathrm{E}(X) - 1 = \frac 1 p - 1 = \frac{1-p}{p}</math>. It can also be shown in the following way:

: <math>
\begin{align}
\operatorname E(Y) & =p\sum_{k=0}^\infty(1-p)^k k \\
& = p (1-p) \sum_{k=0}^\infty (1-p)^{k-1} k\\
& = p (1-p) \left(-\sum_{k=0}^\infty \frac{d}{dp}\left[(1-p)^k\right]\right) \\
& = p (1-p) \left[\frac{d}{dp}\left(-\sum_{k=0}^\infty (1-p)^k\right)\right] \\
& = p(1-p)\frac{d}{dp}\left(-\frac{1}{p}\right) \\
& = \frac{1-p}{p}.
\end{align}
</math>

The interchange of summation and differentiation is justified by the fact that convergent [[power series]] [[uniform convergence|converge uniformly]] on [[compact space|compact]] subsets of the set of points where they converge.