Editing Negative binomial distribution (section)

===Expectation of successes===

The expected total number of failures in a negative binomial distribution with parameters {{math|(''r'', ''p'')}} is {{math|''r''(1 − ''p'')/''p''}}. To see this, imagine an experiment simulating the negative binomial is performed many times. That is, a set of trials is performed until {{mvar|r}} successes are obtained, then another set of trials, and then another etc. Write down the number of trials performed in each experiment: {{math|''a'', ''b'', ''c'', ...}} and set {{math|''a'' + ''b'' + ''c'' + ... {{=}} ''N''}}. Now we would expect about {{math|''Np''}} successes in total. Say the experiment was performed {{mvar|n}} times. Then there are {{math|''nr''}} successes in total. So we would expect {{math|''nr'' {{=}} ''Np''}}, so {{math|''N''/''n'' {{=}} ''r''/''p''}}. See that {{math|''N''/''n''}} is just the average number of trials per experiment. That is what we mean by "expectation". The average number of failures per experiment is {{math|1=''N''/''n'' − ''r'' = ''r''/''p'' − ''r'' = ''r''(1 − ''p'')/''p''}}. This agrees with the mean given in the box on the right-hand side of this page.

A rigorous derivation can be done by representing the negative binomial distribution as the sum of waiting times. Let <math>X_r \sim\operatorname{NB}(r, p)</math> with the convention <math>X</math> represents the number of failures observed before <math>r</math> successes with the probability of success being <math>p</math>. And let <math>Y_i \sim Geom(p)</math> where <math>Y_i</math> represents the number of failures before seeing a success. We can think of <math>Y_i</math> as the waiting time (number of failures) between the <math>i</math>th and <math>(i-1)</math>th success. Thus
:<math>
X_r = Y_1 + Y_2 + \cdots + Y_r.
</math>
The mean is 
:<math>
E[X_r] = E[Y_1] + E[Y_2] + \cdots + E[Y_r] = \frac{r(1-p)}{p},
</math>
which follows from the fact <math>E[Y_i] = (1-p)/p</math>.