Good evening,
I have been trying to figure out how to calculate the expected value of a random variable which is very similar to the hypergeometric distribution. The problem reads as follows:
A population of 𝑏 animals has had a number 𝑎 of its members captured, marked and released. Let X be the number of animals it is necessary to recapture (without re-release) in order to obtain 𝑚 marked animals. Determine E(X).
This seems similar to the hypergeometric distribution, which would fix n and vary m in order describe the event that in n captured animals m had the feature of being marked.
I believe the PMF for this distribution is the same as for the hypergeometric distribution except it is parametrized by n rather than m. It is given by:
$P_X(n) = \frac{\textrm{(ways to select m animals from a marked animals)}*\textrm{(ways to select n-m remaining animals from b-a unmarked animals)}}{\textrm{ways to select n animals from b total animals}}$
$P_X(n) = \frac{{a \choose m}{{b-a} \choose {n-m}}}{b \choose n}$
Since the PMF only makes sense on the interval m $\leq$ n $\leq$ b, from the definition of expectation:
$E(X) = \sum_{n=m}^{b} n*\frac{{a \choose m}{{b-a} \choose {n-m}}}{b \choose n}$
I don't really know where to go from here. I know arguments can be made for using the linearity of expectations to yield the expectation of the hypergeometric distribution. However, they use the fact that parameter n is fixed.
Does any one have any tips on what process to take to solve this, as well as the name for such a distribution?
EDIT: I believe the name of this distribution is the negative hypergeometric distribution
Best Answer
You got the distribution wrong. You wrote down the probability that exactly $m$ out of $n$ uniformly randomly chosen animals are marked. But you want the probability that you need to capture $m$ animals, so you need to exclude the case that you already had $m$ animals before you captured the last one. The probability to capture the $m$-th marked animal on the $n$-th capture is
$$ \frac{\binom a{m-1}\binom{b-a}{n-m}}{\binom b{n-1}}\cdot\frac{a-m+1}{b-n+1}\;, $$
where the first factor is the probability of capturing exactly $m-1$ marked animals when you capture $n-1$ animals, and the second factor is the probability of capturing a marked animal on the $m$-th capture. However, you can use
$$ \binom a{m-1}(a-m+1)=\binom{a-1}{m-1}a $$
to rewrite this such that it has the same essential structure as what you had. In any case, the article you've found in the meantime seems to contain all you need to find the expected value.