Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Using the notation in the Wikipedia article on the hypergeometric distribution, I'm curious how one would obtain the maximum likelihood estimate for parameter $m$, the number of white marbles, given $T$ trials from the same urn. For convenience, I'll copy/paste the notation from the article:

Suppose you are to draw $n$ marbles without replacement from an urn containing $N$ marbles in total, $m$ of which are white. The hypergeometric distribution describes the distribution of the number of white marbles drawn from the urn, $k$.

Again, assuming I conduct $T$ trials, at each trial, I take $n$ balls from the urn, and $k_i$ is the number of white balls at trial $i$. Define $K = (k_1,\ldots,k_T)$. Then the likelihood function $L$: $$L(m; K, N, n) = \prod_i^T \frac{\binom{m}{k_i}\binom{N-m}{n-k_i}}{\binom{N}{n}}$$

Taking a hint from this post, I first tried to solve the inequality: $$L(m;K,N,n) \geq L(m-1;K,N,n)$$ when $T=1$. From this I obtained $$m \leq \frac{Nk+k}{n}$$ so the MLE should be $$m = \left\lfloor \frac{Nk+k}{n} \right\rfloor$$

Now, I'm stuck when I try to generalize to $T \geq 2$.

I first tried doing the same as above and I ended up with the following unwieldy inequality: $$\prod_i^T \frac{m}{m-k_i} \geq \prod_i^T \frac{N-m+1}{N-m-n+k_i+1}$$ which I'm not sure how to solve.

Then I tried to take the log of the likelihood and differentiate as if $m$ were defined over positive reals and I ended up with an equally unwieldy equation to solve: $$\sum_i^T \left(\Psi(m+1) - \Psi(m-k_i+1) - \Psi(N-m+1) + \Psi(N-m-n+k_i+1)\right) = 0$$ where $\Psi$ is the digamma function (i.e. the derivative of the log-gamma function).

My intuition tells me the solution to either of the above would look something like this: $$m = \left\lfloor \frac{(N+1)\sum_i^T k_i}{Tn} \right\rfloor$$ but I have no idea how to get here.

The motivation for this problem is pure curiosity, since I've never seen a MLE for the hypergeometric distribution in terms of $m$.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

Here is an approximate solution. The Poisson approximation to the hypergeometric disribution valid for $\frac{m}{N}<<1$ and $n>>1$, has the form:

$P(K = k|n, M, N) = \frac{exp(-\frac{nm}{N}) (\frac{nm}{N})^k}{k!}$

The likelihood function becomes

$L(m;n,N) = \frac{exp(-\frac{Tnm}{N}) (\frac{nm}{N})^{\sum_i^T k_i}}{\prod_i^T k_i!}$

which can be easily solved to obtain:

$ m = \frac{N\sum_i^T k_i}{Tn} $

share|improve this answer
add comment

The condition you obtained $$ \prod_{i=1}^T \frac{m}{m-k_i} \geq \prod_{i=1}^T \frac{N-m+1}{N-m-n+k_i+1} $$ is correct. Rewriting it as $$ \prod_{i=1}^T \left(1+\frac{k_i}{m-k_i}\right) \geq \prod_{i=1}^T \left(1+\frac{n-k_i}{N-m-n+k_i+1}\right) $$ makes obvious the fact that the LHS is a decreasing function of $m$ and the RHS is an increasing function of $m$. Let $k^*$ denote the minimum value of $k_i$ over $i$. When $m\to k^*$, $m>k^*$, the LHS goes to infinity and the RHS stays finite. When $m\to N-n+k^*+1$, $m<N-n+k^*+1$, the LHS stays finite and the RHS goes to infinity. These observations prove that there exists a unique maximum likelihood estimator of $m$.

Its value is close to the root of a polynomial in $m$ whose degree is a priori at most $2T$. The terms of degree $2T$ cancel and a careful inspection shows that the coefficient of $m^{2T-1}$ is $nT$, hence not zero (this fact alone is sufficient to prove that there exists at least a solution since the degree of the polynomial is odd).

Finally one must find the unique root in the relevant range of a polynomial of degree $2T-1$ hence a closed form formula for the maximum likelihood estimator of $m$ is unlikely when $T\ge3$, the cases $T=1$ (linear polynomial) and $T=2$ (degree $3$ polynomial) being solvable.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.