Prove the loglikelihood is strictly concave for ABO allele frequency blood type data

I am working through the problems in Kenn Lange's book Numerical Analysis for Statisticians. I am going to try and do all of the problems in the book, though none of them are specifically assigned for homework. I could not figure out how to solve this problem and was hoping someone could help. It does not seem particularly hard, but I was not sure how to do it correctly. The problem is #7 in chapter 13 on the EM algorithm.

Suppose the phenotypic counts in the ABO allele frequency estimation example satisfy $n_A + n_{AB} > 0, n_B + n_{AB} > 0, \text{and } n_O > 0$. Show that the loglikelihood is strictly concave and possesses a single global maximum on the interior of the feasible region.

The question here deals with blood types. So there are 4 blood phenotypes: A, B, AB, and O, as everyone knows. Then there are 6 different genotypes--set of 2 alleles--to produce these phenotypes: A/A, A/O produce A blood type, B/B, B/O produce B blood type, A/B produces the AB blood type, and O/O produces the O blood type.

I was not sure how to prove the concavity of the likelihood and incorporate the constraints on the numbers of individuals with A, AB, B, and O blood. The inequality constraints don't seem particularly useful, since they just force the counts to be positive, which is we would expect the data to demonstrate anyway.

I imagine I would use the Lagrange multiplier method with inequality constraints to incorporate the data. But I was not sure how to provide the concavity of the solution. Any help is appreciated.

• Thanks @Glen_b. I added the self-study tag as you asked. I modified the question a bit to match the self-study tag description, though I might add more a bit later. – krishnab May 3 '15 at 2:00
• Do you know any methods for showing that a function is concave? Could you use the definition of concavity? Could you use a theorem whose conclusion is that a function is concave? Could you compute the Hessian of your log-likelihood? – Brian Borchers May 3 '15 at 3:15

Okay, here is the answer I came up with, but I was hoping someone could check to make sure it is correct.

So the approach laid out above was essentially correct. To prove concavity I just needed to prove that the second derivative of the Lagrangian is less than zero. That establishes concavity. The existence of a unique global maximum, that just seems to result from the fact that the second derivative is always less than zero over the domain, and thus there are no other critical points possible.

Recall that we are working with the Hardy-Weinberg law of population genetics. So this essentially sets the frequency for each allele type. So for each genotype:

AA: $p_A^2$,

AO: $2p_Ap_O$,

AB: $2p_Ap_B$,

BB: $p_B^2$,

BO: $2p_Bp_O$,

OO: $p_O^2$,

So here is the basic layout of the math. First, I need to layout the log likelihood of the distribution of alleles.

$$ln f(X|p) = n_{AA}ln(p_A^2) + n_{AO}ln(2p_Ap_O) + n_{BB}ln(p^2_B) + n_{BO}ln(2p_Bp_O) + n_{AB}ln(2p_Ap_B) + n_Oln(p^2_O)$$

Now to maximize the the likelihood subject to the constraint that $\sum{p_i} = 1$, we use the lagrange multiplier method. So we write the Lagrangian as:

$$L(p_i, \lambda) = n_{AA}ln(p_A^2) + n_{AO}ln(2p_Ap_O) + n_{BB}ln(p^2_B) + n_{BO}ln(2p_Bp_O) + n_{AB}ln(2p_Ap_B) + n_Oln(p^2_O) - \lambda(p_A + p_B + p_O - 1)$$

Now we can look at the derivatives of the Lagrangian. I make the argument with respect to $p_A$, but the same results follow for $p_B, p_O$ by symmetry. The first derivative of the Lagrangian with respect to $p_A$ is:

$$\frac{\partial L}{\partial p_A} = \frac{2n_{AA} + n_{AO} + n_{AB}}{p_A}$$

The second derivative of the Lagrangian with respect to $p_A$:

$$\frac{\partial^2 L}{\partial p_A \partial p_A} = -\frac{2n_{AA} + n_{AO} + n_{AB}}{p_A^2} < 0$$

Note if we take the second derivative of the Lagrangian where $p_i \neq p_j$ then the result is zero.

Since the second derivative is negative over the entire domain of $p_i \in (0,1])$, there has to be a unique maximal point.

Please let me know if I need to fix anything?