# Building a single likelihood for the sum of two events

Consider an unknown number of adult deer, $N$, which can be split into two groups - males, $M$, and females, $F$, where $N = F + M$. These adults were hunted during the year (a lethal form of sampling without replacement), with each group having its own individual-level probability of being killed by a hunter, $p_f$ and $p_m$, respectively. For example, male deer in Michigan have a 40% of being killed by a hunter during the hunting season while females only have a 16% chance of being killed. The number of adults that were killed can be denoted as $m$ and $f$, respectively.

I want to build a likelihood for the total number of adults, $N$, given that I know $p_m$, $p_f$, $m$, and $f$, to be used in a larger joint likelihood. If I were interested in building a likelihood for the two different groups of adults, I could write: $$L(M \mid p_m,m) = \binom{M}{m} \cdot (p_m)^{m} \cdot (1-p_m)^{M-m}$$ $$L(F \mid p_f,f) = \binom{F}{f} \cdot (p_f)^{f} \cdot (1-p_f)^{F-f}$$ What I want, however, is $$L(N \mid p_m,p_f,m,f) = \text{?}$$ The full likelihood already has a large number of parameters to estimate, so I'd rather just have a single parameter for the total number of adults, as opposed to having a separate parameters to estimate for males and females. Basically, the likelihood equation cannot contain $M$ or $F$.

Any help would be greatly appreciated!

• First of all p_f=1-p_m. So off the bat you can express these likelihoods in terms of just p_f, m, and f. Also I am puzzled because I don't see any mention of n=f+m. Without n I don't think the problem is well-defined, since there has to be a number of samples that you will take in order to know when to stop. Note the difference between n, f, and m. The difference here is that m and f are known after you sample while n has to be specified before. Dec 23, 2016 at 20:52
• @MichaelChernick, $p_f$ does not equal $1-p_m$. These two quantities are entirely independent - there's a probability of being sampled if you're female, and a separate probability of being sampled if you're male. Further, because $N=F+M$, you can define the total number of sampled individuals as $n=f+m$ Dec 23, 2016 at 20:57
• Wait a minute! What other sex is there that you can sample? I know that n=f+m. I even said it. It seems that before you take the sample you have to decide on n and f and m are the result of the sample. I hope you can at least see that you haven't explained how you are doing the sampling. Dec 23, 2016 at 22:16
• @MichaelChernick, Sorry for the confusion, let me try to set up the problem a bit better. I am combining several pieces of information in a likelihood framework to estimate several parameters, among them being the total number of adults in a certain population. Other sources of data provide information on how many male and female individuals were sampled (at random without replacement) from this population, and at what rate. I'm trying to use those data to estimate the total number of adults. Does this help? Dec 23, 2016 at 22:35
• I am sorry what you are telling me is what I have already assumed. I understand that you want to estimate N without using F and M and you are hoping to get a likelihood to compute an MLE for N. So you take the first sample and it turns out to be a male. Now you have one less male. So after that you have one less male. Are you estimating p_m and p_f or assuming you know them. I can't see that if p_m is the probability for a male and p_f for female how can p_m+p_f be less than 1. That would mean there is p_?=1-p_m-p_f >0. But you have no category for?. Dec 24, 2016 at 0:07

After discussing this with some colleagues, the (sort-of) solution turns out to be to use a multinomial:

$$L(N \mid p_m,p_f,m,f) = \binom{N}{m,f} \cdot (p_m)^{m} \cdot (p_f)^{f} \cdot (p_{bar})^{N-m-f},$$

where $p_{bar}$ is the probability of not being killed by a hunter. This probability is unknown, but can be calculated as the probability of not being hunted given that you are a male times the probability that you are a male plus the probability of not being hunted given that you are a female times the probability that you are a female.

$$p_{bar} = P(not.hunted|male) \cdot P(male) + P(not.hunted|female) \cdot P(female)$$ $$p_{bar} = (1-p_m) \cdot P(male)+(1-p_f) \cdot P(female)$$

Unfortunately, we do not know the probability of being a male or female, so we are left with an unknown quantity - the ratio of males in the population, $R$:

$$p_{bar} = (1-p_m) \cdot R + (1-p_f) \cdot (1-R)$$

And the only way to calculate $R$ is to know either $M$ or $F$. So the short of it is, as far as I can tell, there is no way how to eliminate both $M$ and $F$ from the likelihood.