# Poisson Process Method of Moments

Disclaimer: This is a homework problem

A School of Ornithology researcher wants to estimate the number of red-tailed hawks in Ithaca. She radio tags 10 birds, and then sets up a feeding station with automatic camera

The researcher believes that each individual bird's visits to the feeder can be modeled as a Poisson process with some unknown rate $$\lambda$$. Over the first five weeks, she observes an average of 28.8 birds (tagged and untagged) visiting the feeder, with an average of 6 tagged birds per week. Use the method of moments to obtain an estimate for the total population.

The method of moments is a way to estimate the parameters by gathering a system of equations using empirical moments and setting them equal to the actual moments. Here, we only need one equation for one unknown: $$\lambda$$.

The way I understand this problem, we have $$m_1 = \frac{X_1+X_2+X_3+X_4+X_5}{5} = 6$$. Then we want $$\mu_1 = E[X] = \frac{1}{\lambda} = m_1 = 6 \implies \lambda = \tfrac 16$$. However, I am unsure of my reasoning here as this does not take into account the untagged birds. I am also not sure if I am understanding the empirical moment correctly from this problem.

Roughly, assuming a constant population and its random mixing between the time of tagging and the time of observing hawks at the feeder, the proportion $$10/N$$ of tagged hawks in the population should be estimated by the proportion $$6/28.8$$ at the feeder. So we estimate $$\hat N = 288/6 = 48$$ hawks in the population.
Note: This 'Lincoln-Peterson' method fails if no tagged hawks are seen at the feeder. See Wikipedia and other references on mark-recapture or capture-recapture estimation` for somewhat more satisfactory methods.
• Can you elaborate why it should be estimated by $6/28.8$? – hkj447 Oct 18 at 23:18
• I believe I have figured it out: $m_1=E[X_{tagged}] = \sum_{i=1}^{10} E[X_i] = 10\lambda = 6 \implies \lambda = \frac{6}{10}$. Now we estimate the population of hawks by assuming we have a sum of $N$ Poisson processes. Using $\lambda = \frac{6}{10}$, we get $E[X_{pop}] = N\lambda = 28.8 \implies N = \frac{288}{6} = 48$ hawks – hkj447 Oct 19 at 0:16