Question:
Edwards and Eberhardt (1967) conducted a live-trapping study on a confined population of known size. In their study, wild cottontail rabbits were penned in a 4-acre rabbit-proof enclosure. Live trapping was conducted for 18 consecutive nights. Recorded capture frequencies were as follows: 43 rabbits were caught at exactly 1 night, 16 rabbits at 2 nights, 8 at 3 nights, 6 at four nights, 2 at 6 nights and 1 rabbit was caught at 7 nights. In other words, we have observations from $X_1, · · · X_m$ count variables of m = 76 different, caught rabbits. So, in this sample, 43 of the $x_1, · · · , x_m$ were ones, 16 twos, etc. Every count can take a maximum of T = 18 captures. The number of of zero counts, i.e. the number of rabbits that were not caught, is unobserved. We denote this additional sample as $X_{m+1}, · · · , X_n$. Assuming independence of trapping occasions and homogeneity of capture probability θ at each occasion each $X_i$ would follow a binomial distribution. We assume here that a Poisson approximation is justified. That is:
$P(X_i = x) = \frac{\exp(−θT)(θt)^x}{x!}$ where i = 1, · · · , n.
Note that the observed sample is zero-truncated so that the observed sample can be considered as arising from the zero-truncated Poisson density:
$$\exp(−θT)(θT)^x \over (1 − \exp(−θT))x!$$ for $i = 1, · · · , m$, $\;θ ∈ (0, 1)$ and $x ∈ \{0, 1, · · · , T\}$.
Determine the log-likelihood function for the observed sample.
My solution:
I have already attempted this and got: $\sum_{i=1}^m n_i\log\frac{\exp(−θT)(θT)^x}{x!}\frac{1}{1 − \exp(−θT)}$
Is this correct? If not, what should I have got? What have I done wrong?