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I have two series of $m$ observations each $n_i,X_i$ where $ i=1,....m$, and a related probability $p$. $n_i$ is large and $p$ is small, so we can assume that $X_i \sim Poi(n_ip)$. What is the maximum likelihood estimator of p in this situation and what is the expectation and variance of the estimator?

Thanks in advance for any help!

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    $\begingroup$ Are all $2m$ observations jointly independent? The $n_i$ are presumed known and nonrandom? $\endgroup$
    – cardinal
    Oct 8 '13 at 16:40
  • $\begingroup$ Well $X_i \leq n_i$, but otherwise yes. Think of it as $n_i$ being the number of animals counted, and $X_i$ being the number of the counted animals that share a certain trait. $p$ is then the probability of observing an animal with that trait. $\endgroup$
    – L1meta
    Oct 8 '13 at 16:55
  • $\begingroup$ This looks like a standard textbook problem. Is this for some subject? $\endgroup$
    – Glen_b
    Oct 8 '13 at 20:03
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Trying to cover two bases here; if this is an actual problem you face, I can give you the MLE easily enough. On the other hand if you're supposed to derive an MLE for some subject, this becomes a self-study question and we should look toward giving hints and guidance, so I won't outright derive it for you.

If you treat it as Poisson, $n_i$ is effectively an exposure.

Let $x_i$ be the observed values of the random variables, $X_i$.

The ML estimate of $p$ within each group is just $\sum_i x_i/\sum_i n_i$.

If the population $p$ is the same in both groups, then if $N_1$ is the sum of $n_i$ for group 1 and $N_2$ correspondingly for group 2, and if $\hat p_1$ is the estimate of $p$ for group 1 and $\hat p_2$ for group 2, then the overall ML estimate of $p$ is $\hat p = (N_1 \hat p_1 + N_2 \hat p_2)/(N_1+N_2)$.

However, there's no obvious reason not to treat it as explicitly binomial rather than approximate it as Poisson. Computers do all the work these days.

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  • $\begingroup$ Thanks for this! :) It's not explicitly homework, it's one of the annoying "blah blah left to the reader as an exercise", and I really want to know the full derivations of it all as we will be tested in this stuff on the exam. $\endgroup$
    – L1meta
    Oct 8 '13 at 20:35
  • $\begingroup$ Ah, thanks for clarifying. That definitely counts under the self-study tag. Please see the self-study tag wiki info; there are some obligations placed on both askers and answerers there. In particular, what have you tried and what were your problems with it? (Since it's now clear that it's self-study I have taken the liberty of adding the tag) $\endgroup$
    – Glen_b
    Oct 8 '13 at 21:40
  • $\begingroup$ Ah, I see. Thanks for the heads up. :) Well to be honest I haven't really tried too many different things. I'm very perplexed as to how to approach estimating "a part" of the parameter of the distribution. Which I guess is why I'm on here in the first place. :p $\endgroup$
    – L1meta
    Oct 8 '13 at 22:09
  • $\begingroup$ I'm not sure what you mean by 'a part' there. If $\mu_i = n_i p$ can you write the likelihood for $\mu_i$? Can you see - by putting $n_i p$ instead of $\mu_i$ - how to write the likelihood for the $i^\text{th}$ observation? [I'd suggest you add an index for the group ('series'), by the way, or you'll get tangled up by this use of the same symbols for two different things.] $\endgroup$
    – Glen_b
    Oct 8 '13 at 22:17
  • $\begingroup$ Well, the MLE for $\lambda$ in a 'normal' poisson distribution is $$\hat{\lambda} = \frac{\sum x_i}{n}\,.$$ However I'm not sure how to untangle the parts I need. $\endgroup$
    – L1meta
    Oct 9 '13 at 8:34

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