Suppose I have two sets of samples $\{a_i\}$ for $i=1, \ldots, n_a$, and $\{b_j\}$ for $j=1,\ldots, n_b$, drawn respectively from two Poisson random variables $A \sim\textrm{Poiss}(\lambda_a)$ and $B \sim \textrm{Poiss}(\lambda_b)$ with the constraint $$ \frac{\lambda_a}{c_a}=\frac{\lambda_b}{c_b}=\mu, $$ where $c_{a}$ and $c_{b}$ are known constants. What is the analytical expression for the maximum likelihood of $\mu$?
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2$\begingroup$ Why don't you write down the log likelihood and differentiate that with respect to $\mu$? Equating that to zero and solving will show you the correct solution. $\endgroup$– whuber ♦Commented Mar 24, 2014 at 21:58
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Following the comment by whuber, I wrote down the log likelihood and differentiated with respect to $\mu$. In general, if $\{x_i\}$ is a set of samples drawn from Poisson random variables $X_i\sim \textrm{Poiss}(c_i \mu)$ for $i=1,\ldots,n$ and $\{c_i\}$ are known constants, the maximum likelihood estimator of $\mu$ is given by
$$\hat\mu_{\rm MLE}=\frac{\sum_{i=1}^nx_i}{\sum_{i=1}^nc_i}$$