MLE for scaled Poisson distributions

Question

Let's say I have a quantity, $x$, which I can only measure by drawing from a Poisson distribution with parameter $x$. The MLE for repeated draws from this distribution is just the sample mean. However, if I can arrange to instead draw from 'scaled' distributions in which the parameters are multiplied by a known set of numbers, can I exploit this fact to produce a better estimator?

Answer 1

You have $X_i\sim\operatorname{Poisson}(c_i\lambda)$ and $X_1,\ldots,X_n$ are independent.

\begin{align} L(\lambda) = {} & \prod_{i=1}^n \frac{(c_i\lambda)^{x_i} e^{-c_i\lambda}}{x_i!} = \prod_{i=1}^n \frac{c_i^{x_i}}{x_i!} \lambda^{x_i} e^{-c_i\lambda} = \left( \prod_{i=1}^n \frac{c_i^{x_i}}{x_i!} \right) \left( \prod_{i=1}^n \lambda^{x_i} e^{-c_i\lambda} \right). \\[10pt] \ell(\lambda) = {} & \text{constant} + \sum_{i=1}^n (x_i \log \lambda - c_i\lambda) \\ & \text{where “constant'' means not depending on } \lambda. \\[10pt] \ell\,'(\lambda)= {} & \sum_{i=1}^n \left( \frac{x_i}\lambda - c_i \right) = \frac{n\overline x} \lambda - n\overline c \\ & \text{where } n\overline x = \sum_{i=1}^n x_i \text{ and } n\overline c = \sum_{i=1}^n c_i \\[15pt] \ell\,'(\lambda) = {} & n\left( \frac{\overline x} \lambda - \overline c \right) \begin{cases} >0 & \text{if } 0\le\lambda < \overline x/\overline c, \\[5pt] =0 & \text{if } \lambda = \overline x / \overline c, \\[5pt] <0 & \text{if } \lambda > \overline x/\overline c. \end{cases} \end{align} Therefore the MLE is $\overline x/\overline c.$