Maximum Likelihood Estimate with different parameters

estimationmaximum likelihoodpoisson distributionstatistical-inferencestatistics

Suppose that X and Y are independent Poisson distributed values with means $\theta$
and $2\theta$, respectively. Consider the combined estimator of $\theta$
$$
\tilde{\theta} = k_1 X + k_2 Y
$$
where $k_1$ and $k_2$ are arbitrary constants.

  1. Find the condition on $k_1$ and $k_2$ such that $\tilde{\theta}$ is an unbiased estimator of $\theta$.

  2. For $\tilde{\theta}$ unbiased, show that the variance of the estimator is minimized by taking $k_1 = 1/3$ and $k_2 = 1/3$.

  3. Given observations $x$ and $y$ find the maximum likelihood estimate of $\theta$ and hence show that $\tilde{\theta}$ is also the maximum likelihood estimator.

I have gotten (1) and (2) okay, but it's (3) I am having trouble with, I'd be okay if $X$ and $Y$ had the same parameter but I'm having trouble with $X$ and $Y$ having different parameters, any help would be appreciated.

NOTE

For (1) I got $k_1 = 1 – 2k_2$.

For (2) I found the variance of $\tilde{\theta}$, then differentiated and let equal to zero to minimize – therefore we get (after subbing in $k_2 = 1 – k_1/2 $) $$3k_1-1=0,$$ which when subbing in $1/3$, we see it is minimised.

Thank you.

Best Answer

Write down the likelihood of observing $x$ and $y$. $$P(X=x, Y=y) = P(X=x) P(Y=y) = e^{-\theta} \frac{\theta^x}{x!} e^{-2\theta} \frac{(2\theta)^y}{y!}.$$ Choose $\theta$ to maximize this quantity; this is your maximum likelihood estimator.

By taking logarithms and ignoring constants, it is equivalent to choose $\theta$ maximizing $-3\theta + (x+y) \log \theta$. Setting the derivative to zero yields $-3 + (x+y)/\theta = 0$ and yields the same estimator you had in (b).

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