I have the following question for homework:

  1. Suppose that $X$ and $Y$ are independent Poisson distributed values with means $\theta$ and $2\theta$, respectively. Consider the combined estimator of $\theta$ $$ \tilde\theta = k_1 X + k_2Y $$ where $k_1$ and $k_2$ are arbitrary constants.

(a) Find the condition on $k_1$ and $k_2$ such that $\tilde \theta$ is an unbiased estimator of $\theta$.

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

(c) 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 know the answer for part a) is $$k_1+ 2k_2 = 1$$

I don't know where to go with part b). So far I have as far as

$$Var(\hat\theta)= (k_1)^2Var(X) + (k_2)^2Var(Y)$$

Any help would be appreciated thanks in advance!


1 Answer 1


$\newcommand{\Var}{\mathrm{Var}}$Hints: You effectively need to minimise the function $$f(k_1, k_2):= \Var(X)k_1^2 + \Var(Y)k_2^2$$ subject to the constraint $$k_1 + 2k_2 =1.$$ To do this, you can eliminate one of the variables from $f(k_1, k_2)$ by using the constraint $k_1+2k_2=1$. This will leave you with a single variable quadratic, which you should know how to minimise.

Also, note that $\Var(X)$ and $\Var(Y)$ are just constants that you can express in terms of $\theta$.


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