# Minimum variance of $k_1X+k_2Y$ where $X,Y$ are independent Poisson

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!

$$\newcommand{\Var}{\mathrm{Var}}$$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 $$\theta$$.