0
$\begingroup$

Suppose that X and Y are independent Poisson distributed values with means 3θ and θ, respectively. Consider the combined estimator of θ ˜θ = k1X + k2Y where k1 and k2 are arbitrary constants. (a) Find the condition on k1 and k2 such that ˜θ is an unbiased estimator of θ. (b) For what values of k1 and k2 will the combined estimator ˜θ = k1X + k2Y be an unbiased estimator with smallest variance amongst all such linear combinations? (c) Given observations x and y find the maximum likelihood estimate of θ. I've got part (a) which is k1+k2=1 and (b) which is k1=3/4 and k2=1/4. I just cant get part c..

$\endgroup$
0
$\begingroup$

The idea is to write out a likelihood function and find the $\theta$ in the parameter space that maximizes this likelihood given the observations. The likelihood is proportional to the joint density: $$\mathcal L(\theta \mid x,y) \propto f_{X,Y}(x,y \mid \theta) \overset{\text{ind}}{=} e^{-3\theta} \frac{(3\theta)^x}{x!} e^{-\theta} \frac{\theta^y}{y!}, \quad \theta > 0, \quad x, y \in \{0, 1, 2, \ldots \}.$$ Thus, as a function of $\theta$, the likelihood is proportional to $$\mathcal L(\theta \mid x,y) \propto e^{-4\theta} \theta^{x+y}.$$ Note we have discarded any factors that are not functions of $\theta$. The log-likelihood is then $$\ell(\theta \mid x,y) = -4\theta + (x+y) \log \theta,$$ and its derivative is $$\frac{\partial \ell}{\partial \theta} = -4 + \frac{x+y}{\theta}.$$ Thus $\ell$ is maximized at a critical point $\hat\theta$ satisfying $\partial \ell/\partial \theta = 0$, or $$\hat \theta = \frac{x+y}{4},$$ and we can check that this is indeed a maximum.


This result should call into question your calculation for part (a): note that the expectation of $X$ is $\operatorname{E}[X] = 3\theta$, and the expectation of $Y$ is $\operatorname{E}[Y] = \theta$, thus the expectation of a linear combination of the two is $$\operatorname{E}[k_1 X + k_2 Y] = k_1 \operatorname{E}[X] + k_2 \operatorname{E}[Y] = k_1 (3\theta) + k_2 \theta = (3k_1 + k_2)\theta,$$ and in order for this to equal $\theta$, you must have $3k_1 + k_2 = 1$, not $k_1 + k_2 = 1$. This would also suggest your calculation of the minimum variance unbiased estimator among this family of estimators is not correct.

$\endgroup$
  • $\begingroup$ @Pkr96 No, that is not correct, since $k_1 = 1/4$ and $k_2 = 3/4$ does not satisfy the constraint $3k_1 + k_2 = 1$. $\endgroup$ – heropup Feb 22 '17 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.