# Maximum Likelihood Estimate with different parameters

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.

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).
The likelihood of $$\theta$$ is given by $$L(\theta|x,y) = f_{X,Y}(x,y|\theta) = f_X(x|\theta) f_Y(y|\theta) = \frac{\theta^x}{x!} \exp(-\theta)\cdot \frac{(2\theta)^y}{y!} \exp(-2\theta)\propto \theta^{x+y}\cdot\exp(-3\theta)$$ Maximising this quantity, is done by taking the logarithm and its derivative: $$\frac{\partial}{\partial \theta} \log L(\theta|x,y) = \frac{x+y}{\theta} - 3$$ Putting this equal to $$0$$ leads to the maximum likelihood estimator $$\hat\theta = \frac{1}{3} x + \frac{1}{3} y$$. Of course, you need to check that this indeed maximises the likelihood which follows from the second derivative being negative.