# Maximum likelihood estimator for a Poisson random variable given that the parameter is discrete.

Let $x_1 = x_2=x_3 = 1, x_4 = x_5 = x_6 = 2$ be a random sample from a Poisson random variable with mean $\theta$, where $\theta\in \{1,2\}$. Then, the maximum likelihood estimator of $\theta$ is equal to ______.

Source.

Please someone explain. I calculated that the MLE is $1.5$, but I am not confident with my answer. I just summed up the numbers and divided it by $6$.

• We strongly encourage that you format your questions and work in progress. Formatting tips here. – Em. Feb 9 '16 at 19:32

If $\theta \in \{1, 2\}$, then how can you conclude the maximum likelihood is $\hat \theta = 1.5$? That's not even one of the choices you are allowed.
Here is the simplest thing you can do: the joint probability mass of the sample $\boldsymbol x = (1, 1, 1, 2, 2, 2)$, given $\theta$, is $$f_{\boldsymbol X}(\boldsymbol x \mid \theta) = \Pr[X = 1]^3 \Pr[X = 2]^3 = \left( e^{-\theta} \frac{\theta^1}{1!} \right)^3 \left( e^{-\theta} \frac{\theta^2}{2!}\right)^3.$$ This is proportional to your likelihood. Now, for which value of $\theta \in \{1,2\}$ is the resulting likelihood greater? That is to say, if $$\mathcal L(\theta \mid \boldsymbol x) = f_{\boldsymbol X}(\boldsymbol x \mid \theta),$$ which $\theta$ will give you a larger $\mathcal L$? That choice will be your MLE given the sample. It's either $\hat \theta = 1$, or $\hat \theta = 2$. Or, if both give you the same value for $\mathcal L$, then either choice will be the MLE.
• @user311790 All you need to do is plug in $\theta = 1$ on the right-hand side of the first formula, then compare that number with the one you get if you plug in $\theta = 2$ instead. Which $\theta$ gives you a larger value? That's it. – heropup Feb 9 '16 at 20:21