# 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 $6$.

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## 1 Answer

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.

• i don't have a statistics background.i see you have tried hard and enough to explain this but still could you tell me the final answer.i am sorry for asking you to this .but please if you could.. – user311790 Feb 9 '16 at 19:47
• @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
• Yes i got the answer its theta =2..thnku very much – user311790 Feb 9 '16 at 20:23