# MLE of poisson random variable.

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...

What I know is that, the MLE of poisson distribution is given by $$\hat{\theta}_{MLE}=\sum_{i=1}^n\frac{X_i}{n} .$$ If we evaluate here then $\hat{\theta}_{MLE}$ is coming $1.5$, which is not there in the range of $\theta$. Then how will I find the MLE in this case?

• You want to find the likelihood when $\theta=1$ and the likelihood when $\theta=2$. Then compare – Henry Jan 28 '17 at 10:12
• Maybe there is a typo and it should be $\theta\in (1,2)$ – callculus Jan 28 '17 at 10:21
• @callculus Its not a typo, it is given in the problem. – Sachchidanand Prasad Jan 28 '17 at 10:23
• I mean typo in the problem. – callculus Jan 28 '17 at 10:24
• Okay but answer is given 2. And it is the problem asked in some competitive exam, GATE. So there is a very less probability that this is a typos. – Sachchidanand Prasad Jan 28 '17 at 10:26

You have to decide under which one of the two possible parameters your sample is more probable (literally). Under $\theta = 1$ you have $$\prod_{i=1}^3P(X_i=1)\prod_{i=4}^6P(X_i=2)=(e^{-1}/1!)^3(e^{-1}/2!)^3.$$ Under $\theta = 2$ you have $$\prod_{i=1}^3P(X_i=1)\prod_{i=4}^6P(X_i=2)=(e^{-2}2^1/1!)^3\times (e^{-2}2^2/2!)^3$$ so check which expression is greater.
• You have forgotten to write $x!$ in the denominator, that will give you $e^{-6}/8$, and $64 e^{-6}$. – Sachchidanand Prasad Feb 4 '17 at 0:41