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?

  • 2
    $\begingroup$ You want to find the likelihood when $\theta=1$ and the likelihood when $\theta=2$. Then compare $\endgroup$
    – Henry
    Jan 28, 2017 at 10:12
  • $\begingroup$ Maybe there is a typo and it should be $\theta\in (1,2)$ $\endgroup$ Jan 28, 2017 at 10:21
  • $\begingroup$ @callculus Its not a typo, it is given in the problem. $\endgroup$ Jan 28, 2017 at 10:23
  • $\begingroup$ I mean typo in the problem. $\endgroup$ Jan 28, 2017 at 10:24
  • $\begingroup$ 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. $\endgroup$ Jan 28, 2017 at 10:26

1 Answer 1


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.

  • $\begingroup$ You have forgotten to write $x!$ in the denominator, that will give you $e^{-6}/8$, and $64 e^{-6}$. $\endgroup$ Feb 4, 2017 at 0:41
  • $\begingroup$ You are right. My bad. I've edited the answer. $\endgroup$
    – V. Vancak
    Feb 4, 2017 at 1:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .