1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ You want to find the likelihood when $\theta=1$ and the likelihood when $\theta=2$. Then compare $\endgroup$ – Henry Jan 28 '17 at 10:12
  • $\begingroup$ Maybe there is a typo and it should be $\theta\in (1,2)$ $\endgroup$ – callculus Jan 28 '17 at 10:21
  • $\begingroup$ @callculus Its not a typo, it is given in the problem. $\endgroup$ – Sachchidanand Prasad Jan 28 '17 at 10:23
  • $\begingroup$ I mean typo in the problem. $\endgroup$ – callculus Jan 28 '17 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. $\endgroup$ – Sachchidanand Prasad Jan 28 '17 at 10:26
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You have forgotten to write $x!$ in the denominator, that will give you $e^{-6}/8$, and $64 e^{-6}$. $\endgroup$ – Sachchidanand Prasad Feb 4 '17 at 0:41
  • $\begingroup$ You are right. My bad. I've edited the answer. $\endgroup$ – V. Vancak Feb 4 '17 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.