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Suppose $X$ is a random variable of distribution with probability density function

$$ f(x) = \begin {cases} \theta (1-\theta) & \text{}\ x=-1 \\ \theta^x (1-\theta)^{2-x} & \text{} \ x=0,1,2 \end {cases}, \qquad\qquad0<\theta<1 $$

Determine the MLE of parameter $\theta$

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Seems to be a slight modification of the previous question. Again f is a discrete probability distribution and not a density. –  Michael Chernick Sep 11 '12 at 17:23
$L(\theta)$ is the same when $x=-1$ as when $x=1$. –  Michael Hardy Sep 11 '12 at 18:03
....so the absolute value of the observed $x$ is a sufficient statistic. Figure out what its distribution is. –  Michael Hardy Sep 11 '12 at 18:03
I understand you refuse absolutely to conform your other question to the norms of the site... Voting to close. –  Did Sep 11 '12 at 19:00
It seems clear that you are upset with this OP and rightly so. I no longer give him the benefit of the doubt but when i made the comment I assumed that he might not know the rules and might not understand your criticism of him. But he continued to do this after my comment. So there can be no excuse. you did the right thing did (Is did short for didier?) –  Michael Chernick Sep 14 '12 at 21:44
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closed as not a real question by Did, t.b., William, Thomas, J. M. Sep 29 '12 at 10:38

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1 Answer

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Let k$_1$= # of x=1 k$_2$=# of x=2 k$_3$=# of x=0 and k$_4$=# of x=-1.

The likelihood function is then

[θ (1- θ)]$^k$1 [θ$^2$]$^k$2 [(1-θ)$^2$]$^k$3 [θ (1-θ)]$^k$4

So ln(likelihood)=(k$_1$ + 2k$_2$+k$_4$) ln θ +(k$_1$ + 2k$_3$+k$_4$)ln(1-θ)

Taking the derivative with respect to θ of ln(likelihood) and setting it to 0 gives:

(k$_1$ + 2k$_2$+k$_4$)/θ = (k$_1$ + 2k$_3$+k$_4$)/(1-θ) so

mle for θ = (k$_1$ +2k$_2$+k$_4$)/[2(k$_1$ + k$_2$+ k$_3$+k$_4$)]

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