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suppose $X_1,X_2\ldots,X_n$ be a random sample of distribution with probability density function $$f(\theta, x) = \begin{cases} \theta &\text{if } x=-1 \\ (1-\theta)^2 \theta^x & \text{if } x=0,1,2,\ldots. \end{cases}$$ if $r_n$ be the ratio members of sample are equal to $(-1)$, how can i find MLE of parameter $\theta$

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the likelihood is $\prod_{X_i \ne -1}(1-\theta)^2\theta^{X_i}\prod_{X_i = -1}\theta $ and its really not so bad – mike Apr 26 '12 at 14:45
up vote 0 down vote accepted

$$P(X_1,\ldots, X_n|\theta) = \theta^{\#[X = -1]} (1 - \theta)^{2n(1 - r_n)} \theta^{\sum_{i \geq 0} \#[X = i]i} = \theta^{2\#[X = -1]} (1 - \theta)^{2n(1 - r_n)} \theta^{\sum_{i \geq -1} \#[X = i]i}$$ Therefore, $$P(X_1,\ldots, X_n|\theta) = \theta^{2nr_n} (1 - \theta)^{2n(1 - r_n)} \theta^{n\bar{X}} = (1 - \theta)^{2n(1 - r_n)} \theta^{2nr_n + n\bar{X}}$$ and $$\log P(X_1,\ldots, X_n|\theta) = 2n(1 - r_n)\log(1 - \theta) + (2nr_n + n\bar{X})\log(\theta)$$

Now, maximize this w.r.t $\theta$ subject to the usual constraints of probabilities between 0 and 1 and PDF summing to 1.

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Sorry but this formula for the likelihood is wrong and maximizing it will not yield the MLE. – Did Apr 29 '12 at 8:50
@Didier Feel free to point out the error. – TenaliRaman Apr 29 '12 at 17:40
Feel free to read my answer. – Did May 3 '12 at 14:52
@Didier How is my answer different from yours? – TenaliRaman May 3 '12 at 15:16
Hint: check the powers of $(1-\theta)$. – Did May 3 '12 at 18:22

Hint: Call $n$ the size of the sample, $s$ its sum, and $z$ the number of $-1$s in the sample. Then the likelihood $L(\theta)$ of the sample is a function of $(\theta,n,z,s)$. Solving the equation $L'(\hat\theta)=0$ yields the MLE $\hat\theta$ as a (rational) function of $(n,z,s)$, or, if one prefers, as a (rational) function of $(r,m)$ with $r=z/n$ the proportion of $-1$ and $m=s/n$ the mean of the sample.

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