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A single observation is made from a poisson distribution with unknown mean $\lambda \geq 0$ However any value greater than 2 has been rounded down to 2. This we have the observed value of a single random variable X having distribution depending on $\lambda $ given by; $\ P(X=0) = e^{-\lambda}. P(X=1) = \lambda e^{-\lambda} P(X=2) = 1 - (1+\lambda)e^{-\lambda}$
Parameterise the distribution by $\ \theta = e^{-\lambda} \in (0,1] $ Show that there is a unique unbiased estimator of $\theta$.

So I parameterise it; $\ P(X=0) = \theta$ $\ P(X=1) = -\theta log\theta$ $\ P(X=2) = 1-(1-log\theta)\theta$

But I have no idea how to show there is a unique unbiased estimator. Also this is not a homework question, it is a practice paper question.

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Just a suggestion, I haven't checked if it works: $\theta$ is a function of $X$. Let $\hat \theta$ and $\tilde \theta$ be two estimators of $\theta$, and enumerate $X$ ($\hat \theta$ and $\tilde \theta$ have unique values for all 3 possible X). –  akkkk May 27 '12 at 7:46
    
Oops, what I meant, of course, was that any /estimator/ of $\theta$ is a function of $X$. –  akkkk May 27 '12 at 8:09
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Go back to the definitions: an estimator is a function of the observation, hence let us call $u_0$ the estimate if the observation is $0$, $u_1$ if the observation is $1$ and $u_2$ if the observation is $2$. There is no bias if $$ u_0\mathrm e^{-\lambda}+u_1\lambda\mathrm e^{-\lambda}+u_2(1-(1+\lambda)\mathrm e^{-\lambda})=\mathrm e^{-\lambda}. $$ This identity should hold for every $\lambda$ and the estimates $u_0$, $u_1$ and $u_2$ should be independent on $\lambda$, hence one asks that $$ u_0+u_1\lambda+u_2(\mathrm e^{\lambda}-1-\lambda)=1, $$ uniformly over $\lambda\gt0$. Thus, $\underline{\qquad\qquad\qquad}$.

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I don't understand, $\ u2(1−(1+\lambda)e^{-\lambda}$ = $\ u2(1-e^{-\lambda}-\lambda e^{-\lambda}$ ? So how have you divided by $\ e^{-\lambda}$ ? –  Rosie May 27 '12 at 9:20
    
Yes. Or, equivalently, multiplied everything by $e^\lambda$. –  Did May 27 '12 at 10:26
    
I still don't understand how you have divided by e^-lambda? you would end up with (u2/ $\ e^{-\lambda})-1-\lambda $ –  Rosie Jun 2 '12 at 19:07
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@SeyhmusGüngören Oh yeah? Please share your thoughts. –  Did Jan 9 '13 at 11:25
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@SeyhmusGüngören An estimate is the image of the observations by some function. The function itself is an estimator. Hence the phrase you object to is correct while the subsequent uses of estimator, which you did not object to, can be replaced, if one is nitpicky, by estimate. –  Did Jan 9 '13 at 15:31
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