Estimator: Rao-Blackwell Watch

lilman91
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X_1,...,X_n iid discrete random variables with

P(X_1=r)=p^{r-1}(1-p) for r=1,2,....

Determine a one-dimensional sufficient statistic for p and then by first finding a simple unbiased estimate for p, determined an unbiased estimate for p which is a function of T.

My attempt:

Clearly the sufficient statistic we are looking for is T=\sum_{i=1}^{n}X_i. If I denotes the indicator function, then we can see that \tilde{\theta}=1-I_{X_1=1} is an unbiased estimator for p because E(\tilde{\theta})=1-P(X_1=1) = 1-(1-p)=p .

So using Rao-Blackwell, we want to find \displaystyle E(\tilde{\theta}|\sum_{i=1}^nX_i  =T) = 1- P(X_1=1 | \sum_{i=1}^nX_i=T) = 1 - \frac{P(X=1)P(\sum_{i=2}^nX_i=T)  }{P(\sum_{i=1}^nX_i=T)}

Now the pgf of X_1 is \frac{(1-p)t}{(1-pt)} so the pgf of \sum_{i=1}^nX_i is \frac{(1-p)^nt^n}{(1-pt)^n}

If we expand out that pgf using binomial expansion and consider the coefficient of t^T, we find that \displaystyle P(\sum_{i=1}^nX_i = T) =p^{T-n} (1-p)^n \frac{n(n+1)(n+2)....(T-1)}{(T-n)!}

By replacing n with n-1 we easily get an expression for P(\sum_{i=2}^nX_i = T) . And putting that into what we have above gives:

E(\tilde{\theta}|\sum_{i=1}^nX_i  =T)= 1-  \frac{(n-1)p}{T-n+1}

My problem is that this estimator should NOT depend on p but I don't see where I've gone wrong. There should be another p to cancel it out but I can't find where it should be! Any help would be greatly appreciated.
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lilman91
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anything?
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Mark13
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(Original post by lilman91)
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I think the problem is on the line starting "Using Rao-Blackwell..". The conditional probability should be

\displaystyle \mathbb{P}(X_1=1 | \sum_{i=1}^n = T)=\frac{\mathbb{P}(X_1=1) \mathbb{P} (\sum_{i=2}^n = T-1)}{\mathbb{P}(\sum_{i=1}^n = T)}
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lilman91
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(Original post by Mark13)
I think the problem is on the line starting "Using Rao-Blackwell..". The conditional probability should be

\displaystyle \mathbb{P}(X_1=1 | \sum_{i=1}^n = T)=\frac{\mathbb{P}(X_1=1) \mathbb{P} (\sum_{i=2}^n = T-1)}{\mathbb{P}(\sum_{i=1}^n = T)}

oh haha - stupid! thanks
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