let X be a geometric(p)
find unbiased estimate c for p.
please help.
can und an unbiased estimate for 1/p but don't know how to find for p.
please help. thanks.
Lets start with the probability table for x. This is a geometric(p) distribution so the probabilities are:
$\displaystyle \begin{array}{c|ccccccc}x&1&2&3&...\\prob&p&(1-p)p&(1-p)^2 ...\\\hline\end{array}$
Notice that we have a value of p occuring in the probability table at x=1. We will exploit this to get an unbiased estimate later.
suppose you have 1 realisation of x and your estimator is c(x)
For your estimator to be unbiased you require:
$\displaystyle E(c(x)) = p$
$\displaystyle \sum c(x) \times p(1-p)^{x-1} =p$
where the summation is taken over all possible values of x (1,2,3,4,5,6...)
Now, consider the following function:
c(x)=1 if x=1
c(x)=0 otherwise
The expected value of our function is then
$\displaystyle \sum c(x) \times p(1-p)^x =p$
$\displaystyle =(f(1)\times p) + (f(2)\times p(1-p)) + (f(3)\times p(1-p)^2) +... $
$\displaystyle =(1 \times p) + (0 \times p(1-p)) + (0 \times p(1-p)^2) +... $
$\displaystyle =p$
as required
So your estimator is $\displaystyle c(x)$ where c(x) was defined above.
In reality, this estimator is no practical use. But it is an unbiased which is all the question asked for. I did assume that your sample is a single realisation from the distribution. if you have a sample with multiple data points, just discard all but one of them and the proceedure still works.
PS: not bad for an economist, eh?
Thank you MathEagle and SpringFan25 !
MathEagle: Computing the expectation of E(1/x) seems quite hard since
E(1/x)= sum[[p(1-p)^(x-1)]/x]. Not sure how to solve this.
SpringFan25: I didn't think of that, very smart . However, I don't think this is going to be very applicable in my problem.
I am trying to get the UMVUE for p when X~geometric(p).
X is a complete sufficient statistic.
if use I(c(x)=1) as my unbiased estimate,
then T*(x)=E( I(c(x))|x) is UMVUE for p
but E(I(c(x)=1)|x)=P(c(x)=1)... which will give me some function of p?
but p is what we are trying to estimate, so this is probably not a good solution
yea, I was thinking about some way to take the derivative to obtain
that sum. But it doesn't look good.
Maybe try to integrate wrt p
$\displaystyle \int \sum_{x=1}^{\infty} (1-p)^{x-1} dp = -\sum_{x=1}^{\infty} {(1-p)^x\over x}$
So $\displaystyle \sum_{x=1}^{\infty} {(1-p)^x\over x}=-\ln p$
and $\displaystyle p\sum_{x=1}^{\infty} {(1-p)^x\over x}=-p\ln p$ which doesn't help.
Isn't $\displaystyle I(X=1)$ the UMVUE then?
Do you have any idea of how to find an unbiased estimate for the negative binomial?
wikipedia says a UMVUE for p for Y~negative binomial is
im trying to prove this result.
since the geometric is a special case of the negative binomial, do you think I should get a similar result for the geometric?