Thread: unbiased estimate for p in geometric(p)

1. unbiased estimate for p in geometric(p)

let X be a geometric(p)
find unbiased estimate c for p.

can und an unbiased estimate for 1/p but don't know how to find for p.

2. Since E(X)=p.
Obtain E(1/X), which won't be p, but it's a start.

3. 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?

4. 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

5. yes, my estimate is definately not the UMVUE, but you didn't ask for that :P

6. 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?

7. I think you are right!

I(x=1) would be umvue because it is unbiased and a function of the complete sufficient statististic. (and therefore I don't need to take E(I(x=1)|x).

8. horay then

9. neg binomial

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?

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unbiased estimator of p geometric distribution

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