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.

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- May 29th 2010, 09:07 AMlyannunbiased estimate for p in geometric(p)
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. - May 29th 2010, 02:22 PMmatheagle
Since E(X)=p.

Obtain E(1/X), which won't be p, but it's a start. - May 29th 2010, 06:54 PMSpringFan25
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? - May 29th 2010, 07:42 PMlyann
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 - May 29th 2010, 07:58 PMSpringFan25
yes, my estimate is definately not the UMVUE, but you didn't ask for that :P

- May 29th 2010, 08:00 PMmatheagle
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? - May 30th 2010, 02:38 AMlyann
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). - May 30th 2010, 02:50 AMSpringFan25
horay then

- May 30th 2010, 08:33 AMlyannneg 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

http://upload.wikimedia.org/math/b/9...da4b0ecb78.png

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?