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The number of persons coming through a blood bank until the first person with type A blood is found is a random variable Y with a geometric distribution , i.e.:

p(y) = (1-p)y-1 (p)
0 ≤p ≤ 1

If p denotes the probability that any one randomly selected person will posses type A blood, then E(Y)= 1/p and V(Y) = (1-p)/p² . Find a function that is an unbiased estimator of V(y). .

This is what I have gotten so far

Chris
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  • Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need. – Did Sep 25 '13 at 15:40
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    http://math.stackexchange.com/q/505140/ Same class? – Did Sep 26 '13 at 07:22

2 Answers2

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Hint: Use E[Y^2]=E[Y]^2 +V[Y] to find the estimator of the variance. You can utilize the fact that you can write V[y] as 1/p^2 - 1/p

Hope this helps. You are on the right track.

SamHaim
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QUESTION: Y is Geometric. Find an unbiased estimator of V(Y)

The immediate simple solution is:

For any distribution, an unbiased estimator of the population variance is the sample variance (2nd h-statistic): $\frac{1}{n-1}\sum _{i=1}^n \left(Y_i-\bar{Y}\right){}^2$

wolfies
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