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The number of person coming through a blood bank until the first person with type A is found is a random variable Y with gemetric 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)/p^2. Find a function of Y that is n unbiased estimator of V (y).

I know how to justfy which estimators are unbised when they are given, but do t know how to find unbiased estimators.

Thank you in advance!

Lovelydia
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1 Answers1

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Because $\text{Var}(Y)=\frac{1}{p^2}-\frac{1}{p}$, and $E(Y)=\frac{1}{p}$, we will guess that there might be an unbiased estimator of shape $aY^2+bY$.

Calculate $E(aY^2+bY)$, to see whether the guess was reasonable.

We have $E(Y^2)=\text{Var}(Y)+(E(Y))^2$. Thus $E(Y^2)=\frac{2}{p^2} -\frac{1}{p}$. It follows that $$E(aY^2+bY)=\frac{2a}{p^2}-\frac{a}{p}+\frac{b}{p}.$$ This is equal to $\frac{1}{p^2}-\frac{1}{p}$ if $a=\frac{1}{2}$ and $b=-\frac{1}{2}$. Thus $\frac{1}{2}(Y^2-Y)$ is an unbiased estimator of the variance.

André Nicolas
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