The number of persons coming through a blood bank until the first person with type A blood is a random variable Y with a geometric distribution. If p denotes the probability that any one randomly selected person will possess type A blood, then E[Y ] = 1/p and V [Y ] = (1-p)/p2.
Find a function of Y that is an unbiased estimator of V [Y ].
Then suggest how to form a two-standard error bound on the error of estimation when Y is used to estimate 1/p.

a. ½(Y² - Y) is an unbiased estimator of the variance.

b. 2 Standard Error = 2√(Y²/2 - Y/2)

Explanation:

Given

E(Y) = 1/p

V(Y) = (1 - p)/p²

Simplifying V(Y)...

V(Y) = 1/p² - 1/p

Because V(Y) = 1/p² - 1/p and E(Y) = 1/p,

We'll guess that there might be an unbiased estimator of shape

aY² + bY.

Solving E(aY² + bY)...

First, E(Y²) = V(Y) + (E(Y))²

Substituting the values of V(Y) and E(Y);

E(Y²) = 1/p² - 1/p + (1/p)²

E(Y²) = 1/p² - 1/p + 1/p²

E(Y²) = 2/p² - 1/p

With the above,

E(aY² + bY) is then equal to

E(aY² + bY) = 2a/p² - a/p + b/p

The above equation is equal to V(Y), if and only if

a = ½ and b = -½

Checking

Let E(aY² + bY) = V(Y)

i.e.

2a/p² - a/p + b/p = 1/p² - 1/p

Multiply through by l

2a/p - a + b = 1/p - 1

Comparing right hand side to left hand side

2a/p = 1/p Equation 1

And

- a + b = - 1 Equation 2

Solving Equation 1 (Multiply both sides by p)

2a = 1

So, a = ½

Substitute ½ for a in Equation 2

-½ + b = -1

b = -1 + ½

b = -½

End

Thus ½(Y² - Y) is an unbiased estimator of the variance.

b.

2 Standard Error is given by

2√V(Y)

= 2√(1/p² - 1/p)

= 2√(Y²/2 - Y/2)

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