# statistical estimators

• Feb 2nd 2010, 11:40 PM
nycprincess
statistical estimators
I would really appreciate some help!

Suppose that a certain drug is to be administered to two different types of patients A & B. It is known that the mean response of patient of type A is the same as the mean response of patients of type B, but the common value Q of this mean is unknown and must be estimated.

It is also known that the variance of the response of patients of type A is four times as large as variance of the response of patients of type B.

Let X1...Xm be responses of random sample of m patients of type A.
Let Y1...Yn be responses of random sample of n patients of type B.

Estimator Qhat = aX(bar-m_ + (1-a)Y(bar-n)

a) For what values of a, m, and n is Qhat an unbiased estimator of Q?
b) For fixed values of m and n, what value of a yields an unbiased estimator with minimum variance?

Thank you (sorry the question is so long).
• Feb 2nd 2010, 11:56 PM
matheagle
I don't understand what ...'common value Q of this mean' means.
• Feb 3rd 2010, 12:06 AM
nycprincess
Quote:

Originally Posted by matheagle
I don't understand what ...'common value Q of this mean' means.

in the problem the Q is a theta. It's supposed to be the mean of type A & B together.
• Feb 3rd 2010, 12:13 AM
matheagle
theta is just a greek letter, your highness.
do you mean that the expected value from each distribution is Q?
As in E(X)=E(Y)=Q?
• Feb 3rd 2010, 12:18 AM
matheagle
IF $E(X_i)=Q=E(Y_j)$ then

$E\left(a\bar X_n+(1-a)\bar Y_m\right)=aQ+(1-a)Q=Q$ for all a,n,m.

$V\left(a\bar X_n+(1-a)\bar Y_m\right)=a^2{V(X)\over n}+(1-a)^2{V(Y)\over m}$

$=a^2{4V(Y)\over n}+(1-a)^2{V(Y)\over m}$

$=V(Y)\left({4a^2\over n}+{(1-a)^2\over m}\right)$

So, minimize $\left({4a^2\over n}+{(1-a)^2\over m}\right)$ wrt a.