My Math Forum Estimator for mean

 January 2nd, 2016, 11:07 PM #1 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Estimator for mean Not looking for the answer here, but just some guidance on how to start. If I have two independent samples from the same normal population where u is unknown: 1. X1,..,Xn ~ N(u,σ1^2), mean $\displaystyle \bar{X}$, σ1^2 is known 2. Y1,...,Yn ~ N(u,σ2^2), mean $\displaystyle \bar{Y}$, σ2^2 is known $\displaystyle un = a\bar{X} + (1-a)\bar{Y}, 0<= a <= 1$ I want to show that un, which is an unbiased estimator for u, is a minimum variance unbiased estimator when $\displaystyle a = \frac{σ2^2}{σ1^2 + σ2^2}$ I have tried the following using the Cramer-Rao lower bound to prove estimator is MVUE using var(un). I(u) = 1 However, I can not get any of the terms to cancel; is it the correct strategy? Last edited by skipjack; January 5th, 2016 at 08:58 AM.
 January 5th, 2016, 08:10 AM #2 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 Just compute Var(un), which is a function of a, and find its minimum, the value of a where the derivative is null. Last edited by skipjack; January 5th, 2016 at 08:58 AM.
January 6th, 2016, 01:26 PM   #3
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 Originally Posted by mehoul Just compute Var(un), which is a function of a, and find its minimum, the value of a where the derivative is null.
To work out the var(un), I did:

$\displaystyle var(un) = var(a\bar{X} + (1-a)\bar{Y})$
$\displaystyle var(un) = a^2var(\bar{X}) + (1-a)^2var(\bar{Y})$
$\displaystyle var(un) = a^2\frac{σ1^2}{n} + (1-a)^2\frac{σ2^2}{n}$

To find the minimum of var(un), what should I differentiate with respect to?

January 6th, 2016, 01:52 PM   #4
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 Originally Posted by calypso To work out the var(un), I did: $\displaystyle var(un) = var(a\bar{X} + (1-a)\bar{Y})$ $\displaystyle var(un) = a^2var(\bar{X}) + (1-a)^2var(\bar{Y})$ $\displaystyle var(un) = a^2\frac{σ1^2}{n} + (1-a)^2\frac{σ2^2}{n}$ To find the minimum of var(un), what should I differentiate with respect to?
a

 January 6th, 2016, 02:23 PM #5 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Great, thanks when I differentiate with respect to a, I get the correct value for a. How do I know this is a MVUE? I thought I needed to use Cranmer-Rao lower bound for this.
January 7th, 2016, 01:44 PM   #6
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 Originally Posted by calypso Great, thanks when I differentiate with respect to a, I get the correct value for a. How do I know this is a MVUE? I thought I needed to use Cranmer-Rao lower bound for this.
The expression is a simple quadratic polynomial and the coefficient of the square term is positive. In that case the function is a parabola and the minimum as at the point where the derivative = 0. The probability theory background is irrelevant.

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