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 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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Quote:
 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. Tags estimator Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post calypso Advanced Statistics 5 December 29th, 2015 05:11 PM areth Probability and Statistics 0 January 12th, 2015 02:08 AM nadroj Advanced Statistics 7 September 12th, 2014 12:59 AM tchicory Advanced Statistics 1 August 10th, 2012 05:12 AM PerMortenIvar Advanced Statistics 1 March 8th, 2012 12:14 PM

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