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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 $\displaystyle un = a\bar{X} + (1a)\bar{Y}, 0<= a <= 1$ 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 CramerRao 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  
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0  Quote:
$\displaystyle var(un) = var(a\bar{X} + (1a)\bar{Y})$ $\displaystyle var(un) = a^2var(\bar{X}) + (1a)^2var(\bar{Y})$ $\displaystyle var(un) = a^2\frac{σ1^2}{n} + (1a)^2\frac{σ2^2}{n}$ To find the minimum of var(un), what should I differentiate with respect to?  
January 6th, 2016, 01:52 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,524 Thanks: 587  Quote:
 
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 CranmerRao lower bound for this.

January 7th, 2016, 01:44 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,524 Thanks: 587  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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