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January 2nd, 2016, 11:07 PM   #1
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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.
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January 5th, 2016, 08:10 AM   #2
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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.
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January 6th, 2016, 01:26 PM   #3
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Quote:
Originally Posted by mehoul View Post
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?
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January 6th, 2016, 01:52 PM   #4
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Quote:
Originally Posted by calypso View Post
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
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January 6th, 2016, 02:23 PM   #5
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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.
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January 7th, 2016, 01:44 PM   #6
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Quote:
Originally Posted by calypso View Post
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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