Minimum variance unbiased estimator

[dvvv]

Homework Statement

Let \bar{X}1 and \bar{X}2 be the means of two independent samples of sizes n and 2n from an infinite population that has mean μ and variance σ^2 > 0. For what value of w is w\bar{X}1 + (1 - w)\bar{X}2 the minimum variance unbiased estimator of μ?
(a) 0
(b) 1/3
(c) 1/2
(d) 2/3
(e) 1

Homework Equations

If θ~ is unbiased for θ and
Var(θ~)= 1/E[(d log_e f (x)/dθ)^2] = 1/E[(dl(θ)/dθ)^2]
then θ~ is a minimum variance unbiased estimator of θ.

The Attempt at a Solution

E[w\bar{X}1 + (1 - w)\bar{X}2] = wμ + (1-w)μ = μ
So it's an unbiased estimator of μ.

I tried calculated the variance but I guess it's wrong.
Var[w\bar{X}1 + \bar{X}2 - w\bar{X}2] = w^2.σ^2/n + σ^2/n + w^2.σ^2/n = σ^2/n(2w^2 +1)

I think I have to use the formula above but I don't know how.
Thanks.

 

[Ray Vickson]

Homework Statement

Let \bar{X}1 and \bar{X}2 be the means of two independent samples of sizes n and 2n from an infinite population that has mean μ and variance σ^2 > 0. For what value of w is w\bar{X}1 + (1 - w)\bar{X}2 the minimum variance unbiased estimator of μ?
(a) 0
(b) 1/3
(c) 1/2
(d) 2/3
(e) 1

Homework Equations

If θ~ is unbiased for θ and
Var(θ~)= 1/E[(d log_e f (x)/dθ)^2] = 1/E[(dl(θ)/dθ)^2]
then θ~ is a minimum variance unbiased estimator of θ.

The Attempt at a Solution

E[w\bar{X}1 + (1 - w)\bar{X}2] = wμ + (1-w)μ = μ
So it's an unbiased estimator of μ.

I tried calculated the variance but I guess it's wrong.
Var[w\bar{X}1 + \bar{X}2 - w\bar{X}2] = w^2.σ^2/n + σ^2/n + w^2.σ^2/n = σ^2/n(2w^2 +1)

I think I have to use the formula above but I don't know how.
Thanks.

Note: use brackets, since otherwise your expressions are ambiguous. Better yet, use LaTeX, as you did in the first part of your post.

Your variance formula is incorrect. Since \bar{X}_1 and \bar{X}_2 are independent we have
\text{Var}(a \bar{X}_1 + b \bar{X}_2) = a^2 \text{Var}(\bar{X}_1) + b^2 \text{Var}(\bar{X}_2)
for any constants a, \: b. Use a = w and b = 1-w.

I don't know why you wanted to write (1-w)X as X - wX and then apply the variance formula, but you did it incorrectly. Using V(.) for the variance of a random varable, we have (using the fact that X and X are dependent): V(X - wX) = 1^2 V(X) + w^2 V(X) - 2 \cdot 1\cdot w\: \text{Cov}(X,X), and, of course, \text{Cov}(X,X) = V(X).

RGV

 

[dvvv]

I don't know why I did that either.

So is it right to say:
\text{Var}( \bar{X}_1) = \text{Var}(\bar{X}_2) = σ^2/n ?

How do I work out what w is?

 

[Ray Vickson]

In terms of sigma, what would be the variance of a sample mean of size 10 (that is, n=10)? What about for a sample of size 20? Now go back and read (carefully) the original question. Do you honestly still need me to answer?

RGV

 

[dvvv]

In terms of sigma, what would be the variance of a sample mean of size 10 (that is, n=10)? What about for a sample of size 20? Now go back and read (carefully) the original question. Do you honestly still need me to answer?

RGV

I guess it would be (σ^2)/10 and (σ^2)/20, so it's (σ^2)/n and (σ^2)/2n for \bar{X}_1 and \bar{X}_2, respectively.

I subbed that into
\text{Var}(a \bar{X}_1 + b \bar{X}_2) = a^2 \text{Var}(\bar{X}_1) + b^2 \text{Var}(\bar{X}_2)
and subbed in a and b, and I got
(σ^2(w^2+1))/2n

I still don't know how to get w, sorry...

 

[Ray Vickson]

I guess it would be (σ^2)/10 and (σ^2)/20, so it's (σ^2)/n and (σ^2)/2n for \bar{X}_1 and \bar{X}_2, respectively.

I subbed that into
\text{Var}(a \bar{X}_1 + b \bar{X}_2) = a^2 \text{Var}(\bar{X}_1) + b^2 \text{Var}(\bar{X}_2)
and subbed in a and b, and I got
(σ^2(w^2+1))/2n

I still don't know how to get w, sorry...

No wonder: you have made a serious algebraic blunder, so you end up with the wrong expression.

RGV

 

[dvvv]

Tried again and got:
((3 w^2 - 2 w + 1) σ^2)/(2 n)
which is correct according to wolfram alpha http://www.wolframalpha.com/input/?i=w^2(σ^2/n)+(1-w)^2(σ^2/2n)

What now?

 

[Ray Vickson]

Tried again and got:
((3 w^2 - 2 w + 1) σ^2)/(2 n)
which is correct according to wolfram alpha http://www.wolframalpha.com/input/?i=w^2(σ^2/n)+(1-w)^2(σ^2/2n)

What now?

You should not need a powerful tool to do such simple, high-school algebra, but since you have already done it, OK. The question asked you to find the value of w that minimizes the variance. So, that is what you need to do. At this point I am signing off this thread.

RGV

 

[dvvv]

I just used it to confirm I was right. Thanks for your help.