Is W an Unbiased Estimator of Mu?

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SUMMARY

The discussion centers on the unbiasedness of the estimator W, defined as W = (1/8)Y_1 + (1/8)Y_2 + (1/4)Y_3 + (1/2)Y_4, in relation to the population mean mu. The expected value E(W) is confirmed to equal mu, establishing W as an unbiased estimator. The variance V(W) is calculated using the weights and the variance sigma^2 of the IID random variables Y_i. The preference between the estimators Y^bar and W is determined based on their variances, with the estimator yielding the lower variance being favored.

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Homework Statement



Let Y_1, Y_2, Y_3, Y_4 be IID RV from a population with mean mu and variance sigma^2. Let Y^bar = .25(Y_1+Y_2+Y_3+Y_4) denote the average of these four RV's.

1)What are the expected value and variance of Y^bar in terms of mu and sigma^2

2)Consider a different estimator of mu... W = (1/8)Y_1 + (1/8)Y_2 + (1/4)Y_3 + (1/2)Y_4
This is an example of a weighted average of Y_i. Show that W is also an unbiased estimator of mu and find V(W)

3)Based on your answer, which estimator of mu do you prefer, Y^bar or W.

Homework Equations





The Attempt at a Solution



Welp I'm not really sure where to start. I know that Y^bar is a sample mean from 4 samples while mu is the population mean based on every sample, but I do not know how to represent one in terms of the other
 
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joemama69 said:

Homework Statement



Let Y_1, Y_2, Y_3, Y_4 be IID RV from a population with mean mu and variance sigma^2. Let Y^bar = .25(Y_1+Y_2+Y_3+Y_4) denote the average of these four RV's.

1)What are the expected value and variance of Y^bar in terms of mu and sigma^2

2)Consider a different estimator of mu... W = (1/8)Y_1 + (1/8)Y_2 + (1/4)Y_3 + (1/2)Y_4
This is an example of a weighted average of Y_i. Show that W is also an unbiased estimator of mu and find V(W)

3)Based on your answer, which estimator of mu do you prefer, Y^bar or W.

Homework Equations





The Attempt at a Solution



Welp I'm not really sure where to start. I know that Y^bar is a sample mean from 4 samples while mu is the population mean based on every sample, but I do not know how to represent one in terms of the other

Doesn't your text have formulas for the expected value and variance of a sum (linear combination) of IID random variables?
 
LCKurtz said:
Doesn't your text have formulas for the expected value and variance of a sum (linear combination) of IID random variables?

Sure does, but how to I express mean and variance IN TERMS of mu and sigma^2

Heres what I got outa the book...

so Y^bar = Sum(Y_i/n)

E(Y^bar) = E(Sum(Y_i/n)) = (1/4)Sum(E(Y_i)) = (1/4)Sum(mu)

V(Y^bar) = V(Sum(Y_i/n)) = (1/16)V(Sum(Y_i)) = (1/16)(Sum(Y_i) + 2 SumSumcov(Y_i, Y_j)) = (1/16)Sum(V(Y_i)) = (1/16)Sum(sigma^2)

So that's part one. Part two I am lossed with the weighted average parts. How would I go about that
 
joemama69 said:
Sure does, but how to I express mean and variance IN TERMS of mu and sigma^2

Heres what I got outa the book...

so Y^bar = Sum(Y_i/n)

E(Y^bar) = E(Sum(Y_i/n)) = (1/4)Sum(E(Y_i)) = (1/4)Sum(mu)

Hopefully you are using proper notation on your worksheet. When you write (1/4)sum(mu) what you really mean is$$
\frac 1 4\sum_{i=1}^4\mu$$What does that equal when you write it out?
V(Y^bar) = V(Sum(Y_i/n)) = (1/16)V(Sum(Y_i)) = (1/16)(Sum(Y_i) + 2 SumSumcov(Y_i, Y_j)) = (1/16)Sum(V(Y_i)) = (1/16)Sum(sigma^2)

So that's part one. Part two I am lossed with the weighted average parts. How would I go about that

Same comment about the variance sums. And do you need the covariance for IID random variables?
 
Ya sorry about the notation, I'm not up to speed with this syntax

So now my question is regarding part two... How would I find E(W) and V(W) when W = the sum of the weighted averages of Y (as apposed to part 1 where Y^bar was just the sum of the Y's divided by n = 4). Basically I'm confused what to do with the weights i.e. 1/8, 1/8, 1/4, 1/2?
 
joemama69 said:
Ya sorry about the notation, I'm not up to speed with this syntax

So now my question is regarding part two... How would I find E(W) and V(W) when W = the sum of the weighted averages of Y (as apposed to part 1 where Y^bar was just the sum of the Y's divided by n = 4). Basically I'm confused what to do with the weights i.e. 1/8, 1/8, 1/4, 1/2?
You have ##W=\sum c_iY_i## with ##Y_i## IID. Look in your text for the mean and variance formulas for such sums.
 

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