# Estimate variance of the mean

1. Nov 13, 2013

### emperorvinayak

1. The problem statement, all variables and given/known data
Suppose that:
E(X1) = μ, Var(X1) = 7,
E(X2) = μ, Var(X2) = 13,
E(X3) = μ, and Var(X3) = 20,

and consider the point estimates:

μˆ1 = X1/3 + X2/3 + X3/3
μˆ2 = X1/4 + X2/3 + X3/5
μˆ3 = X1/6 + X2/3 + X3/4 + 2

(a) Calculate the bias of each point estimate. Is any one of
them unbiased?
(b) Calculate the variance of each point estimate. Which
one has the smallest variance?
(c) Calculate the mean square error of each point estimate.
Which point estimate has the smallest mean square
error when μ = 3?

2. Relevant equations
Var(μ) = $\frac{1}{n-1}$ $\sum$(xi -$\bar{X}$)2 from i=1 to n for unbiased
and
Var(μ) = $\frac{1}{n}$ $\sum$(xi -$\bar{X}$)2 from i=1 to n for biased

3. The attempt at a solution

I found out part a) pretty easily: I just replaced all the Xn values with μ and if it returned μ again, it was unbiased.

The problem I'm having with, is part B. I just don't know what to plug into the forumulae displaed above!
I tried plugging in the variance for each of the X values, subtracting what I assumed to be the mean of those values and squaring what I got. Then I divided it by n for unbiased (b and c)
This is what I did for b) as an example:

$\frac{1}{3}$($\frac{7}{3}$+$\frac{13}{3}$+$\frac{20}{3}$)2

But the answer I'm getting is wrong. It was a wild shot anyway. I tried watching a few videos on Youtube to understand the concept and I think I get it. But it's the Variance OF the mean that's really bothering me.
After that, c) should be easy since all I have to do is to subtract the square of the bias from the variance I'm getting.

2. Nov 13, 2013

### Ray Vickson

If $X_1, X_2, X_3$ are dependent, you cannot say what the above variances are without also knowing the covariances between different $X_j$. If they are independent, you have not used known elementary results about combining variances in linear combinations. Back to square one.

3. Nov 13, 2013

### emperorvinayak

Oh my goodness!! Thank you so much for "you have not used known elementary results about combining variances in linear combinations."!

I looked that up and got the answer.. So it's nothing but $\frac{7}{16}+\frac{13}{9}+\frac{20}{25}$

I don't have the exact answer to this but I tried a similar problem that had the solutions at the back and it's correct! All I had to do was to square the constants being multiplied by the Xn variables and add them up.

Thanks again for your time I really appreciate it! :)