Optimizing Point Estimates: Bias and Variance Analysis for Mean Estimation

[emperorvinayak]

Homework Statement

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 anyone 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?

Homework Equations

Var(μ) = \frac{1}{n-1} \sum(x_i -\bar{X})² from i=1 to n for unbiased
and
Var(μ) = \frac{1}{n} \sum(x_i -\bar{X})² from i=1 to n for biased

The Attempt at a Solution

I found out part a) pretty easily: I just replaced all the X_n 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.

 

[Ray Vickson]

Homework Statement

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 anyone 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?

Homework Equations

Var(μ) = \frac{1}{n-1} \sum(x_i -\bar{X})² from i=1 to n for unbiased
and
Var(μ) = \frac{1}{n} \sum(x_i -\bar{X})² from i=1 to n for biased

The Attempt at a Solution

I found out part a) pretty easily: I just replaced all the X_n 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.

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.

Never mind Youtube; try reading some actual articles on-line or in your textbook.

 

[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 X_n variables and add them up.

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