Biased and Unbiased Estimators

[Michelle]

Hi, I'm not sure how to find the bias estimator.
For example,
If we let X and Y be a random sample from a population with mean=1,
compute the bias for the following estimators of the population mean.
1)0.91X+0.09Y.

We knowthat the bias estimator is Bias=E-theta

I'm not sure how to use this formula with equation 1.

Thank You.

 

[JohnM]

Assuming bias is a measure of the "distance" from the estimation to the true value of the parameter, maybe you need to use the Euclidean distance formula if you're working with two variables (or "dimensions")?

d = sqrt(x^2 + y^2)

I'm really not completely sure here......

 

[quark]

I'm not sure either.

It seems that E(X)=E(Y)=1, E(0.91X+0.09Y)=0.91*E(X)+0.09*E(Y)=1, so 0.91X+0.09Y is unbiased.

 

[JohnM]

Now I see it - the problem doesn't imply that the estimator is biased, it just asks you to estimate the bias, which of course could be 0...

so I think the answer is 0

 

[Michelle]

Now I see it - the problem doesn't imply that the estimator is biased, it just asks you to estimate the bias, which of course could be 0...

so I think the answer is 0

Hi~
The asnwer is zero...but i'm not sure how you got it...

 

[JohnM]

Since X and Y are both random samples from a population with mean = 1, E(X)=1 and E(Y)=1 (expected values)

So, the estimator = .91X + .09Y = .91(1) + .09(1) = .91 + .09 = 1

Since the actual population mean =1, then the bias =0 since pop-est = 1-1=0

 

[Michelle]

Since X and Y are both random samples from a population with mean = 1, E(X)=1 and E(Y)=1 (expected values)

So, the estimator = .91X + .09Y = .91(1) + .09(1) = .91 + .09 = 1

Since the actual population mean =1, then the bias =0 since pop-est = 1-1=0

hi!!Oh i seee!!
Thank You very very much ,,now i see how it wokrs.
)