Calculating the bias of variance estimator

#1
Dear all,
I have been given two variance (S^2) estimators, S1 and S2. I am not including the square (^2) as it may cause confusion. And I am to determine the bias of these estimators. Now, given that estimator S1 has the same equation as sample variance, it should therefore be classed as 'Unbiased'. However, estimator S2 on the other hand has an equation which differs to S1, and has been proven to be bias (please correct me if i'm wrong).
The issues that I am facing are that given that I am to calculate the bias of both estimators S1 and S2, I would state that S1 has a bias of 0. However, when calculating the mean [E(S1)] and subtracting it from the population variance [Var.P], which is the formula for calculating bias of an estimator (as shown in the attachment), I do not obtain a result = 0. E.g. the population variance was something along the lines of 120, and the estimator S1 = 120.087627... Therefore does this mean that the estimator S1 is bias, even though the formula is unbias? Also, if the estimator is classed as Unbiased, then in order to calculate the bias of Estimator S2, do I proceed with [E(S2) - Var.P] or with [E(S2) - E(S1)] since S1 in this case is unbias and approx equal to Var.P

Any help would be much appreciated. Thanks
 

Dason

Ambassador to the humans
#4
I guess I'm confused since it looks like in your derivation you show that the bias of S1^2 is 0 but in your initial post you say you show it wasn't 0?
 
#5
I guess I'm confused since it looks like in your derivation you show that the bias of S1^2 is 0 but in your initial post you say you show it wasn't 0?
In my initial post, I explained that I had derived S1 to have a bias of 0. However, according to the numerical data that I had, the mean of the values of S1 were not equivalent to the population variance, and so I am confused as to whether the estimator would be classed as Biased or Unbiased.

Would it be reasonable to state that the variance estimator S1 is unbiased, but has a rather small degree of error? Of which the error would be equivalent to the difference between the population variance and the mean of the values of estimator S1.
 

Dason

Ambassador to the humans
#6
However, according to the numerical data that I had, the mean of the values of S1 were not equivalent to the population variance
Please explain what you mean by this in more detail.
 
#7
Please explain what you mean by this in more detail.
Okay.

So I was given a population of size N = 1000 and from that I obtained 30 random samples of size n = 56 (so there were 30 samples, but each sample had a size of n = 56). I then calculated the variance estimator S1 and S2 for each of the 30 random samples. Thus I was left with 30 values of the estimate S1 and 30 values of the estimate S2. From there, I then calculated the mean of these 30 values for S1, to obtain E(S1). Likewise, I did the same for S2 to obtain E(S2). Then (according to the bias formula) I subtracted the POPULATION VARIANCE from the mean value of estimator S1 [E(S1)]. I was left with a value similar to: 0.087627 which is not equal to zero. And so my question is "What is the bias of estimator S1?", because theoretically, as shown in the derivation, S1 should have a bias of 0. So is it normal that there may be slight deviations when calculating (in numerical format) the variance estimate S1? Also, is it normal for an UNBIASED estimator to have a 'standard error'? which in this case would be +0.087627
 

Dason

Ambassador to the humans
#8
Note that what you did was calculate an estimate of the bias. Try doing it again for 30 different samples. You'll get another (slightly different) estimate of the bias. Do you see the issue? The "bias" you calculated mathematically is THE bias. You can't get a perfect value from a sample (hence why we calculate the expected value and variance of estimators in the first place) so you really shouldn't expect to get a sample estimate of the bias that is exactly 0.

When you take a sample of 30 from a population with a mean of 0 are you questioning whether the mean really is 0 when the sample mean is .08?
 
#9
Note that what you did was calculate an estimate of the bias. Try doing it again for 30 different samples. You'll get another (slightly different) estimate of the bias. Do you see the issue? The "bias" you calculated mathematically is THE bias. You can't get a perfect value from a sample (hence why we calculate the expected value and variance of estimators in the first place) so you really shouldn't expect to get a sample estimate of the bias that is exactly 0.

When you take a sample of 30 from a population with a mean of 0 are you questioning whether the mean really is 0 when the sample mean is .08?
I understand what you are implying. However, I guess that I am questioning whether the mean sample variance can be equal to .08 yet have a bias of 0. Because the sample variance is effectively equivalent to the population variance. It uses the exact same method of calculation and so, if it was applied to the entire population, the results would be equivalent (or so I believe). Therefore, I'm asking, given a sample of 30, can the sample variance estimator still be classed as unbias even though it has produced a slight variation due to the sample size. Nonetheless it uses the same method of calculation and I don't think that it would therefore be 'biased' as it is not influenced in anyway (as shown through the derivation) to produce a result of which is any other than that of the population variance. Whereas the estimator S2 is derived to be [(55/27)*Populationvariance] which explains the difference in values for S2.
 

BGM

TS Contributor
#10
Because the sample variance is effectively equivalent to the population variance. It uses the exact same method of calculation and so, if it was applied to the entire population, the results would be equivalent (or so I believe).
It is not true - unless you have an equalliy-likely discrete model and you do a complete census which collect all the population data at once.

And I guess you may have some misinterpretation about the bias as well.