1

In an exercise they asked me: "Why could we use the following correction factor? $\text{varianceX} = \frac{n-1}{n}*\text{varianceY}$

What I said was basically, because the unbiased sample variance have a factor of $\frac{n}{n-1}$, we could multiply the variance by $\frac{n-1}{n}$ to cancel the correction of Bessel and get the biased variance.

But what would be the utility of calculating a biased variance?, maybe I'm wrong about something...

OiciTrap
  • 316
  • This might be a duplicate of this question. – robjohn Apr 28 '15 at 06:35
  • I understand the source of bias, what I don't understand is the utility of the biased variance, because in the task I was asked the factor was $\frac{n-1}{n}$, not $\frac{n}{n-1}$, so varianceX should be the "uncorrected" sample variance, not the corrected. – OiciTrap Apr 28 '15 at 12:32
  • I would assume that $\text{varianceY}$ is a population variance and $\text{varianceX}$ is a sample variance or an estimated sample variance. – robjohn Apr 28 '15 at 13:30

1 Answers1

1

Well, after reading a lot on wikipedia I think this is the answer:

In calculating the expected value of the sample variance a factor equal to $\frac{n-1}{n}$ is obtained, which underestimates the expected variance so usually when calculating the sample variance we multiply it by the factor $\frac{n}{n-1}$ which is generally known as the variance unbiased sample variance or corrected sample variance.

The problem with this is that to correct the bias produces a large MSE, so one can choose a scaling factor that behave better than the variance of the corrected sample. This is always scaling down, choosing a 'a' greater than n-1, such that:

$S^2_a = \frac{n-1}{a}S^2_{n-1}$

In my case $a = n$.

OiciTrap
  • 316