# Question about vaiance, population and sample

1. Sep 21, 2005

### robert Ihnot

In the October edition of the magazine, Active Trader, a reader writing in Chat Room, "Deviating from deviation?" asks that in explaining last month the viaiance, why did you not in your example divide by two?

{(8-9)^2 + (9-9)^2 +(10-9)^2}/3 = .667.

The explanation given is nothing more than,'That's how it is done,' and completely ignores, adding, "We're not math majors," the difference between the sample diviation and the population deviation. (There is no explanation of where the above example come from, and probably it is nothing but an equation invented by the writers.)

Elementary statistic books do a very poor job of explaining WHY that difference occurs, saying such as "It eliminates bias," or even "It makes the theory work out better, and isn't worth going into."

Does anyone have a good explanation of why there is that distinction, and assuming it is a sample deviation, why is it better to divide by 2 than by 3?

Last edited: Sep 21, 2005
2. Sep 21, 2005

### hypermorphism

In the estimators section of your statistics text, you should get, either as a problem or example, a simple calculation that shows the "divide by n" estimator of population variance for variance of samples is biased, while "divide by n-1" is unbiased. Are you looking for an intuitive answer ?

3. Sep 21, 2005

### robert Ihnot

Well, I have made up several examples about dice, but when the number of trials falls, say three throws of the dice, this greatly changes the variance.

This is my example, the population is the six sides of a dice, mean is 3.5 on a throw, variance is 2.92. Now if we throw three times, and get a perfectly reasonable outcome: 2,3,4. The mean is 3, and dividing by 2 the variance is 1, where as dividing by 3 it would have been 2/3. In neither case are we near 2.94. Thanks, bob

4. Sep 21, 2005

### EnumaElish

No, not "on average." Your example is conditional on a given sample. That's not a good basis to verify the expected value of any variance estimator.

5. Sep 21, 2005

### Hurkyl

Staff Emeritus
Run this experiment a million times, and look at the average value for the variance that you compute.

6. Sep 21, 2005

### robert Ihnot

Hurkyl: Run this experiment a million times, and look at the average value for the variance that you compute.

If it is so run, there will not be much difference between dividing by 1,000,000 or 999,999.

7. Sep 21, 2005

### EnumaElish

Correct, both "1/n" and "1/(n-1)" are unbiased estimators.

8. Sep 22, 2005

### Hurkyl

Staff Emeritus
You entirely misunderstand:

You described an experiment where you roll a die three times, and then compute two different estimates for the variance, one where you divide by 3, and one where you divide by 2.

Now, you perform that experiment a million times, and you get a million estimates where you divided by 3, and a million estimates where you divided by 2.

You can then find the average of the divide by 3 estimates, and the average of the divide by 2 estimates. One of them will be (very close to) the actual variance. One will not.

They cannot possibly both be unbiased, unless the variance is zero.

if s/n is, on average (where s is the calculation in the numerator), the variance v, and so is s/(n-1), then we have:

E = nv
E = (n-1)v
0 = v

:tongue2:

9. Sep 22, 2005

### EnumaElish

Hurkyl, you are correct. My bad. What I meant was, although the "1/n" estimator is biased, it is consistent.

10. Oct 10, 2005

### robert Ihnot

Well, I got an answer here in this statistic book, "Principals of Statistics," MG Bulmer, dover paperback, 1967, p130:

"It may seem surprising that the Expected value of the sample variance is slightly less than the population variance. The reason is that sum of the squared deviations of a set of observations from their mean is always less than the sum the squared deviations from the population mean."

11. Oct 10, 2005

### EnumaElish

Good work; now you can give advice to the needy in these forums.

12. Oct 11, 2005

### robert Ihnot

It may seem surprising that the Expected value of the sample variance is slightly less than the population variance.

If we look at $$F(x)=\sum_{i=1}^{i=n}(a_i-x)^2$$

By taking the derivative and setting it equal to 0, we arrive at the minimal value of the function: $$nx=\sum_{i=1}^{i=n}a_i$$

Thus letting x take on the value of the mean of the sample gives us the minimal value for the variance.

Last edited: Oct 11, 2005
13. Oct 14, 2005

### robert Ihnot

It may seem surprising that the Expected value of the sample variance is slightly less than the population variance.

On page 129-130, Principles of Statistics, we have this problem gone into, though here a few additional details are presented. He writes:

$$(S^2)= \sum(x_i-X)^2=\sum(x_i-\mu)^2-N(X-\mu)^2$$
for the above S^2 is as defined, N is the number of samples, mu is the mean, X is the sample mean, each X_i represents a variable that takes on various sample values.

Now the point is to find the expectation, E. We have:
$$E(\sum(x_i-\mu)^2 = N\sigma^2$$, where sigma is the STD.

For the second term, E(x)=mu, and we have $$N*E(X-\mu)^2=N*(EX^2-(EX)^2)$$
The later term after N is $$V(X)=\frac{V(NX)=\sum V(X_i)=N\sigma^2}{N^2}$$

Thus returning to the original equation we have:
$$E(S^2)=N\sigma^2-\sigma^2=(N-1)\sigma^2.$$

Author adds: "Because of this fact S^2 is often divided by N-1 instead of N in order to obtain an unbiased estimate..."

Last edited: Oct 14, 2005