Marcos

Posted on Tuesday, 04 November, 2003 - 06:37 pm:

My knowledge of statistics is fairly limited but I'm curious as to why when we replace N with (N-1) in the denominator of the variance formula we get a better estimate of the variance of the population.

Thanks,
Marcos

Kerwin Hui

Posted on Tuesday, 04 November, 2003 - 07:02 pm:

The idea is that we have used up one degree of freedom in computing the sample mean and using it to estimate the population variance. One rarely gets sample mean=population mean, and so we have to make allowance for that.

We can compute the expected value of S_XX , the sum of squares of deviation. We have
LaTeX Image

Now the first term has expected value N*variance of the population, and each summand of second term has expected value=population variance/N, so we get the expected value of S_XX is (N-1) times the population variance, and so (1/(N-1))*S_XX is an unbiased estimate of the population variance.

Kerwin

Marcos

Posted on Tuesday, 04 November, 2003 - 07:15 pm:

Hrm...

Basically I don't get why the expected value of $(μ - \bar{X})^{2} = σ^{2} / N$

Thanks,
Marcos

Kerwin Hui

Posted on Tuesday, 04 November, 2003 - 07:34 pm:

Are you happy with the basic rules for manipulating mean and variances? i.e.

$E (a X) = a E (X)$ , for all $X$ , $Y$
$E (X + Y) = E (X) + E (Y)$ , for all $X$ , $Y$
$Var (a X) = a^{2} Var (X)$ , for all $a$
$Var (X + Y) = Var (X) + Var (Y)$ , for independent $X$ , $Y$
(In general, we will have a covariance term coming in here)

Each

X_{i}

is just a random variable of mean

μ

and variance

σ^{2}

. Since the

X_{i}

are independent, their sum has mean

N μ

and variance

N σ^{2}

. Hence

\bar{X}

, being an

N

th of the sum of the

X_{i}

, has mean

μ

and variance

(1 / N^{2}) N σ^{2} = σ^{2} / N

, which gives the statement

E ((μ - \bar{X})^{2}) = σ^{2} / N

Kerwin

Marcos

Posted on Tuesday, 04 November, 2003 - 07:48 pm:

Thanks, I get it now (at least enough to satisfy myself)

Marcos