Re: Biased and unbiased std dev

"Vinay" <vinkolar@xxxxxxxxx> wrote in news:1121288120.240622.52330
@g43g2000cwa.googlegroups.com:

> I was wondering if someone can explain in simple terms why std dev s=
> sqrt(sum_(x_i-\bar{x})/n) is biased. I understand that 's' should be
> invalid when n=1 and hence they divide by (n-1). But is it just to
> avoid this case? Or is there more intersting fact for that.
>
> Why does dividing by n always give biased estimates? Is this because
> there are n-1 intervals between x_1 to x_n?

As Bob has pointed out, it's only the variance that you can get an
unbiased estimate of by dividing by (n-1); taking the square root biases
the result.

The reason for the bias is that when you take the average sum of squares
around the *sample* mean rather than the population mean, you end up
subtracting out the variance of the sample mean itself (which will be
sigma^2/N, where sigma^2 is the population variance) and thus wind up
with a value that's too small by a factor of (n-1)/n. So you multiply
the result by n/(n-1) and come up with the standard formula.

A more intuitive explanation is that the mean of a set of data is, by
definition, the value that minimizes the sum of squared deviations around
it. For a sample, the sample mean will have a smaller sum of squared
deviations around it than any other value. But if you're trying to
estimate the *population* variance, you're really interested in the
deviations around the *population* mean. Since the population mean will
not in general be *exactly* equal to the sample mean, the sum of squared
deviations around it will be greater than the sum of squared deviations
around the sample mean, and therefore the latter will be biased low.
.

Relevant Pages

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