Re: standard deviation and N-1

From: Rob Johnson (rob_at_trash.whim.org)
Date: 07/08/04 

Date: Thu, 8 Jul 2004 11:40:01 +0000 (UTC)

In article <Pine.GSO.4.05.10407072147070.8955-100000@hercules.acsu.buffalo.edu>,
Victoria Florsheim <vf2@buffalo.edu> wrote:
>In high school, I learned that the formula for standard deviation has n in
>the denominator, but in college the book has N-1 in the denominator. What
>is the reason for this?
>
>So far, I found this in my book (By Yates, Moore, McCabe):
>Why do we average by dividing by n-1 rather than n? Because the sum of
>deviations is always zero, the last deviation can be found once we know
>the other n-1. So we are not averaging n unrelated numbers. Only n-1 of
>the squared deviations can cary freely, and we average by dividing by the
>total by n-1. The n-1 is called the degrees of freedom of the variance or
>standard deviation.
>
>
>I sort of understand that, but could someone explain in simpler terms and
>expand on that? I'm still a little puzzled as to why n-1.

The point is that the sample mean, m_s, is not the distribution mean,
m_d. Suppose the distribution variance is v_d. n m_s is the sum of n
variates (the sample). Recall that the mean and variance of a sum of
variates are the sums of the means and variances of the variates. That
is, the mean of the sum of the sample is n m_d and the variance of the
sum of the sample is n v_d. In other words,

                   2
    E[ (n m - n m ) ] = n v [1]
           s d d

or equivalently,

               2 1
    E[ (m - m ) ] = - v [2]
         s d n d

Write the distribution variance as

    v
     d

            n
         1 --- 2
    = E[ - > (x - m ) ]
         n --- k d
           k=1

            n
         1 --- 2 2
    = E[ - > ( (x - m ) + 2(x - m )(m - m ) + (m - m ) ) ]
         n --- k s k s s d s d
           k=1

            n
         1 --- 2 1
    = E[ - > (x - m ) ] + - v
         n --- k s n d
           k=1

                1
    = E[ v ] + - v [3]
          s n d

where v_s is the sample variance. Solving [3] for v_d, we get

          n
    v = --- E[ v ] [4]
     d n-1 s

This is why, to compute the distribution variance, we multiply the
sample variance by n/(n-1). Thus, it appears as if we are dividing by
n-1 instead of n.

Rob Johnson <rob@trash.whim.org>
take out the trash before replying

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