Increase sample standard deviation

iocal said:

Hello,

I have stumbled upon an incomprehensible statement, the statement is:

" In order to increase the standard deviation of a set of numbers, you must add a value that is more than one standard deviation away from the mean"

Can that be proven algebraicaly? Thanks.

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DrPhil said:

Sure, with some effort.

Start with a distribution of N data, mean = mu = Sum(x)/N, Variance = sigma^2 = Sum(x^2)/N - mu^2

Add one more datum, X to the distribution.
The new mean is mu' = (N*mu + X)/(N + 1)
The new mean of the squares = . . .
New Variance = . . .

For what values of X is new Variance > old Variance?

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iocal said:

After doing the math, what I saw is that the new value of X, let us call it \(\displaystyle X_{N+1} \) has to satisfy:

\(\displaystyle \displaystyle \left| X_{N+1} -\mu \right| >\sqrt{ \frac {N+1}{N} \sigma^2} \) in order for the variance to increase.

For large \(\displaystyle N\) the RHS is close to \(\displaystyle \sigma \) which I gather is what we needed to show. Thank you very much.

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