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Re: "The median may not be more than one standard deviation away from the mean"

Cross-posted to 2 groups.

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On 17/10/16 22:16, Paul N wrote:
> On comp.programming, Malcom McLean wrote "Anyway, I've consulted with
> the bulging brains on Math Exchange, and the median may not be more
> than one standard deviation away from the mean."
>
> I've not done much maths for a long while, but this piqued my
> interest, and I have managed to come up with a proof of it. It may
> perhaps be a bit unconventional, and I'd be interested in comments,
> not only whether it is right, but also whether it is
> over-complicated, naive or anything.
>
> Malcom was talking about statistics, but I more usually deal with
> probability and so my proof is in that area. I presume it could be
> used in the context of statistics simply by letting each datum be an
> equally likely event.
>
> So - suppose we have a variable X with a distribution P - it can be
> discrete, continuous, or a bit of both. Suppose it has a median m,
> and a mean u. (I'm already using m for median, and u looks more like
> a mu anyway...) Further suppose that u > m, with x = u - m.
>
> We can divide the distribution of X up into three "tranches", as
> follows:
>
> P1 = Prob( X <= m )
>
> P2 = Prob(m < X < u)
>
> P3 = Prob(X >= u)
>
> This gives P1 + P2 + P3 = 1, and also P1 >= 1/2 as m is the median.
> From this it follows that P1 >= P3.
>
> Now suppose we were calculating the mean of the distribution. The
> first tranche are all at least x short of u, and so give in total a
> "shortfall" of at least P1 x. The second tranche also all lie below
> u. So for u to be the mean, the third tranche must contribute at
> least P1 x above the mean. If we let u2 be the mean, over the third
> tranche only, of X-u, we have that P3 u2 >= P1 x.
>
> Now, the variance of any distribution is "the mean of the square
> minus the square of the mean", and since the variance cannot be
> negative, this means that the mean of the square must be at least the
> square of the mean. So if we let v2 be the mean value of (X-u)^2 over
> the third tranche, this result gives us v2 >= u2^2.
>
> Now the variance of the original distribution will be at least P1 x^2
> from the first tranche, something non-negative from the second
> tranche, and P3 v2 from the third tranche.
>
> Var >= P1 x^2 + P3 v2 >= P1 x^2 + P3 u2^2 >= P1 x^2 + (P1^2/P3) x^2
> >= P1 x^2 + P1 x^2 >= x^2
>
> So the standard deviation is at least x.
>
> Applying the same logic to -X allows the proof to work where u < m as
> well.
>
> I was hoping to cross=post this to both comp.programming and
> sci.math, but I don't think Google allows such things.)

No comments to make - but I have cross-posted to sci.math in accordance 
with your wishes.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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Thread

"The median may not be more than one standard deviation away from the mean" Paul N <gw7rib@aol.com> - 2016-10-17 14:16 -0700
  Re: "The median may not be more than one standard deviation away from the mean" Don Reble <djr@nk.ca> - 2016-10-17 17:39 -0600
  Re: "The median may not be more than one standard deviation away from the mean" Richard Heathfield <rjh@cpax.org.uk> - 2016-10-18 00:55 +0100
  Re: "The median may not be more than one standard deviation away from the mean" root <NoEMail@home.org> - 2016-10-18 16:43 +0000

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