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

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  "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

#6977 — "The median may not be more than one standard deviation away from the mean"

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.)

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#6978

> the median may not be more than one standard deviation away from the mean.

    An oldie but a goodie.
    My prob&stats textbook [1] shows more: a random variable is more
    than X standard deviations above the mean at most 1/(1+X*X) of the
    time. This inequality is similar to the Bienayme-Chebyshev
    inequality; maybe they knew it, too.

    [1] Olkin, Gleser, Derman, "Probability Models and Applications",
        p157. Macmillan, 1980. ISBN 0-02-389230-7.

-- 
Don Reble  djr@nk.ca

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#6979

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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#6980

Paul N <gw7rib@aol.com> 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.

P1+P2+P3 >=1

>
> 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.)

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