can SD be greater than AM?

[Pushpender Nath]

Ladies and Gentlemen
Good Day!
I have a question in my mind. I want to know "Whether Standard Deviation can
be greater than Arithmatic Mean?"
I asked this to many fellows. Some say yes , some say no. The fellows who
say yes, could never give me any example.
I suppose, SD can never be greater than AM. Am I right? If not then please
give me some data example.

--
Regards

Pushpender Nath
[phone]

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[Hector Maletta]

         It happens all the time, Push, especially when there are extreme
values far apart. Example:
         1000 cases with the value x=0.01, one case with the value x=100.
Approx results: Mean=0.11, SD=2.23.
         It may also happen in normal distributions. Take a standard normal
distribution of a variable Z with mean=0 and SD=1. Shift it slightly to the
right by making the variable W=Z+0.1. Now you have mean=0.1 and SD=1.
         Hector

         -----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of
Pushpender Nath
Sent: 06 January 2008 11:41
To: [hidden email]
Subject: can SD be greater than AM?

         Ladies and Gentlemen
         Good Day!
         I have a question in my mind. I want to know "Whether Standard
Deviation can
         be greater than Arithmatic Mean?"
         I asked this to many fellows. Some say yes , some say no. The
fellows who
         say yes, could never give me any example.
         I suppose, SD can never be greater than AM. Am I right? If not then
please
         give me some data example.

         --
         Regards

         Pushpender Nath
         [phone]

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[Art Kendall-2]

An AM can be a negative number.  An SD cannot.

There is example syntax below the sig block.

Save all your current work, then open a new instance of SPSS. Make sure
that you put warnings, etc. into the output file. <edit> <options>
<viewer>. Cut-and-paste then run the syntax.
try it again using this data
10 10
10 30
1 21

and again using this data

10 10
10 30
1  10000

Art Kendall
Social Research Consultants



data list list/ wgt (f3) score(f5).
begin data
10 -10
10 10
1 1
end data.
weight by wgt.
descriptives score / statistics= all.

Pushpender Nath wrote:

> Ladies and Gentlemen
> Good Day!
> I have a question in my mind. I want to know "Whether Standard Deviation can
> be greater than Arithmatic Mean?"
> I asked this to many fellows. Some say yes , some say no. The fellows who
> say yes, could never give me any example.
> I suppose, SD can never be greater than AM. Am I right? If not then please
> give me some data example.
>
> --
> Regards
>
> Pushpender Nath
> [phone]
>
> =====================
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> [hidden email] (not to SPSSX-L), with no body text except the
> command. To leave the list, send the command
> SIGNOFF SPSSX-L
> For a list of commands to manage subscriptions, send the command
> INFO REFCARD
>
>
>

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[Baker, Harley]

One might consider that the z-score transformation both demonstrates and proves that the SD can be greater than the mean. After such a transformation the mean is always 0 and the SD is always 1.

Harley


Dr. Harley Baker
Associate Professor and Chair, Psychology Program
Chief Assessment Officer for Academic Affairs
California State University Channel Islands
One University Drive
Camarillo, CA 93012

805.437.8997 (p)
805.437.8951 (f)

[hidden email]



From: Pushpender Nath
Sent: Sun 1/6/2008 2:41 AM
To: [hidden email]
Subject: can SD be greater than AM?


Ladies and Gentlemen
Good Day!
I have a question in my mind. I want to know "Whether Standard Deviation can
be greater than Arithmatic Mean?"
I asked this to many fellows. Some say yes , some say no. The fellows who
say yes, could never give me any example.
I suppose, SD can never be greater than AM. Am I right? If not then please
give me some data example.

--
Regards

Pushpender Nath
[phone]

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====================To manage your subscription to SPSSX-L, send a message to
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[Richard Ristow]

At 05:41 AM 1/6/2008, Pushpender Nath wrote:

>I have a question in my mind. I want to know "Whether Standard
>Deviation can be greater than Arithmetic Mean?"
>I asked this to many fellows. Some say yes , some say no. The
>fellows who say yes, could never give me any example.
>I suppose, SD can never be greater than AM. Am I right? If not then
>please give me some data example.

It's very easy indeed. There's no required relationship whatever.
Take any sample or population, subtract any constant a from all
values, and if the old arithmetic mean (AM) was A, the new one is
A-a, without changing the SD at all. You can make the AM as much
smaller than the SD, or as much larger, as you like.

Now, this can easily result in negative data values. Your advisors
maybe are thinking of data with only positive values, for which it
isn't quite as easy. But you can still get a SD much larger than the
mean. Here's a log-normal distribution:

SET RNG = MT       /* 'Mersenne twister' random number generator  */ .
SET MTINDEX = 2069 /*  Providence, RI telephone book              */ .

INPUT PROGRAM.
.  STRING  City     (A10).
.  NUMERIC Person   (N3).
.  NUMERIC Income   (DOLLAR10).
.  LEAVE   City
            Person .

.  COMPUTE    City    = 'Barrington'.
.  LOOP Person  = 1 TO 500.
.     COMPUTE Income = 25E3 *  2**RV.NORMAL(0,1.4825*SQRT(2.0)).
.     END CASE.
.  END LOOP.
END FILE.
END INPUT PROGRAM.


DESCRIPTIVES
   VARIABLES=Income
   /STATISTICS=MEAN STDDEV MIN MAX .

Descriptives
|---------------------------|---------------------|
|Output Created             |06-JAN-2008 21:02:49 |
|---------------|---|-------|----------|----------|--------------|
|               |N  |Minimum|Maximum   |Mean      |Std. Deviation|
|---------------|---|-------|----------|----------|--------------|
|Income         |500|$322   |$1,868,068|$76,739.48|$168,062.673  |
|---------------|---|-------|----------|----------|--------------|



But you needn't be that fancy; you can do it with a simple two-point
distribution:

If Z is x with probability p, y with probability q, where
(1) p+q=1

then, skipping most of the algebra,
(2) AM=px+qy
(3) SD**2=VAR=pq(x-y)**2

If y=1, AM=2,
(4) x-1=1/p
and
(5) VAR=(1-p)/p (see below).

So, keeping AM=2, the variance and the SD can be made arbitrarily
large, by decreasing p (the probability of choosing x) and increasing
x correspondingly.
.................................
Here's the algebra for (5), using y=1 plus (1) through (4):

(3) Var=pq(x-y)**2

By (1), and because y=1, this gives
(6) Var=p(1-p)(x-1)**2

Since (4) gives x-1=1/p, this becomes
(5) Var=p(1-p)(1/p)**2=(1-p)/p   QED

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[Pushpender Nath]

Ladies and Gentlemen
Thanks a lot for the answers to my question. I am pretty clear now, and in a
position to answer others satisfactorily.

Regards

On 1/7/08, Richard Ristow <[hidden email]> wrote:

>
> At 05:41 AM 1/6/2008, Pushpender Nath wrote:
>
> >I have a question in my mind. I want to know "Whether Standard
> >Deviation can be greater than Arithmetic Mean?"
> >I asked this to many fellows. Some say yes , some say no. The
> >fellows who say yes, could never give me any example.
> >I suppose, SD can never be greater than AM. Am I right? If not then
> >please give me some data example.
>
>

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[Simon, Steve, PhD]

Pushpender Nath writes:

> I have a question in my mind. I want to know "Whether Standard
Deviation
> can be greater than Arithmetic Mean?"

Several people have already pointed out the obvious counterexamples,
such as a negative mean and standardized data.

A more interesting question is whether the standard deviation can be
larger than the mean for data that is non-negative. Here you have to
work a bit harder to find an example. Data with an outlier though will
serve here. As you push the outlier further and further away from the
data, the standard deviation increases faster than the mean does. So a
data set with values 1,2,4 doesn't work (mean=2.3,sd=1.5)but 1,2,8 works
(mean=3.7, sd=3.8) and the gap widens as you get more extreme. 1,2,16
yields mean=6.3, sd=8.4 while 1,2,32 yields mean=11.7, sd=17.6.

Many researchers in Industrial Hygiene (IH) summarize their data using a
coefficient of variation (also known as the relative standard deviation)
which is simply the standard deviation of a data set divided by the
mean. And the IH community can provide plenty of examples of real data
sets where the coefficient of variation is greater than 1.

Steve Simon, [hidden email], Standard Disclaimer
CMH (Kansas City) is hiring a second statistician. See
www.childrensmercy.org/stats/JobOpening.asp for details.
Evidence Based Medicine gives my book 4/4.5 stars out of five!
Full text is at http://ebm.bmj.com/cgi/content/full/12/2/59

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[Marta Garcia-Granero]

This data (from a published dataset, therefore they are real) show a
highly skewed variable (log-normal) with SD greater than AM:

DATA LIST FREE/bilirrubina (F8).
BEGIN DATA
 25 178 15 10  90  30 280  20   58   8
 20  75 85 12 158  10   6  22  103  16
170  28 240  8  16  20  80 420  36  24
 70  22  14  5 130  54  25  22  15  22
 50  24 143 42   4  18  44 220  54  38
135  24  78 24   4 152  68  45  38  18
120  18  30 20 360  26  12  16 310  72
 48  96  32 22  55  12  10  62  46  35
 15 192  20 65  42   6  60  34  14 115
  9 164  12 40  18 530  32 104  28   6
 81  10  26 94  46  40  14  40
END DATA.
VAR LEVEL bilirrubina(SCALE).

HTH,
Marta

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[Richard Ristow]

At 10:28 AM 1/11/2008, Simon, Steve, PhD wrote:

>Several people have already pointed out the obvious counterexamples,
>such as a negative mean and standardized data. A more interesting
>question is whether the standard deviation can be larger than the
>mean for data that is non-negative.

Which, as previously noted, it can. It's usually a distribution
that's mostly concentrated around low values, with a long
low-probability upward 'tail'. As Marta noted, you can get this with
a log-normal distribution; see her example, or the one I posted on the 6th.

Or to be really simple, you can have a two-point distribution:

If Z is
0          with probability p,
x=1/(1-p)  with probability 1-p

then AM(Z)=1, SD(Z)=SQRT(p/(1-p))  (algebra below)

and the SD can be made as large as you like, by bringing p closer to
1 and increasing x correspondingly.

If you want positive values, rather than non-negative, add 1 to Z;
the mean is then 2 and the SD as above.
...........................
Algebra:
AM = p*0 + (1-p)(1/(1-p)) = 1; easy enough.

SD = SQRT(Variance); let variance = V

The variance is the mean of the square minus the square of the mean,
V=AM(Z**2)-AM(Z)**2
where
AM(Z**2)=p*0**2 + (1-p)(1/(1-p))**2=1/(1-p)
AM(Z)**2=1**2=1

So
V=1/(1-p)-1= (1-(1-p))/(1-p)=p/(1-p)

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[Bob Schacht-3]

At 02:13 PM 1/12/2008, Richard Ristow wrote:
>At 10:28 AM 1/11/2008, Simon, Steve, PhD wrote:
>
>>Several people have already pointed out the obvious counterexamples,
>>such as a negative mean and standardized data. A more interesting
>>question is whether the standard deviation can be larger than the
>>mean for data that is non-negative.

I'm sorry, I haven't been following this discussion closely, but has anyone
pointed out yet that in a normal distribution, the mean and the SD are, by
definition, independent? Consequently, the original question seems to
challenge this basic feature of a normal distribution.

That is not to say that in normal distributions, the mean and SD often seem
to behave in certain ways. But if the SD is dependent on the mean, then it
is not truly a normal distribution.

Bob in HI

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[Richard Ristow]

At 08:02 PM 1/12/2008, Bob Schacht wrote:

>At 02:13 PM 1/12/2008, Richard Ristow wrote:
>>At 10:28 AM 1/11/2008, Simon, Steve, PhD wrote:
>>
>>>Several people have already pointed out the obvious
>>>counterexamples, such as a negative mean and standardized data. A
>>>more interesting question is whether the standard deviation can be
>>>larger than the mean for data that is non-negative.
>
>Has anyone pointed out yet that in a normal distribution, the mean
>and the SD are independent? Consequently, the original question
>seems to challenge this basic feature of a normal distribution.

Yes; that's been discussed, not for normal distributions in
particular but for general distributions.

Indeed, you can specify any (real) mean, and any (real) non-negative
standard deviation, for a normal distribution. So you can get any
relationship between AM and SD that you like, so long as it doesn't
require a negative SD.

And generalizing, you can specify the mean for any distribution that
has one, by adding or subtracting a desired number to all values.

However, with a normal distribution or others, you'll commonly get a
distribution that's negative with high probability. That raised the
question that got more discussion by Simon, Garcia-Granero, and
Ristow: how do you get SD>AM in a distribution restricted to
non-negative values? And that led to the examples given, of
log-normal distributions (Ristow, Garcia-Granero), or a two-point
distribution in which one of the values is large and has low
probability (Ristow).

Cheers!
Richard

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