🚀 go-pugleaf

so you can get it down for your assignment.... right?

Jon Dodd <jvd@physiol.ox.ac.uk> wrote in article
<36052C4E.A697E07@physiol.ox.ac.uk>...
> Hi,
> 
> Please could you help me with this simple problem.
> 
> From a normal distribution with mean m and standard deviation sd,  I
> pick a random sample A and then second B. If A>B then the value of A is
> assigned to group X, if B>A then value of A is assigned to group Y. What
> is the resulting mean and standard deviation of the two distributions X
> and Y? I am not mathamatically oriented so a step by step proof would be
> great.
> 
> many thanks
> Jon
> 
> 
> 

Ben wrote:

> >so you can get it down for your assignment.... right?
>
>

 Dear Ben,
No its not an assignment, as you can see I am a physiologist and not a
mathematician and the problem is just the first part of a neuronal decision
process that I am thinking about.

I have no background in any of this stuff and no formal teaching in it, I am
just trying to sort something out in my head. I thought that asking the
experts would be a useful and enlightening way to learn but I guess not, all
I have had back so far are sarcastic and derisive comments.

Well I apologise for not giving you lot a difficult and challenging enough
problem, unfortunatly no-one has seen fit to offer me an explanation of this
obviously easy problem but there you go.

Hey I'll trade you for physiology/psychology information if you like!

--
==============================================
Jon Dodd
Department of Physiology; University of Oxford
Parks Road; Oxford; OX1 3PT
tel 01865-282483; fax 01865-272469
http://waller.physiol.ox.ac.uk/~jvd
==============================================

Jon Dodd wrote:
> ...
> From a normal distribution with mean m and standard deviation sd,  I
> pick a random sample A and then second B. If A>B then the value of A is
> assigned to group X, if B>A then value of A is assigned to group Y. What
> is the resulting mean and standard deviation of the two distributions X
> and Y? I am not mathamatically oriented so a step by step proof would be
> great.
> ...


Hi Jon,

For a normal distribution A and B are independent random variables.
Therefore, the likelihood for P(A>B) = 0.5 (i.e. 50%) and 
P(B>A) = 1 - 0.5 = 0.5 (P(u) + P(v) = 1, if v is the complementary event
of u  - think of a coin: P(head) + P(tail) = 1). 
You are just dividing a set of N random numbers into two sets of N/2
elements by random! (However, if you had specified something like A > m
+ sd this would make a different case.) Of course both sets are random
representations of the same distribution with the same mean and standard
deviation - but slightly different sample means and sample standard
deviations as those are only estimators of the "true" parameters of the
distribution. For N -> Infinity: mean(x) = mean(y) =
mean_of_normal_distribution.

I think, you should read one or two basic books on statistics, there are
even some focussing on medical/biological issues.

Regards,
Harald

(BTW, I had the same thought as Ben.)

-- 
Harald Giese
Email: giese@dkrz.de
Phone: +49 (0)40 4123 5796; Fax: +49 (0)40 5605724
Institut fuer Meereskunde der Universitaet Hamburg
(Institute of Oceanography of the University of Hamburg)
Troplowitzstrasse 7, D-22529 Hamburg

Harald Giese wrote:

> For a normal distribution A and B are independent random variables.
> Therefore, the likelihood for P(A>B) = 0.5 (i.e. 50%) and
> P(B>A) = 1 - 0.5 = 0.5 (P(u) + P(v) = 1, if v is the complementary event
> of u  - think of a coin: P(head) + P(tail) = 1).
> You are just dividing a set of N random numbers into two sets of N/2
> elements by random! (However, if you had specified something like A > m
> + sd this would make a different case.) Of course both sets are random
> representations of the same distribution with the same mean and standard
> deviation - but slightly different sample means and sample standard
> deviations as those are only estimators of the "true" parameters of the
> distribution. For N -> Infinity: mean(x) = mean(y) =
> mean_of_normal_distribution.

Dear Harald,

Thankyou for your reply. I don't think you understood my problem correctly or
perhaps I misinterpreted your reply.

I have 1 normal distribution (in fact I have 2 identical normal distributions
but I posed the problem this way to make it simpler). I take random sample A
and then random sample B. if A>B then the value of A goes to group X. if A<B
then value A (value A again not B) goes to Y. Thus it is intuitive that X
will have a greater mean than Y. They will differ from the original by the
same amounts and be mirror images of each other about the mean.

I have done some more thinking about this and the distributions of X and Y
obviously must sum to the original distribution. If you draw this you get 2
non normal distributions with differences from the mean of about 0.5 sd. My
question again though is how do you go through a mathematical proof showing
the exact values of the means and the common new sd? Is there an equation
giving the shape of the new distributions? Is it just multiplying a
cumulative gaussian (prob that A>B) by the original distribution - if so how
do you do this?

many thanks

Jon

Jon Dodd wrote:
> ...
> Dear Harald,
> 
> Thankyou for your reply. I don't think you understood my problem correctly or
> perhaps I misinterpreted your reply.


I have not read your original question carefully, so I my answer was
useless, sorry!


Now my second attempt:

1. the normal probability density function (normPDF) for A (and B) is 
  normPDF[z] := 1/Sqrt[2 Pi sd^2] * Exp[(-1/2)* (z-m)^2/sd^2]

2. the probability for any A being greater then B (A and B from the same
normPDF) is:
  Integrate[normPDF[z], with z running from -Infinity to B]
This gives the cumulative probability density function (normCDF) for 
A > B:
normCDF[z] := 1/2 * (1-Erf[(x-m)/(Sqrt[2]*sd)]) with Erf being the
so-called "error function" (see any statistic book for details).
Some values: 
  normCDF[-Infinity] = 0
  normCDF[0] = 0.5
  normCDF[+Infinity] = 1
Essentially, normCDF rises from 0 to 1 within (m - 4*sd) and (m + 4*sd). 

3. the conditional probability density function X(A|A>B) (PDF for A
given A>B) is just the product:
  condPDF[z] := normPDF[z]*normCDF[z]
( that for A<B is simply  normPDF[z]*(1-normCDF[z]) )

Now the problem: 
  if you assume the mean of a density distribution to be the "value of
maximum likelihood" i.e. the maximum of the function, then you have to
differentiate condPDF[z] for z (no problem) and solve the result for its
root: Derivative[condPDF[z] for z] = 0 .
The latter makes the problem if you look for an analytical result,
rather than a numerical.

I tried it with Mathematica - an algebra software (www.wolfram.com) -
but that failed for the analytical solution.

Regards,
Harald


-- 
Harald Giese
Email: giese@dkrz.de
Phone: +49 (0)40 4123 5796; Fax: +49 (0)40 5605724
Institut fuer Meereskunde der Universitaet Hamburg
(Institute of Oceanography of the University of Hamburg)
Troplowitzstrasse 7, D-22529 Hamburg

Jon Dodd wrote in message <360661F6.4DDCADE1@physiol.ox.ac.uk>...
....
>I have 1 normal distribution (in fact I have 2 identical normal
distributions
>but I posed the problem this way to make it simpler). I take random sample
A
>and then random sample B. if A>B then the value of A goes to group X. if
A<B
>then value A (value A again not B) goes to Y. Thus it is intuitive that X
>will have a greater mean than Y. They will differ from the original by the
>same amounts and be mirror images of each other about the mean.
>
>I have done some more thinking about this and the distributions of X and Y
>obviously must sum to the original distribution. If you draw this you get 2
>non normal distributions with differences from the mean of about 0.5 sd. My
>question again though is how do you go through a mathematical proof showing
>the exact values of the means and the common new sd? Is there an equation
>giving the shape of the new distributions? Is it just multiplying a
>cumulative gaussian (prob that A>B) by the original distribution - if so
how
>do you do this?
>


The distributions of X and Y are the distribution of the largest and the
smallest observation in a sample of 2 from a normal distribution
respectively. Properties of these distributions may be found under the
heading "order statistics" in statistical literature.

Jos Jansen.