The reason why I need to understand this is because:

S=R*R'  and  R=R'  and  w'*R'=(R*w)'  and   sqrt(x'x)=norm(x)

w'*S*w < r^2     w'*R*R'*w< r^2    w'*R'*R*w<r^2     (R*w)'*(R*w)< r^2
sqrt((R*w)'*(R*w))< sqrt(r^2)    norm(R*w)< r


Hence, we are using the fact that R=R' which means that we should be able to show
that that is actually the case.

Christopher,
thanx for asking the question. It made me start to think about the answer.
If we assume that the winner shares the winnings ie more winners equals
less winnings per persons then the question which numbers people most
frequently choice becomes highly important.

You want to pick a number combination that people less frequently picks
inorder to maximize your chances of winning alone...

Right on the money Christopher :-)

sounds great Christopher :-) I have a bit on my plate as of now but I would
love to check into it if I had the time. Dont let go of it and let me know
how you are coming along :-)

on a similar note. One thing that I find really annoying is that you have to specify
all the local variables in a procedure otherwise you get a tone of error messages.
for example X:=proc() local A,B,C,D,E,F ;

For a procedure with a lot of different variables this soon becomes really really boreing.
Why not by default assume that all variables inside a procedure are local unless otherwise
stated ?

I retract this question because I notice that is based on an error:

"Correlation of two returns when both returns are one standard deviation above their mean"

It would be very strange if the correlation would not be positive since both returns
per definition are positive (one standard deviation above the mean) he he.
To bad I did not see this earlier. My fault.

I will change the code so it looks at returns that are above mean+1*stdev and below
mean-stdev hence the correlation will not only be positve ie [ -5, 5], [6,7], [-3, -4] etc
I think this is more correct.

In the maple file it shoud read (it went away for some reason)

ii) Sort out the return of both series when both :

returns> mean+threshold*standarddeviation of the individual series
returns< mean-threshold*standarddeviation of the individual series

@epostma  thank you for taking the time to respond.

I have done experiments where I have simulated two random return series ie normal
with a given cross correlation ie 0.8 by using Cholesky decomposition. See attached maple file.
I have then sorted out the observations when both returns are larger than the
expected return + 1* standard deviation of the return for each series. We can
then calculate the cross correlation of return for the "sorted out returns" and
compare it to the cross correlation of the intire two series.

It turnes out that the cross correlation becomes lower the further out we get
in the tail and the standard deviation of the cross correlation coefficient is increasing.
I now want to explain this phenomenon by using algebra ie probability
density functions etc etc . I need some help here.

In regards to the second question; I think we can assume bivariate normal distributions.
In regards to the first question; I am not sure I understand what you mean but it sounds good :-)
I think I want conditional correlation.


The black thick line is the expected correlation ie average of all 5 or 6 simulations.

Map_-ecorr.mw

 

Thank you for that. I was not aware of such a new feature.
However, it would be nice to be able to understand what is going on and
how exactly the solution is found ie the algorithm.
I hate these kind of black box things

Thank you for that. I was not aware of such a new feature.
However, it would be nice to be able to understand what is going on and
how exactly the solution is found ie the algorithm.
I hate these kind of black box things

ha, ha that is cool !!!!  You are a genius Christopher2222 or maybe not a
true genious (I have to little information to judge that) but certainly a creative
person loyal to the cause. And, yes to be a genious you need to be creative !

Thanx Robert :-) Works great! 
You are stable as usual!

Thanx Robert :-) Works great! 
You are stable as usual!

Ok, nice! Now the only thing left is trying to understand what is happening.
I assume sqrt(x^2 + y^2) is the cone. The objective function x^2 + y^2 + z^2
has 3 dimensions so it might be difficult to 3dplot it. Could you please explain
a little more in detail what is going on. Does there exist a way we can show
the solution graphically? Maybe something like this :

ap := Array(1 .. 1, 1 .. 3):
ap[1, 1] := plots[implicitplot3d](sqrt(x^2+y^2+z^2) = 1, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1, axes = boxed):
ap[1, 2] := plots[implicitplot3d](x+y+z = 5, x = -3 .. 3, y = -3 .. 3, z = -3 .. 3, axes = boxed):
ap[1, 3] := plot3d(sqrt(x^2+y^2), x = -3 .. 3, y = -3 .. 3, axes = boxed):
plots[display](ap);

but maybe a contor plot is more prefered to show the interaction point

Ok, nice! Now the only thing left is trying to understand what is happening.
I assume sqrt(x^2 + y^2) is the cone. The objective function x^2 + y^2 + z^2
has 3 dimensions so it might be difficult to 3dplot it. Could you please explain
a little more in detail what is going on. Does there exist a way we can show
the solution graphically? Maybe something like this :

ap := Array(1 .. 1, 1 .. 3):
ap[1, 1] := plots[implicitplot3d](sqrt(x^2+y^2+z^2) = 1, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1, axes = boxed):
ap[1, 2] := plots[implicitplot3d](x+y+z = 5, x = -3 .. 3, y = -3 .. 3, z = -3 .. 3, axes = boxed):
ap[1, 3] := plot3d(sqrt(x^2+y^2), x = -3 .. 3, y = -3 .. 3, axes = boxed):
plots[display](ap);

but maybe a contor plot is more prefered to show the interaction point