Given a multivariate normal distribution (mu, sigma), is there an
explicit formula for the expectation of the magnitude of the vectors
from that distibution? That is

f~N(u,S)

/
|f(x)|x|dx   where x is a vector, so its a multiple integral
/

wicksom-ga,

I'll give this a go. This definitely can be classified as a "question
answered by an enthusiast" on the pricing guide
(https://answers.google.com/answers/pricing.html) - I want to know the
answer myself, so I'll walk you through my thought process.

If x is a vector then |x| = (x.x)^1/2 where x.x = sum(xi˛).

In other words if x is an n vector then |x|˛ is a chi-squared
distribution with n degrees of freedom.

Not surprisingly, if |x|˛ is a chi-squared distribution, then |x| is a
chi-distribution (yes one actually does exist, you just don't hear
much about it).

Knowing that much all I had to do was either figure out the moment
generating function for a chi-distribution... or... find someone who
had done the dirty work for me.

A great place to find almost all the math answers you are looking for
is Mathworld (www.mathworld.com). Here's the link to the page on the
chi-distribution :

http://mathworld.wolfram.com/ChiDistribution.html

The chi distribution has a mean given by :

sqrt(2) * Gamma(.5*(n+1)) / Gamma(.5*n)

where Gamma is the gamma function (see
http://mathworld.wolfram.com/GammaFunction.html)

Hope this helps

calebu2-ga

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Clarification of Answer by calebu2-ga on 18 Dec 2002 13:56 PST

I should point out that this is for a standard multinormal distributed
x. If u<>0 and S<>I (identity), then you have x.x being noncentral
chi-squared. The math gets a little more complex for figuring out the
mean (lets put it this way, there are only two pages listed when you
search for "noncentral chi distribution" on google).

Useful resources :

Discussion list about the |x| distribution
http://www.hsph.harvard.edu/cgi-bin/lwgate/STATALIST/archives/statalist.0201/Author/article-479.html

The noncentral chi-SQUARED distribution from mathworld
http://mathworld.wolfram.com/NoncentralChi-SquaredDistribution.html

Search strategy :
"noncentral chi distribution"

Regards

calebu2-ga

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Request for Answer Clarification by wicksom-ga on 18 Dec 2002 14:31 PST

calebu2-ga

Thanks for the very useful insight that it can be thought of as a
non-central chi-distribution. Do you know what the problem would 
be called if S<>I, the variables are correlated?

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Clarification of Answer by calebu2-ga on 18 Dec 2002 15:44 PST

A useful way of thinking about a multinormal distribution is in terms
of a 3D density plot. All normal distrubutions have "iso-density"
ellipses around the mean.

A standard normal distribution is just a series of concentric spheres
- each one representing a larger and larger confidence interval around
the mean (zero).

If you ignore the mean for a second - all that adding correlation to
the covariance matrix does is stretch and rotate the distribution -

if X ~ N(0,I) and Y ~ N(0,S) you can generate from Y by taking an
observation from X and multiplying it by S^(1/2).

If S is just a rotation of I, then |x| will not change - as rotation
does not affect length. If however S is not a rotation (ie it's
eigenvalues are not 1) then |x| will be inextricably related to the
distortion of the variances. My guess is that it will be strongly
related to the average eigenvalue (the "true variances" after
correlation has been "rotated" away").

How you proceed with x ~ N(u,S) is something that you are going to
have to sit down and figure out with diagrams. To understand this will
require drawing lots of triangles and figure out how |x| relates to
|y| = |x - u|.

calebu2-ga