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From: wiso on 22 Apr 2010 14:56
Suppose we have a 3D unitary vector x with a random isotropic
direction. If we generate a random walk starting from zero summing n
of this random variable it's easy to proof that the expected value of |
xn|^2 (where xn is the sum of n unitary vectors) is

<|x1|^2> = 1
<|x2|^2> = <|x1|^2 + |x|^2 + 2 * |x| |x1| * cos theta> = 1 + 1 + 0 =
2
where theta is the angle between the two vectors. The last term is
zero because \int cos(theta) d\omega / 4pi = 0 where omega is the
solid angle

In general:
<|xn|^2> = n

but what about the variance after n steps, V( |xn|^2)? I can't proof
anything:

V( |x1|^2 ) = 0
V( |x2|^2 ) = V( |x1|^2 + |x|^2 + 2 |x1||x| cos theta) = 1 * 1 * 4
V(cos(theta))
V(cos theta) = \int cos^2 d\omega / 4pi = 1/3, then
V( |x2|^2 ) = V( |x1|^2 + |x|^2 + 2 |x1||x| cos theta) = 4/3
is it right? Then?

V( |x3|^2 ) = V( |x2|^2 + |x|^2 + 2 |x| |x2| cos theta) =
= 4/3 + 0 + 4 V( |x| |x2| cos theta) = ?
From: Robert Israel on 22 Apr 2010 19:52
wiso <giurrero(a)gmail.com> writes:

> Suppose we have a 3D unitary vector x with a random isotropic
> direction. If we generate a random walk starting from zero summing n
> of this random variable it's easy to proof that the expected value of |
> xn|^2 (where xn is the sum of n unitary vectors) is
>
> <|x1|^2> = 1
> <|x2|^2> = <|x1|^2 + |x|^2 + 2 * |x| |x1| * cos theta> = 1 + 1 + 0 =
> 2
> where theta is the angle between the two vectors. The last term is
> zero because \int cos(theta) d\omega / 4pi = 0 where omega is the
> solid angle
>
> In general:
> <|xn|^2> = n
>
> but what about the variance after n steps, V( |xn|^2)? I can't proof
> anything:
>
> V( |x1|^2 ) = 0
> V( |x2|^2 ) = V( |x1|^2 + |x|^2 + 2 |x1||x| cos theta) = 1 * 1 * 4
> V(cos(theta))
> V(cos theta) = \int cos^2 d\omega / 4pi = 1/3, then
> V( |x2|^2 ) = V( |x1|^2 + |x|^2 + 2 |x1||x| cos theta) = 4/3
> is it right? Then?
>
> V( |x3|^2 ) = V( |x2|^2 + |x|^2 + 2 |x| |x2| cos theta) =
> = 4/3 + 0 + 4 V( |x| |x2| cos theta) = ?

For convenience write x_n = [x,y,z], x_{n+1} = [x+dx, y+dy, z+dz]
where x,y,z are independent of dx,dy,dz. Note that dx, dy, dz have
marginal densities 1/2 on [-1,1].

E[|x_n|^4] = E[(x^2 + y^2 + z^2)^2] = 3 E[x^4] + 6 E[x^2 y^2]
(by symmetry)
E[(x+dx)^4] = E[x^4 + 4 x^3 dx + 6 x^2 (dx)^2 + 4 x (dx)^3 + (dx)^4]
= E[x^4] + 2 E[x^2] E[(dx)^2] + E[(dx)^4]
= E[x^4] + 2/3 n + 1/5
Thus E[x^4] = n (5 n - 2)/15

E[(x+dx)^2 (y+dy)^2] = E[x^2 y^2] + 2 E[x^2] E[(dy)^2] + E[(dx)^2 (dy)^2]
= E[x^2 y^2] + 2 n/9 + 1/15
Thus E[x^2 y^2] = n (5 n - 2)/45

And so E[|x_n|^4] = n(5 n - 2)/3
Since E[|x_n|^2] = n, that makes V(|x_n|^2) = 2 n (n-1)/3.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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