High order joint moments calculation


by Wu Xiaobin
Tags: calculation, joint, moments, order
Wu Xiaobin
Wu Xiaobin is offline
#1
Dec7-12, 01:58 AM
P: 28
Recently I have been working on classical Gaussian electrical field and I come through this joint
moments calculation.
Suppose we got the joint density function as:
[itex]
p(s_i,s_j)=\frac{1}{2\pi d}\exp{[\frac{1}{2d}(<s_i>s_i+<s_j>s_j)]}K_0(\frac{1}{2d}\sqrt{<I>^2-s_k^2}\sqrt{s_i^2+s_j^2})
[/itex]
[itex]<I>,<s_i>,<s_j>,<s_k>[/itex] are the known mean of [itex]I,s_i,s_j,s_k[/itex].
the high order moment of [itex]<s_i^n>[/itex] can be calculated as:
[itex]
<s_i^n>=\frac{n!}{2^{n+1}\sqrt{<I>^2-<s_j>^2-<s_k>^2}}[(\sqrt{<I>^2-<s_j>^2-<s_k>^2}+<s_i>)^{n+1}+(-1)^n(\sqrt{<I>^2-<s_j>^2-<s_k>^2}-<s_i>)^{n+1}]
[/itex]
with the above result, The paper I referred to give the following conclusion which I can't catch up with:
[itex]
<s_i^ns_j^m>=(\frac{2d}{<s_j>})^m[\frac{\partial^m<s_i^n(x,s_j)>}{\partial x^m}]_{x=1}
[/itex]
where
[itex]
<s_i^n(x,s_j)>
[/itex]
is given by multiplying [itex]<s_j>^2[/itex], in the square brackets in [itex] <s_i^n>[/itex]
Can anyone tell me why the author calculates the joint moments in that way.

Look forward to your reply
Sincerely yours
Jacky Wu
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