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High order joint moments calculation 
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#1
Dec712, 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>^2s_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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