Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

High order joint moments calculation

  1. Dec 7, 2012 #1
    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
     

    Attached Files:

  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted