+ Reply to Thread
Results 1 to 2 of 2

Thread: cumulants

  1. #1
    Points: 632, Level: 12
    Level completed: 64%, Points required for next Level: 18

    Posts
    24
    Thanks
    8
    Thanked 2 Times in 2 Posts

    cumulants




    I don't know much about cumulants. What I know is that they can be generated by using the cumulant generating function g(t) = \log(E[\exp(tX)]).
    I'm studying the so-called Independent Component Analysis and my book says that if the pdfs involved are symmetric then the odd order cumulants are zero. Can you tell me why?

  2. #2
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: cumulants


    Let me try.

    When the pdf f_X is symmetric about 0, we have

    f_X(x) = f_X(-x) ~~ \forall x \in \mathbb{R}

    Next we consider the odd moments

    m_{2k-1} = E[X^{2k-1}]

    = \int_{-\infty}^{+\infty} x^{2k-1}f_X(x)dx

    = \int_{-\infty}^0 x^{2k-1}f_X(x)dx + \int_0^{+\infty} x^{2k-1}f_X(x)dx

    Consider the former integral:

    \int_{-\infty}^0 x^{2k-1}f_X(x)dx

    = \int_{-\infty}^0 x^{2k-1}f_X(-x)dx

    Next let y = -x; then when x \to -\infty, y \to +\infty; x = 0, y = 0; dx = -dy

    = \int_0^{+\infty} (-y)^{2k-1}f_X(y)dy

    = -\int_0^{+\infty} y^{2k-1}f_X(y)dy

    which cancel with the latter integral. Essentially, f_X(x) is an even function and x^{2k-1} is an odd function, therefore their product, the integrand x^{2k-1}f_X(x) is also an odd function which should integrate to 0 as expected.

    The moment-generating function in this case can be expressed as

    E[e^{tX}] = 1 + \sum_{k=1}^{+\infty} m_{2k} \frac {t^{2k}} {(2k)!}

    since all the odd moments vanish.

    Denote h(t) =  \sum_{k=1}^{+\infty} m_{2k} \frac {t^{2k}} {(2k)!}

    such that E[e^{tX}] = 1 + h(t)

    http://en.wikipedia.org/wiki/Mercator_series

    From the above result we know that the cumulant generating function is

    g(t) = \ln E[e^{tX}] = \ln(1 + h(t)) = \sum_{n=1}^{+\infty} \frac {(-1)^{n+1}} {n}h(t)^n

    As we know all the coefficients of odd powered terms in h(t) are 0, it should be the same for h(t)^n and thus g(t);
    and by definition the cumulants \kappa_n satisfies

    g(t) = \sum_{n=1}^{+\infty} \kappa_n \frac {t^n} {n!}

    By comparing the coefficients, we can immediately claim that all the odd cumulants \kappa_{2k-1} vanish.

    There maybe some shorter and smarter ways to show this without going through the technical details in the above arguments. (There are always some technical conditions need to be considered whenever you are dealing with sum to infinity)

  3. The Following User Says Thank You to BGM For This Useful Post:

    Kiuhnm (08-29-2013)

+ Reply to Thread

           




Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats