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Math Help - Borel-Cantelli lemmas

  1. #1
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    Characteristic Function of a symmetric RV

    A random variable  X is called symmetric if  X and  -X have the same distribution. Show that the characteristic function of a symmetric random variable takes values only in the reals.
    Last edited by funnyinga; August 14th 2009 at 04:08 AM. Reason: Gave a title to a problem I was going to post orginally, but then solved later.
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  2. #2
    MHF Contributor matheagle's Avatar
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    If the distribution is symmmetric then all the odd moments are zero...

    0=E(X)=E(X^3)=...

    Then use e^{\theta}=1+\theta +{(\theta)^2\over 2!}+{(\theta)^3\over 3!}+....

    So E(e^{itX})=1+itE(X) +{E(itX)^2\over 2!}+{E(itX)^3\over 3!}+....

    =1-{t^2E(X^2)\over 2!}+...

    which no longer has those eye's in it, see my point?
    Last edited by matheagle; August 14th 2009 at 10:04 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    If the distribution is symmmetric then all the odd moments are zero...

    0=E(X)=E(X^3)=...
    Can you please explain how this is the case? I can't see it

    Then use e^{\theta}=1+\theta +{(\theta)^2\over 2!}+{(\theta)^3\over 3!}+....

    So E(e^{itX})=1+itE(X) +{E(itX)^2\over 2!}+{E(itX)^3\over 3!}+....

    =1-{t^2E(X^2)\over 2!}+...

    which no longer has those eye's in it, see my point?
    No I can't. I am completely lost with how you did this.

    The characteristic function is  \mathbb{E} e^{itX} = \int_{- \infty}^{\infty} e^{itx} d F(x) , right? I can't see what you have done here....
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  4. #4
    MHF Contributor matheagle's Avatar
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    Calc 2, I obtained the Taylor Series for e^{\theta}

    I then let \theta =itX

    i^2=-1, i^4=1 and so on.

    If the distribution function is symmetric then the odd moments are 0, so those terms drop out of the series.
    In the continuous case f(-x)=f(x) means that f(x) is an even function.
    An even times an odd function is odd.

    So, \int_{-\infty}^{\infty}x^{\rm odd \; power}f(x)dx=0.
    Last edited by matheagle; August 14th 2009 at 11:35 PM.
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