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Math Help - characteristic function (evaluation of an definite integral)

  1. #1
    Super Member Random Variable's Avatar
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    characteristic function (evaluation of an definite integral)

    I want to show that the characteristic function of Cauchy's distribution ( f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0) is  \phi_{X}(t) = e^{-b|t|}

     \phi_{X}(t) = E[e^{itX}] = E[\cos(tX)] + i E[\sin (tX)]


     = \frac{b}{\pi} \int^{\infty}_{-\infty} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx + i \ \frac{b}{\pi} \int^{\infty}_{-\infty}  \frac{\sin(tx)}{b^{2}+x^{2}} \ dx


    The second integral evaluates to zero because the integrand is odd.

    The integrand of the first integral is even, so I need to evaluate  \frac{2 b}{\pi} \int^{\infty}_{0} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx

    Any suggestions?
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    I want to show that the characteristic function of Cauchy's distribution ( f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0) is  \phi_{X}(t) = e^{-b|t|}

    Any suggestions?
    Usual proofs are:

    - Fourier inversion formula. It is easy to compute the Fourier transform of e^{-b|t|} and it relates simply to Cauchy's distribution, so we can have it work backwards using Fourier inversion formula...

    - contour integration in the complex plane. The integral \int_{-R}^R \frac{e^{itx}}{1+t^2}dt is completed into a contour integral using a half circle in the upper half-plane; the integral along the circle is seen to converge to 0 as the radius goes to infinity (dominated convergence theorem) while the value of the contour integral is obtained by computing the residue at the pole i (Residue theorem).
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  3. #3
    Super Member Random Variable's Avatar
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    I hate contour integration.

    But this one is not that bad because there's a nice formula.


    Let  f(z) = \frac{e^{itz}}{b^{2}+z^{2}}

    which has the simple pole  z = bi in the upper half plane


    so  \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = -2 \pi \ Im (Res\{f,bi\})

     Res\{f,bi\} = \lim_{z \to bi} (z-bi) \frac{e^{itz}}{b^{2}+z^{2}}= \lim_{z \to bi} \frac{e^{itz}}{z+bi} = \frac{e^{-bt}}{2bi} = -i \ \frac{e^{-bt}}{2b}

    so  \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = \pi \ \frac{e^{-bt}}{b}


    and  \phi_{X}(t) = e^{-bt}

    From where is the absolute value coming?
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  4. #4
    MHF Contributor

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    Quote Originally Posted by Random Variable View Post

    From where is the absolute value coming?
    It's hidden in your computation. The integral along the upper half-circle converges to 0 only when t>0. Otherwise, you have the consider the lower half-circle hence the pole is -ib (or you can note from the beginning that the characteristic function is even by symmetry of the Cauchy distribution).
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