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Re: Characteristic Function Question

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  • Noel Vaillant
    ... I have never done this calculation before, so my level of confidence is not too high, but here is what I would do in the case when X~N (0,1) [unfortunately
    Message 1 of 21 , Aug 2, 2003
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      > Can someone show me how to obtain the characteristic
      > function for X^2, where X ~ N(mu,sigma^2)?

      I have never done this calculation before, so my level of confidence
      is not too high, but here is what I would do in the case when X~N
      (0,1) [unfortunately I can easily extend it to N(0,s^2), but I don't
      have an answer for the general case N(m,s^2)]

      Given u in R, we want E[exp(iuX^2)] where X~N(0,1), i.e. we want the
      integral:

      I=(1/sqrt(2pi))int_R exp(iux^2-x^2/2)dx
      =(1/sqrt(2pi))int_R exp(-cx^2/2)dx

      where c is the complex number c=1-2iu. Using Fubini theorem, we have:

      I^2=(1/2pi)int_R^2 exp(-c(x^2+y^2)/2)dxdy

      and after switching to polar coordinates and integrating w.r. to the
      angle:

      I^2=int_0^+oo rexp(-cr^2/2)dr=1/c=(1+2iu)/(1+4u^2) (*)

      The complex number 1+2iu has two roots:

      + or - [((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2] (**)

      where q(u)=sqrt(1+4u^2) and sgn(u)=u/|u| (for u<>0) is the sign of u.

      It is not too difficult to convince yourself that for u>0, the
      imaginary part of I should be positive. From this fact and (*) and
      (**) we obtain:

      I=(1/q(u))[((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]

      I am sorry I can't do better that this for the time being.

      Regards Noel.
    • Tony Wong
      The generalized version of this question is in fact to find the characteristic function of Y=X1^2+X2^2+...+Xn^2 where X1, X2,....are independent normal random
      Message 2 of 21 , Aug 2, 2003
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        The generalized version of this question is in fact to
        find the characteristic function of
        Y=X1^2+X2^2+...+Xn^2
        where X1, X2,....are independent normal random
        variable N(Ui, 1) i=1,2,....n. This question arises
        from my course work in graduate multivariate analysis.
        Now here the question is simpler because n=1. So
        direct derivation like the way Noel does makes the
        best sense and is possible. But when n is arbitrary,
        you need more work and that's something you need from
        a book like "Introduction to Multivariate Analysis" by
        T. Anderson. I am not gonna type the long derivation.
        But I believe the solution should be:

        (1-2*i*t*s^2)^(-1/2)*exp{i*t*s^2*mu^2*(1-2*i*t*s^2)^(-1)}


        --- Noel Vaillant <vaillant@...>
        的郵件內容:
        ---------------------------------
        > Can someone show me how to obtain the characteristic
        > function for X^2, where X ~ N(mu,sigma^2)?

        I have never done this calculation before, so my level
        of confidence
        is not too high, but here is what I would do in the
        case when X~N
        (0,1) [unfortunately I can easily extend it to
        N(0,s^2), but I don't
        have an answer for the general case N(m,s^2)]

        Given u in R, we want E[exp(iuX^2)] where X~N(0,1),
        i.e. we want the
        integral:

        I=(1/sqrt(2pi))int_R exp(iux^2-x^2/2)dx
        =(1/sqrt(2pi))int_R exp(-cx^2/2)dx

        where c is the complex number c=1-2iu. Using Fubini
        theorem, we have:

        I^2=(1/2pi)int_R^2 exp(-c(x^2+y^2)/2)dxdy

        and after switching to polar coordinates and
        integrating w.r. to the
        angle:

        I^2=int_0^+oo rexp(-cr^2/2)dr=1/c=(1+2iu)/(1+4u^2) (*)

        The complex number 1+2iu has two roots:

        + or - [((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]
        (**)

        where q(u)=sqrt(1+4u^2) and sgn(u)=u/|u| (for u<>0) is
        the sign of u.

        It is not too difficult to convince yourself that for
        u>0, the
        imaginary part of I should be positive. From this fact
        and (*) and
        (**) we obtain:

        I=(1/q(u))[((q(u)+1)/2)^(1/2)+isgn(u)((q(u)-1)/2)^1/2]

        I am sorry I can't do better that this for the time
        being.

        Regards Noel.





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