Find the dist. Func. of Random Variables from Exponential Dist. by using Char. Func.

  1. 1. The problem statement, all variables and given/known data
    [itex]F_{X}(x)= λe^{-λx} \;for\; x>0 \;\;\;and \;0 \;otherwise[/itex]

    After finding the characteristic function for the Exponential Distribution, which is (I could do this without problem);
    [itex]F_{X}(k)=λ(λ-ik)^{-1}[/itex]

    Now the question is;

    Let [itex]X_1,X_2,\ldots,X_i[/itex] be i.i.d. exponential random variables with parameter λ and let;

    [itex]Y_N=\sum_{i=1}^{N}X_i[/itex] (Sum starts from i=1, I am new to LaTeX, I'm not sure if this is the right way to express the end points of the sum)

    Using the generating function method, show that the pdf of [itex]Y_N[/itex] is given as;

    [itex]f_Y(y) = λ\frac{(λy)^{N-1}}{(N-1)!}e^{-λy}[/itex]

    2. Relevant equations

    [itex]\Gamma(n) = \int_0^{\infty}t^{n-1}e^{-t}dt[/itex]
    Which is (n-1)! for n>0 together with [itex]\Gamma(1/2)=\sqrt{\pi}[/itex]

    Also with a simple substutition of t=az dt = adz
    [itex]\Gamma(n) = a^n\int_0^{\infty}z^{n-1}e^{-za}dz[/itex]

    I used this to show in the same question that
    [itex]<X^n>= n!λ^{-n}[/itex]

    There is a Hint in this part of the question which says (exact copy);
    "You can do this without having to explicitly do the k-integral of the
    inverse Fourier transform. Instead show that this integral can be written as a
    higher-order derivative with respect to a parameter inside a simpler integral,
    whose result you already now"

    3. The attempt at a solution
    From the fact that the sum of random variables applies to generating functions as multipication, it can easily be found that;
    [itex]F_{Y}(k)=λ^N(λ-ik)^{-N}[/itex]

    Now my first problem is taking the inverse Fourier Transform of this guy because I am not sure what the end points of the integral should be. In the first part where I was finding the Characteristic Function of the Exponential Distribution, it was easy to see since [itex]F_X(x)[/itex] was defined to be 0 when x is negative. But now taking the inverse fourier transform, should I leave the limits from -infinity to infinity as it is for the usual Fourier Transform, or should they be from 0 to infinity?

    The answer to this question won't help me solve my problem since I tried with both, but I want to learn how should I be thinking here to get to the right answer.

    So the Fourier Transform looks like;

    [itex]f_Y(y) = λ^N \int_{0 or -\infty}^{\infty}(λ-ik)^{-N}e^{-iky}dk = I[/itex]

    The result is given but I can't get myself to it. I played with this for hours and I am at a point where since I did focus on something for too long, I lost perspective and can't have any new ideas to try.

    We know I am not to solve the integral explicitly, instead change it into something I already now. I tried;

    [itex](λ-ik)^{-N} =\frac{i^N}{N!}\frac{d^N(λ-ik)^{-1}}{dk^N}[/itex]

    I played with the Gamma Function etc. By the way I should say, I don't think we are meant to be familiar with the Incomplete Gamma Function.
    Since I need to get (N-1)! at the bottom of the fraction I know somehow the reciprocal of the Gamma Function is to be found here, but I don't think I am meant to know the reciprocal of the Gamma Function, so I need to somehow get (N-1)! out and from the remaining integral get y^(N-1).

    Also please note that if the initial integral is called I, then;

    [itex]Ie^{λy}=λ^N\int_{0 or -\infty}^{\infty}(λ-ik)^{-N}e^{(λ-ik)y}dk[/itex]

    I of course also tried the substition of u = (λ-ik) du = -idk idu = dk and several other substitions similar to this.

    I would have written a lot more about my hours of attempts, but I really don't think spending hours on typing in LaTeX is necessary at this point :)
    I would really appreciate if someone could push me in the right direction since after hours I am stuck at the same thoughts and can't continue. :)

    P.S. I forgot 1/2pi in the inverse fourier transforms :)

    Thank you for all your help.
     
    Last edited: Oct 15, 2011
  2. jcsd
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