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

by sarperb
Tags: char, dist, exponential, func, random, variables
 P: 2 1. The problem statement, all variables and given/known data $F_{X}(x)= λe^{-λx} \;for\; x>0 \;\;\;and \;0 \;otherwise$ After finding the characteristic function for the Exponential Distribution, which is (I could do this without problem); $F_{X}(k)=λ(λ-ik)^{-1}$ Now the question is; Let $X_1,X_2,\ldots,X_i$ be i.i.d. exponential random variables with parameter λ and let; $Y_N=\sum_{i=1}^{N}X_i$ (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 $Y_N$ is given as; $f_Y(y) = λ\frac{(λy)^{N-1}}{(N-1)!}e^{-λy}$ 2. Relevant equations $\Gamma(n) = \int_0^{\infty}t^{n-1}e^{-t}dt$ Which is (n-1)! for n>0 together with $\Gamma(1/2)=\sqrt{\pi}$ Also with a simple substutition of t=az dt = adz $\Gamma(n) = a^n\int_0^{\infty}z^{n-1}e^{-za}dz$ I used this to show in the same question that $= n!λ^{-n}$ 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; $F_{Y}(k)=λ^N(λ-ik)^{-N}$ 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 $F_X(x)$ 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; $f_Y(y) = λ^N \int_{0 or -\infty}^{\infty}(λ-ik)^{-N}e^{-iky}dk = I$ 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; $(λ-ik)^{-N} =\frac{i^N}{N!}\frac{d^N(λ-ik)^{-1}}{dk^N}$ 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; $Ie^{λy}=λ^N\int_{0 or -\infty}^{\infty}(λ-ik)^{-N}e^{(λ-ik)y}dk$ 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.