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Characteristic function of a Gaussian

  1. Sep 14, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must find the characteristic function of the Gaussian distribution [itex]f_X(x)=\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2} \left [ \frac{(x- \langle x \rangle )^2}{\sigma ^2} \right ]}[/itex]. If you cannot see well the latex, it's the function P(x) in http://mathworld.wolfram.com/NormalDistribution.html.


    2. Relevant equations
    The characteristic function is [itex]\phi _X(k)=\int _{-\infty}^{\infty} e^{ikx}f_X(x)dx[/itex].
    And a hint given (it seems I didn't copy it well): [itex]\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{-a}} e^{-\frac{b^2}{4a}}[/itex]. The -a in the square root bothers me, maybe I got it wrong but Wolfram Alpha isn't helping me either.


    3. The attempt at a solution
    I applied the definition of the characteristic function and reached [itex]\phi _X(k)=\frac{1}{\sqrt{2\pi} \sigma} \int _{-\infty}^{\infty} \frac{e^{ikx}}{e^{\frac{1}{2} \left [ \frac{(x-\langle x \rangle )^2}{\sigma ^2} \right ] }}dx[/itex].
    I define [itex]z=\frac{x-\langle x \rangle }{\sigma }[/itex] which pushes me to [itex]\phi _X(k)=\frac{e^{ik\langle x \rangle}}{\sqrt{2\pi}} \int _{-\infty}^{\infty}e^{ikz\sigma}-\frac{z^2}{2}dz[/itex]. It is the form I could in principle apply the hint given. But as it is here, "a" would be 1/2 and the "-a" in the square root would make no sense... or would if I consider complex numbers?!
    Considering complex numbers (I think it's fair since the characteristic function must be complex for the moments to be defined), I reach that [itex]\phi _X(k)=ie^{ik \langle x \rangle +8k^2 \sigma ^2}[/itex].
    I'd appreciate extremely very much if someone could check my result and correct me if I'm wrong. Thank you.

    Edit: I used the wolfram integrator, the correct evaluation is [itex]\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{a}} e^{\frac{b^2}{4a}}[/itex] giving me [itex]\phi _X(k)=e^{ik \langle x \rangle - \frac{k^2 \sigma ^2}{2}}[/itex]. I think it is a correct result.
     
    Last edited: Sep 15, 2012
  2. jcsd
  3. Sep 15, 2012 #2

    Ray Vickson

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    If ##X \sim \text{N}(\mu,\sigma^2)## then we can write ##X = \mu + \sigma Z,## where ##Z \sim \text{N}(0,1).## So, if ##\tilde{\phi}(w)## is the characteristic function of the unit normal Z, then the characteristic function of X is
    [tex]\tilde{\phi}_X(k) = E \,e^{i k (\mu + \sigma Z)}
    = e^{i k \mu} \tilde{\phi}(\sigma k).[/tex]

    RGV
     
  4. Sep 16, 2012 #3

    fluidistic

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    Hey Ray!
    Ok this mean what I've done has chances to be right if I understand well; your [itex]\mu[/itex] is my [itex]\langle x \rangle[/itex].
    I get the correct first 2 raw moments, which is a good sign.
     
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