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Characteristic function of a Gaussian

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  • #1
fluidistic
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Homework Statement


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.


Homework 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.


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.
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


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.


Homework 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.


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.
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
 
  • #3
fluidistic
Gold Member
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Hey Ray!
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
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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