- #1

fluidistic

Gold Member

- 3,888

- 226

## 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.

Last edited: