# Characteristic function of a Gaussian

Gold Member

## Homework Statement

I must find the characteristic function of the Gaussian distribution $f_X(x)=\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2} \left [ \frac{(x- \langle x \rangle )^2}{\sigma ^2} \right ]}$. 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 $\phi _X(k)=\int _{-\infty}^{\infty} e^{ikx}f_X(x)dx$.
And a hint given (it seems I didn't copy it well): $\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{-a}} e^{-\frac{b^2}{4a}}$. 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 $\phi _X(k)=\frac{1}{\sqrt{2\pi} \sigma} \int _{-\infty}^{\infty} \frac{e^{ikx}}{e^{\frac{1}{2} \left [ \frac{(x-\langle x \rangle )^2}{\sigma ^2} \right ] }}dx$.
I define $z=\frac{x-\langle x \rangle }{\sigma }$ which pushes me to $\phi _X(k)=\frac{e^{ik\langle x \rangle}}{\sqrt{2\pi}} \int _{-\infty}^{\infty}e^{ikz\sigma}-\frac{z^2}{2}dz$. 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 $\phi _X(k)=ie^{ik \langle x \rangle +8k^2 \sigma ^2}$.
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 $\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{a}} e^{\frac{b^2}{4a}}$ giving me $\phi _X(k)=e^{ik \langle x \rangle - \frac{k^2 \sigma ^2}{2}}$. I think it is a correct result.

Last edited:

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

I must find the characteristic function of the Gaussian distribution $f_X(x)=\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2} \left [ \frac{(x- \langle x \rangle )^2}{\sigma ^2} \right ]}$. 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 $\phi _X(k)=\int _{-\infty}^{\infty} e^{ikx}f_X(x)dx$.
And a hint given (it seems I didn't copy it well): $\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{-a}} e^{-\frac{b^2}{4a}}$. 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 $\phi _X(k)=\frac{1}{\sqrt{2\pi} \sigma} \int _{-\infty}^{\infty} \frac{e^{ikx}}{e^{\frac{1}{2} \left [ \frac{(x-\langle x \rangle )^2}{\sigma ^2} \right ] }}dx$.
I define $z=\frac{x-\langle x \rangle }{\sigma }$ which pushes me to $\phi _X(k)=\frac{e^{ik\langle x \rangle}}{\sqrt{2\pi}} \int _{-\infty}^{\infty}e^{ikz\sigma}-\frac{z^2}{2}dz$. 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 $\phi _X(k)=ie^{ik \langle x \rangle +8k^2 \sigma ^2}$.
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 $\int _{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt {\frac{\pi}{a}} e^{\frac{b^2}{4a}}$ giving me $\phi _X(k)=e^{ik \langle x \rangle - \frac{k^2 \sigma ^2}{2}}$. 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
$$\tilde{\phi}_X(k) = E \,e^{i k (\mu + \sigma Z)} = e^{i k \mu} \tilde{\phi}(\sigma k).$$

RGV

Gold Member
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
$$\tilde{\phi}_X(k) = E \,e^{i k (\mu + \sigma Z)} = e^{i k \mu} \tilde{\phi}(\sigma k).$$

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