Characteristic function of a random Gaussian variable

I have to find the characteristic function of a random Gaussian variable of $$\sigma_z (w) = E e^{i w z }$$. This is the variable and I know , from the theory that the characteristic function of this variable is the Fourier transform of the probability density. From theory I read that the probability density is $$\frac{1}{\sigma \sqrt 2 \pi } e^{- \frac{(z - \mu)^2 }{2 \sigma^{2}}}$$. Now I’m blocked because I don’t know how to calculate this :/ Thank you so much

In general, the characteristic function of a random variable is related to the fourier transform of the distribution as follows:

$$\varphi_Z(-\omega) = \mathscr F \big\{ f_Z(z) \big\}$$

Why?

Because, fourier transform of a PDF $$f_Z(z)$$ is:

$$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} f_Z(z)\cdot e^{-j\omega z} \ \mathrm dz$$

And Characteristic function of a PDF is given by:

$$\varphi_Z(\omega) = \mathbf E_Z \big\{e^{j\omega z} \big\} = \int^{\infty}_{-\infty} f_Z(z) \cdot e^{j\omega z} \ \mathrm dz$$

As you can see the difference is just in the argument $$\omega$$. I think now you can get the characteristic function of a Gaussian Random Variable. Just calculate the fourier transform of the PDF and replace $$\omega$$ by $$-\omega$$.

Fourier Transform of Gaussian PDF :

$$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(z-\mu)^2}\cdot e^{-i\omega z} \ \mathrm dz$$

Now, just consider the power of exponential and simplify as follows: $$(\frac{-(z-\mu)^2}{2\sigma^2} - i\omega z )= -\frac{1}{2\sigma^2}(z^2+\mu^2-2\mu z + i2\sigma^2 \omega z)$$ $$= -\frac{1}{2\sigma^2}(z^2 -2(\mu - i\sigma^2 \omega)z + \mu^2 + (\mu - i\sigma^2 \omega)^2 - (\mu - i\sigma^2 \omega)^2)$$ $$= -\frac{1}{2\sigma^2}((z - (\mu - i\sigma^2 \omega))^2 - (\mu - i\sigma^2 \omega)^2 + \mu^2)$$

Now, basically, the integral becomes the following: $$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(z-(\mu - i\sigma^2 \omega))^2}\cdot e^{\frac{1}{2\sigma^2}((\mu - i\sigma^2 \omega)^2 - \mu^2)} \ \mathrm dz$$ The second exponential can come out of the integral since it does not contain $$z$$, then what you are left with is just area under a Gaussian PDF with mean $$(\mu - i\sigma^2 \omega)$$ and variance $$\sigma^2$$. Area under any PDF is equal to $$1$$. So, basically the fourier transform of the Gaussian PDF becomes: $$\mathscr F \big\{ f_Z(z) \big\} = e^{\frac{1}{2\sigma^2}((\mu - i\sigma^2 \omega)^2 - \mu^2)} = e^{-\frac{\sigma^2 \omega^2}{2} - i\mu \omega}$$ Once you have fourier transform, you can get the characteristic function by substituting $$\omega = -\omega$$ as follows: $$\varphi_Z(\omega) = e^{-\frac{\sigma^2 \omega^2}{2} + i\mu \omega}$$

• Assuming $$f_z z$$ as $$\frac{1}{\sigma \sqrt 2 \pi } e^{\frac{-(z-\mu)^2}{2\sigma ^{2}}}$$ now I don’t know how to solve $$\frac{1}{\sigma \sqrt 2 \pi } \int_{\infty}^{-\infty} e^{\frac{-(z-\mu)^2}{2\sigma ^{2}}+iwz}$$ Apr 20 '20 at 14:26
• @ElenaMartini Try to write $e^{\frac{-(z-\mu)^2}{2\sigma^2} + \mathbb i \omega z}$ as sum of a complete square and a residual term without $z$. Apr 20 '20 at 14:36
• @ElenaMartini I have added the derivation in case you need further help. Apr 20 '20 at 15:17
• Thank you so much , now I’ll do all the step from the beginning for better understand Apr 20 '20 at 15:40