DaemonicSigil comments on The central limit theorem in terms of convolutions

[DaemonicSigil]

Thanks for the interesting post. Looking forward to seeing where this series is going. Mostly for my own understanding, I’m going to mess around with fourier transforms and convolutions in this comment.

Proof that the fourier transform of the convolution of 2 functions is the product of their fourier transforms:

$$F[f\star g]=\int dt{e}^{-i\omega t}\int d\tau f\left(\tau \right)g(t-\tau )$$

$$=\int d\tau f\left(\tau \right)e^{-i\omega \tau }\int dt{e}^{-i\omega (t-\tau )}g(t-\tau )$$

Let $u=t-\tau$. Then, holding $\tau$ constant, $du=dt$. So:

$$F[f\star g]=\int d\tau f\left(\tau \right)e^{-i\omega \tau }\int du{e}^{-i\omega u}g\left(u\right)=F\left[f\right]F\left[g\right]$$

Fourier transform of a general Gaussian distribution, $G\left(t\right)=e^{-(\frac{t-\mu }{\sigma }{)}^2}$:

$$F\left[G\right]\left(\omega \right)=\int dt{e}^{-(\frac{t-\mu }{\sigma }{)}^2}{e}^{-i\omega t}$$

Let $u=t-\mu$.

$$=\int du{e}^{-(\frac{u}{\sigma }{)}^2}{e}^{-i\omega (u+\mu )}=\int du{e}^{-(\frac{u}{\sigma }{)}^2-i\omega (u+\mu )}$$

Now we complete the square:

$$=e^{-i\omega \mu }\int du{e}^{-(\frac{u}{\sigma }{)}^2-i\omega u+(\sigma \omega {)}^2-(\sigma \omega {)}^2}$$

$$=e^{-i\omega \mu }{e}^{-(\sigma \omega {)}^2}\int du{e}^{-(\frac{u}{\sigma }+i\sigma \omega {)}^2}$$

Let $v=(\frac{u}{\sigma }+i\sigma \omega )$, so that $\sigma dv=du$.

$$=e^{-i\omega \mu }{e}^{-(\sigma \omega {)}^2}\sigma \int dv{e}^{-v^2}=e^{-i\omega \mu }{e}^{-(\sigma \omega {)}^2}\sigma \sqrt{\pi }$$

We get a result of $\sqrt{\pi }$ for the integral in $v$ using a clever trick [https://​​en.wikipedia.org/​​wiki/​​Gaussian_integral#By_polar_coordinates]. So the result is:

$$F\left[G\right]\left(\omega \right)=\sigma \sqrt{\pi }{e}^{-i\omega \mu }{e}^{-(\sigma \omega {)}^2}$$

[DaemonicSigil]

${\sigma }^2$ isn’t actually the variance here. The variance is ${\sigma }^2/2$. Sorry for the confusing choice of variable name.