Inversion of Cumulant Generating Function

Inversion of Cumulant Generating Function

Example: to get from the cumulant generating function back to the actual pdf

Eg. For normal distribution

$K_X(t)=\mu t + \frac{\sigma^2 t^2}{2}$, therefore $M_X(t)=e^{\mu t + \frac{\sigma^2 t^2}{2}}$

$f(x)=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty}M_X(it)e^{ixt} dt$: Fourier Transform

$f(x)=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{\mu i t + \frac{\sigma^2 i^2 t^2}{2}}e^{ixt} dt$

$=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{\frac{-\sigma^2 t^2}{2}}e^{i(\mu t+xt)} dt$

$=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{\frac{-\sigma^2 t^2}{2}} \left(cos(\mu t + xt) + i sin(\mu t + xt)\right) dt$

The real valued probability density function is:

$=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{\frac{-\sigma^2 t^2}{2}} cos(\mu t + xt) dt$

Extension to more complicated functions

If we extend the cumulant generating function to:

$K_X(t)=\mu t + \frac{\sigma^2 t^2}{2} + \frac{\kappa_3 t^3}{3!} + \frac{\kappa_4 t^4}{4!}$, then

$f(x)=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{\mu i t + \frac{\sigma^2 i^2 t^2}{2} + \frac{\kappa_3 i^3 t^3}{3!} + \frac{\kappa_4 i^4 t^4}{4!} }e^{ixt} dt$

$f(x)=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{-\frac{\sigma^2 t^2}{2}+ \frac{\kappa_4 t^4}{24}}e^{i(\mu t + xt - \frac{\kappa_3 t^3}{6})} dt$

$f(x)=\frac{1}{2 \pi} \displaystyle\int_{-\infty}^{\infty} e^{-\frac{\sigma^2 t^2}{2}+ \frac{\kappa_4 t^4}{24}} \left(cos(\mu t + xt - \frac{\kappa_3 t^3}{6})+i sin(\mu t + xt - \frac{\kappa_3 t^3}{6})\right) dt$