Using Characteristic Functions to Show Levy Sum Closure

# Probability density function of the sum of two independent Levy-distributed random variables?


# Probability density function of the sum of two independent Levy-distributed random variables?












I posted the following questions in math stack exchange https://math.stackexchange.com/posts/2762047/edit Here's the text:

Prove that the sum of two independent Levy-distributed (having parameter $c$) random variables has also Levy distribution with parameter $4c$.

Idea of the proof By Levy-Khitchine theorem one can derive the characteristic function of Levy distribution and then apply the inverse Fourier transform.

My question is, is there a more intuitive a less computational way to deduce the Levy distribution?

Levy distribution:

$p(x) = \sqrt{\frac{c}{2\pi}}\frac{e^{-\frac{c}{2x}}}{x^{3/2}}$

## Answer by LocalVolatility (score 2, accepted)

https://quant.stackexchange.com/a/39554

You approach sounds good, but there is no need to compute the inverse Fourier transform.

The characteristic function of a Levy-distributed random variable with parameter $c$ is given by

\begin{equation} \phi(\omega; c) = \exp \left\{ -\sqrt{-2 \mathrm{i} c t} \right\}. \end{equation}

Thus, the characteristic function of the sum of two i.i.d. Levy-distributed random variables $X$ and $Y$, has the characteristic function

\begin{equation} \mathbb{E} \left[ e^{\mathrm{i} \omega (X + Y)} \right] = \mathbb{E} \left[ e^{\mathrm{i} \omega X} \right] \mathbb{E} \left[ e^{\mathrm{i} \omega Y} \right] = \left( \mathbb{E} \left[ e^{\mathrm{i} \omega X} \right] \right)^2, \end{equation}

where the first equality follows from independence and the second from $X$ and $Y$ being identically distributed. We get

\begin{equation} \ldots = \exp \left\{ -2 \sqrt{-2 \mathrm{i} c t} \right\} = \exp \left\{ -\sqrt{-2 \mathrm{i} (4 c) t} \right\} = \phi(\omega; 4c). \end{equation}

This is sufficient to conclude that $X + Y$ is Levy-distributed with parameter $4c$.