# Sum of random variables and Fourier transform

1. Jul 29, 2009

### jostpuur

If $X_1$ and $X_2$ are independent random variables in $\mathbb{R}^n$, and $\rho_{X_1}$ and $\rho_{X_2}$ are their probability densities, then let $\rho_{X_1+X_2}$ be the probability density of the random variable $X_1+X_2$. Is it true that

$$\hat{\rho}_{X_1+X_2}(\xi) = \hat{\rho}_{X_1}(\xi)\hat{\rho}_{X_2}(\xi),$$

when $\hat{\rho}$ is the Fourier transform

$$\hat{\rho}(\xi) = \int\limits_{\mathbb{R}^n} \rho(x)e^{-2\pi ix\cdot\xi} d^nx?$$

I believe it is true, but am unable to prove it when $n>1$. If $n=1$, then I can show that

$$\rho_{X_1+X_2}(x) = \int\limits_{-\infty}^{\infty}\rho_{X_1}(x-y)\rho_{X_2}(y) dy.$$

This follows from a suitable variable change:

$$P(0\leq X_1+X_2\leq a) = \int\limits_{0\leq x_1+x_2\leq a} dx_1\; dx_2\; \rho_{X_1}(x_1)\rho_{X_2}(x_2) = \int\limits_0^a\Big(\int\limits_{-\infty}^{\infty} \rho_{X_1}(x-y)\rho_{X_2}(y) dy\Big) dx$$

The result $\hat{\rho}_{X_1+X_2} = \hat{\rho}_{X_1}\hat{\rho}_{X_2}$ is then an easy result about convolutions and Fourier transforms.

If $n>1$, then a similar approach would start from the probability

$$P(X_1+X_2\in [0,a_1]\times\cdots\times [0,a_n]) = \int\limits_{\mathbb{R}^n} \Big( \int\limits_{-x_1 + [0,a]^n} \rho_{X_1}(x_1) \rho_{X_2}(x_2) d^nx_2\Big)d^nx_1$$

but now I'm not sure what kind of variable change would work.