- #1
jostpuur
- 2,112
- 19
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
<br /> \hat{\rho}_{X_1+X_2}(\xi) = \hat{\rho}_{X_1}(\xi)\hat{\rho}_{X_2}(\xi),<br />
when \hat{\rho} is the Fourier transform
<br /> \hat{\rho}(\xi) = \int\limits_{\mathbb{R}^n} \rho(x)e^{-2\pi ix\cdot\xi} d^nx?<br />
I believe it is true, but am unable to prove it when n>1. If n=1, then I can show that
<br /> \rho_{X_1+X_2}(x) = \int\limits_{-\infty}^{\infty}\rho_{X_1}(x-y)\rho_{X_2}(y) dy.<br />
This follows from a suitable variable change:
<br /> 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)<br /> = \int\limits_0^a\Big(\int\limits_{-\infty}^{\infty} \rho_{X_1}(x-y)\rho_{X_2}(y) dy\Big) dx<br />
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
<br /> 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<br />
but now I'm not sure what kind of variable change would work.
<br /> \hat{\rho}_{X_1+X_2}(\xi) = \hat{\rho}_{X_1}(\xi)\hat{\rho}_{X_2}(\xi),<br />
when \hat{\rho} is the Fourier transform
<br /> \hat{\rho}(\xi) = \int\limits_{\mathbb{R}^n} \rho(x)e^{-2\pi ix\cdot\xi} d^nx?<br />
I believe it is true, but am unable to prove it when n>1. If n=1, then I can show that
<br /> \rho_{X_1+X_2}(x) = \int\limits_{-\infty}^{\infty}\rho_{X_1}(x-y)\rho_{X_2}(y) dy.<br />
This follows from a suitable variable change:
<br /> 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)<br /> = \int\limits_0^a\Big(\int\limits_{-\infty}^{\infty} \rho_{X_1}(x-y)\rho_{X_2}(y) dy\Big) dx<br />
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
<br /> 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<br />
but now I'm not sure what kind of variable change would work.
