most recent 30 from http://mathoverflow.net 2013-05-22T04:48:50Z http://mathoverflow.net/feeds/question/106827 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106827/fourier-inversion-formula-for-complex-valued-random-variables Joshua Cooper 2012-09-10T16:01:50Z 2012-09-10T18:44:43Z

<p>The characteristic function of a complex-valued random variable $X$ with pdf $\mu$ is given by $$ \phi(t) = \int \exp[i \Re(\bar{t} X)] \; d\mu $$ (or, so says Wikipedia). How does one recover the pdf from $\phi$, i.e., what is the Fourier inversion formula for measures on $\mathbb{C}$? The $\mu$ I am working with is as "nice" as one could ask for.</p> <p>P.S. Where would I find such a result? (Of course, I could try to work out the exact form of Pontryagin Duality for $\mathbb{C}$ from the definitions, but presumably somebody has done this before.)</p>

http://mathoverflow.net/questions/106827/fourier-inversion-formula-for-complex-valued-random-variables/106832#106832 Carlo Beenakker 2012-09-10T17:06:23Z 2012-09-10T17:06:23Z

<p>you have to take the real part of $\bar{t}X$ in the exponent, in order for the integral to make sense; if you then decompose $t=t_1+it_2$ and $X=X_1+iX_2$ into real and imaginary parts, you just have a conventional two-dimensional Fourier transform</p> <p>$\phi(t_1,t_2)=\int_{-\infty}^{\infty}dX_1 \int_{-\infty}^{\infty}dX_2 exp(it_1 X_1+i t_2 X_2) P(X_1,X_2)$</p> <p>the inverse is, as usual,</p> <p>$P(X_1,X_2)=(2\pi)^{-2}\int_{-\infty}^{\infty}dt_1 \int_{-\infty}^{\infty}dt_2 exp(-it_1 X_1-i t_2 X_2) \phi(t_1,t_2)$</p>