October 20th, 2005, 1:38 am
sam, my sketch which is probably wrong:* g(u) is the characteristic function* f(x) is the pdf* F(x,y) is the Pr(x<X<y) your F(x) = P(x>X) = F(x,inf)The density function (pdf) is given as the Fourier inverse of g(u)f(x) = 1/pi int{ exp(-i u x) g(u), u in R }The cumulative F(x) = int{ f(s) ds, s in (x,y) }Which makes F(x) equal to the double integral F(x) = 1/pi int2{ exp(-i u s) g(u), u in R, s in (x,y) }Changing the integration order will give F(x) = 1/pi int2{ exp(-i u s) g(u), s in (x,y), u in R }and applying the integral with respect to s over (x, y) producesint{ exp(-i u s) g(u), s in (x,y) } = int{ exp(-i u s), s in (x,y) } g(u)Nowint{ exp(-i u s), s in (x,y) } = - exp(-i u y)/(i u) + exp(-i u x)/(i u)thusF(x,y) = I(y) + 1/pi int{ exp(-i u x)/(i u) g(u), u in R } where I(y) = 1/pi int{ exp(-i u y)/(i u) g(u), u in R }Your F(x) would be retrieved at the limit y->inf, where automagically I(y)->1/2 because g(u) is the Fourier transform of a pdf. Actually, it can't be anything else I guess... if we try F(-inf)=0, F(+inf)=1 etc.