2
$\begingroup$

Well... I spent an hour trying to figure out how to go from lhs to rhs: $\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\left( \int _{ k }^{ +\infty } e^{ -iux }dx \right) du=\frac { 1 }{ 2 } +\frac { 1 }{ \pi } \int _{ 0 }^{ +\infty } \Re \left[ \frac { \phi _{ T }(u)e^{ -iuk } }{ iu } \right] du$ The $\phi$ being a characteristic function.

What I get is : $\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac { 1 }{ u } \left[ \sin(ux) \right] _{ k }^{ +\infty }du+\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac { i }{ u } \left[ \cos(ux) \right] _{ k }^{ +\infty }du$ which seems to me will go nowhere since I don't see a way to solve the improper integrals with the cos and sin...

Would appreciate a hint, thanks.

  • 0
    It's defined as being the Fourier transform of the density of T; the expectation of exp(iuT).2012-07-23

1 Answers 1

1

For the internal integral you get: $ \int _{ k }^{ +\infty } e^{ -iux }dx=\biggr[\frac{e^{-iux}}{-iu}\biggr]_{k}^{\infty}, $ whatever that means... So you have

$ \begin{eqnarray} \frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\left( \int _{ k }^{ +\infty } e^{ -iux }dx \right) du&=&\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\biggr[\frac{e^{-iux}}{-iu}\biggr]_{k}^{\infty} du\\ &=&\lim_{z\to \infty}\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\biggr[\frac{e^{-iuz}}{-iu}-\frac{e^{-iuk}}{-iu}\biggr] du\\ &=&\lim_{z\to \infty}\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac{e^{-iuz}}{-iu} du\\ &&-\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du \end{eqnarray} $ Let's assume $ \lim_{z\to \infty}\frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac{e^{-iuz}}{-iu} du =\frac12 $ for the moment. Now split the remaining integral in 2 parts $\int_{-\infty}^0 \dots du$ and $\int_0^{\infty} \dots du$ to get: $ \frac { 1 }{ 2\pi } \int _{ -\infty }^{ +\infty } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du= \frac { 1 }{ 2\pi } \int _{ -\infty }^{ 0 } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du+ \frac { 1 }{ 2\pi } \int _{ 0 }^{ +\infty } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du. $ Now substitute $u=-u'$, use $\phi_T(u)=\phi_T(-u)^*$ to get $ \frac { 1 }{ 2\pi } \int _{ -\infty }^{ 0 } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du= \frac { 1 }{ 2\pi } \int _{ 0 }^{ \infty } \phi _{ T }(-u')\frac{e^{iu'k}}{iu'} du'= \frac { 1 }{ 2\pi } \int _{ 0 }^{ \infty } \left(\phi _{ T }(u')\frac{e^{-iu'k}}{-iu'}\right)^* du' $ Write $u$ for $u'$ again since it doesn't matter and combine the integrants

$\frac { 1 }{ 2\pi } \int _{ 0 }^{ \infty } \left(\phi _{ T }(u)\frac{e^{-iuk}}{-iu}\right)^* + \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du= \frac { 1 }{ 2\pi } \int _{ 0 }^{ +\infty }2 \Re \left[ \frac { \phi _{ T }(u)e^{ -iuk } }{ iu } \right] du$