# characteristic function (evaluation of an definite integral)

• Apr 28th 2010, 09:20 AM
Random Variable
characteristic function (evaluation of an definite integral)
I want to show that the characteristic function of Cauchy's distribution ($\displaystyle f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0$) is $\displaystyle \phi_{X}(t) = e^{-b|t|}$

$\displaystyle \phi_{X}(t) = E[e^{itX}] = E[\cos(tX)] + i E[\sin (tX)]$

$\displaystyle = \frac{b}{\pi} \int^{\infty}_{-\infty} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx + i \ \frac{b}{\pi} \int^{\infty}_{-\infty} \frac{\sin(tx)}{b^{2}+x^{2}} \ dx$

The second integral evaluates to zero because the integrand is odd.

The integrand of the first integral is even, so I need to evaluate $\displaystyle \frac{2 b}{\pi} \int^{\infty}_{0} \frac{\cos(tx)}{b^{2}+x^{2}} \ dx$

Any suggestions?
• Apr 28th 2010, 11:32 AM
Laurent
Quote:

Originally Posted by Random Variable
I want to show that the characteristic function of Cauchy's distribution ($\displaystyle f(x) = \frac{1}{\pi} \frac{b}{b^{2}+x^{2}}, \ b>0$) is $\displaystyle \phi_{X}(t) = e^{-b|t|}$

Any suggestions?

Usual proofs are:

- Fourier inversion formula. It is easy to compute the Fourier transform of $\displaystyle e^{-b|t|}$ and it relates simply to Cauchy's distribution, so we can have it work backwards using Fourier inversion formula...

- contour integration in the complex plane. The integral $\displaystyle \int_{-R}^R \frac{e^{itx}}{1+t^2}dt$ is completed into a contour integral using a half circle in the upper half-plane; the integral along the circle is seen to converge to 0 as the radius goes to infinity (dominated convergence theorem) while the value of the contour integral is obtained by computing the residue at the pole $\displaystyle i$ (Residue theorem).
• Apr 28th 2010, 07:45 PM
Random Variable
I hate contour integration.

But this one is not that bad because there's a nice formula.

Let $\displaystyle f(z) = \frac{e^{itz}}{b^{2}+z^{2}}$

which has the simple pole $\displaystyle z = bi$ in the upper half plane

so $\displaystyle \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = -2 \pi \ Im (Res\{f,bi\})$

$\displaystyle Res\{f,bi\} = \lim_{z \to bi} (z-bi) \frac{e^{itz}}{b^{2}+z^{2}}= \lim_{z \to bi} \frac{e^{itz}}{z+bi} = \frac{e^{-bt}}{2bi} = -i \ \frac{e^{-bt}}{2b}$

so $\displaystyle \int^{\infty}_{-\infty} \frac{\cos (tx)}{b^{2}+x^{2}} = \pi \ \frac{e^{-bt}}{b}$

and $\displaystyle \phi_{X}(t) = e^{-bt}$

From where is the absolute value coming? (Wondering)
• Apr 29th 2010, 03:23 AM
Laurent
Quote:

Originally Posted by Random Variable

From where is the absolute value coming? (Wondering)

It's hidden in your computation. The integral along the upper half-circle converges to 0 only when $\displaystyle t>0$. Otherwise, you have the consider the lower half-circle hence the pole is $\displaystyle -ib$ (or you can note from the beginning that the characteristic function is even by symmetry of the Cauchy distribution).