characteristic function of Cauchy distribution

hbghlyj

$\def\R{\mathbb R}\def\E{\mathbb E}$The Cauchy Distribution \(X\) has characteristic function \(\chi_0\) given by$$\chi_0(t) = \exp\left(-\left|t\right|\right)$$for \(t \in \R \).

Proof
By definition, \[ \chi_0(t) = \E(e^{i t X}) = \int_{-\infty}^\infty e^{i t x} \frac{1}{\pi (1 + x^2)} \, dx \] We will compute this integral by evaluating a related contour integral in the complex plane using, appropriately enough, Cauchy's integral formula.
Suppose first that \(t \ge 0\). For \(r \gt 1\), let \(\Gamma_r\) denote the curve in the complex plane consisting of the line segment \(L_r\) on the \(x\)-axis from \(-r\) to \(r\) and the upper half circle \(C_r\) of radius \(r\) centered at the origin. We give \(\Gamma_r\) the usual counter-clockwise orientation. On the one hand we have \[ \int_{\Gamma_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz = \int_{L_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz + \int_{C_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz\] On \(L_r\), \(z = x\) and \(dz = dx\) so \[ \int_{L_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz = \int_{-r}^r \frac{e^{i t x}}{\pi (1 + x^2)} dx \] On \(C_r\), let \(z = x + i y\). Then \(e^{i t z} = e^{-t y + i t x} = e^{-t y} [\cos(t x) + i \sin(t x)]\). Since \(y \ge 0\) on \(C_r\) and \(t \ge 0\), we have \(|e^{i t z} | \le 1\). Also, \(\left|\frac{1}{1 + z^2}\right| \le \frac{1}{r^2 - 1}\) on \(C_r\). It follows that \[ \left|\int_{C_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz \right| \le \frac{1}{\pi (r^2 - 1)} \pi r = \frac{r}{r^2 - 1} \to 0 \text{ as } r \to \infty \] On the other hand, \(e^{i t z} / [\pi (1 + z^2)]\) has one singularity inside \(\Gamma_r\), at \(i\). The residue is \[ \lim_{z \to i} (z - i) \frac{e^{i t z}}{\pi (1 + z^2)} = \lim_{z \to i} \frac{e^{i t z}}{\pi(z + i)} = \frac{e^{-t}}{2 \pi i} \] Hence by Cauchy's integral formula, \[ \int_{\Gamma_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz = 2 \pi i \frac{e^{-t}}{2 \pi i} = e^{-t} \]Putting the pieces together we have \[ e^{-t} = \int_{-r}^r \frac{e^{i t x}}{\pi (1 + x^2)} dx + \int_{C_r} \frac{e^{i t z}}{\pi (1 + z^2)} dz \] Letting \(r \to \infty\) gives \[ \int_{-\infty}^\infty \frac{e^{i t x }}{\pi (1 + x^2)} dx = e^{-t}=e^{-|t|} \] For \(t \lt 0\), we can use the substitution \(u = - x\) and our previous result to get \[ \int_{-\infty}^\infty \frac{e^{i t x}}{\pi (1 + x^2)} dx = \int_{-\infty}^\infty \frac{e^{i (-t) u}}{\pi (1 + u^2)} du = e^t=e^{-|t|} \]

hbghlyj

n1.pdf 11/10/2006, 4:45:10 PM
Example 2.2 (Moments of the Cauchy Distribution).
Same density (2.5) as in the preceding example. Now that we know that it is a probability density, we can determine whether some expectations exist. For what positive $\beta$ do $\mathbb E\left(|X|^\beta\right)$ exist?

First write down the integral that defines this expectation, if it exists,
\[
\mathbb E\left(|X|^\beta\right)=\int_{-\infty}^{\infty} \frac{c|x|^\beta}{1+x^2} d x
\]
call that integrand something
\[
g(x)=\frac{c|x|^\beta}{1+x^2}
\]
then we see that $g$ behaves like $|x|^{\beta-2}$ as $|x|$ goes to infinity. Applying the theorem and the comment about minus infinity following it, we see that $g$ has finite integral if and only if $\beta-2<-1$, which is when $\beta<+1$.

So that is our answer. $E\left(|X|^\beta\right)$ exists when $0<\beta<1$ and does not exist when $\beta \geq 1$.

hbghlyj

Note that $e^{-\abs t}$ is not differentiable at 0. [The mean of Cauchy distribution doesn't exist]
$\phi_{S_n\over n}(t)=\left(\phi(\frac tn)\right)^n=\left(e^{-\abs{\frac tn}}\right)^n=e^{-\abs t}$
By Uniqueness theorem, $\frac{S_n}n$ has the same distribution as $X$.
In particular $\frac{S_n}n\overset{d}\to\text{Cauchy}$
Hence $X$ doesn't satisfy Weak Law of Large Numbers.