Cauchy Distribution

Modeling extreme events and 'fat-tailed' phenomena.

The "Black Swan" Distribution

The Cauchy distribution (also known as the Lorentz distribution) is a continuous probability distribution famous for its heavy, or "fat," tails. This means it assigns a much higher probability to extreme events compared to the normal distribution.

In finance, it's a powerful conceptual tool for modeling phenomena where "black swan" events are more common than traditional models suggest. Its most striking feature is that its expected value (mean) and variance are undefined. No matter how many samples you take, the average will not converge, making it a radical departure from well-behaved distributions like the Normal distribution.

Interactive Cauchy Distribution

Adjust the location (x₀) and scale (γ) parameters to see how the shape of the distribution changes. Note how the tails remain "heavy" even with a small scale parameter.

Location (x₀): 0.0

Scale (γ): 1.0

Mean (

\mu

): 0.00

Variance (

\sigma^2

): Undefined

Core Concepts

Probability Density Function (PDF)

The PDF gives the characteristic bell-shape, but with much heavier tails than the Normal distribution.

f(x; x_0, \gamma) = \frac{1}{\pi\gamma \left[1 + \left(\frac{x-x_0}{\gamma}\right)^2\right]}

$x_0$ is the location parameter, which specifies the location of the peak (the median and mode).
$\gamma > 0$ (gamma) is the scale parameter, which specifies the half-width at half-maximum.

Cumulative Distribution Function (CDF)

F(x; x_0, \gamma) = \frac{1}{\pi} \arctan\left(\frac{x - x_0}{\gamma}\right) + \frac{1}{2}

Key Derivations: The Undefined Moments

Why is the Expected Value (Mean) Undefined?

The mean is undefined because the integral required to calculate it does not converge to a finite value. This is a result of the "fat tails" not decaying quickly enough.

Step 1: Set up the Integral for Expected Value

For a standard Cauchy distribution ( $x_0=0, \gamma=1$ ), the expected value is given by the integral:

E[X] = \int_{-\infty}^{\infty} x \cdot f(x) dx = \int_{-\infty}^{\infty} \frac{x}{\pi(1+x^2)} dx

Step 2: Evaluate the Integral

We can split this integral. The antiderivative of $\frac{x}{1+x^2}$ is $\frac{1}{2}\ln(1+x^2)$ .

E[X] = \frac{1}{\pi} \left[ \frac{1}{2} \ln(1+x^2) \right]_{-\infty}^{\infty}

When we evaluate this at the limits:

\lim_{a \to \infty} \frac{1}{2\pi} (\ln(1+a^2) - \ln(1+(-a)^2)) = \lim_{a \to \infty} \frac{1}{2\pi} (\ln(1+a^2) - \ln(1+a^2))

This results in the indeterminate form $\infty - \infty$ . Because the integral does not converge to a single finite value, the expected value is formally undefined.

Why is the Variance Undefined?

Since the mean is undefined, the variance, which is defined as $E[(X - E[X])^2]$ , must also be undefined.