Deriving the Cauchy Distribution from the Ratio of Normal Random Variables

In probability theory, the Cauchy distribution is a fascinating and important density function with many unique properties. One notable result is that the ratio of two independent normal random variables follows a Cauchy distribution. This article delves into the derivation of this result, providing a detailed and clear explanation suitable for advanced students and practitioners in statistics and probability theory.

Introduction

The Cauchy distribution is a continuous probability distribution that is often used in physics, engineering, and finance because of its heavy tails, which make it more robust to outliers compared to the normal distribution. We will demonstrate that if (X) and (Y) are independent normal random variables with mean 0 and variance (sigma^2), then the ratio (Z frac{Y}{X}) has a Cauchy distribution. This derivation requires a good understanding of normal distributions, the definition of the Cauchy distribution, and properties of integrals.

Step-by-Step Derivation

Step 1: Define the Random Variables

Let (X) and (Y) be independent random variables such that:

[ X sim N(0, sigma^2) quad text{and} quad Y sim N(0, sigma^2) ]

Step 2: Standardize the Random Variables

We can standardize (X) and (Y) by defining:

[ Z_X frac{X}{sigma} sim N(0, 1) quad text{and} quad Z_Y frac{Y}{sigma} sim N(0, 1) ]

This implies:

[ X sigma Z_X quad text{and} quad Y sigma Z_Y ]

Step 3: Express the Ratio

Now we can express the ratio (Z):

[ Z frac{Y}{X} frac{sigma Z_Y}{sigma Z_X} frac{Z_Y}{Z_X} ]

Step 4: Distribution of the Ratio

Since (Z_X) and (Z_Y) are independent standard normal random variables, the ratio (frac{Z_Y}{Z_X}) follows a standard Cauchy distribution. This is a well-known result in probability theory, specifically concerning the distribution of the ratio of two independent standard normal variables.

Step 5: Conclusion

Thus, we conclude that:

[ Z frac{Y}{X} sim text{Cauchy}(0, 1) ]

Therefore, the ratio (frac{Y}{X}) has a Cauchy distribution.

Mathematical Derivation

Let (Z frac{Y}{X}). Observe that the distribution of (Z) must be symmetric, so its density function is even. We will find this density (f_Z(z)) for positive (z) only and then use the fact that it’s an even function.

The distribution function of (Z) is:

[ F_Z(z) P(-z leq Y leq zX) 2P(0 leq Y leq zX) ]

Using (f) and (F) as the density and cumulative distribution functions of the Normal (0, sigma^2) respectively, we can write:

[ F_Z(z) int_{y0}^infty int_{x0}^{zy} 2f(x)f(y) , dx , dy ]

[ F_Z(z) int_{y0}^infty 2 left(F(zy) - frac{1}{2}right)f(y) , dy ]

Distributing the two leaves the difference of two integrals, the second of which is one-half as it’s the integral over the positive domain of a zero-mean normal density.

[ F_Z(z) int_{y0}^infty 2F(zy)f(y) , dy - frac{1}{2} ]

Differentiating this function of (z) after exchanging the order of differentiation and integration gives:

[ f_Z(z) int_{y0}^infty 2yf(zy)f(y) , dy ]

[ f_Z(z) int_{y0}^infty frac{y}{pisigma^2} expleft(-frac{y^2}{2sigma^2z^2}right) , dy ]

[ f_Z(z) -frac{1}{pi z^2} int_{y0}^infty -frac{y}{sigma^2} expleft(-frac{y^2}{2sigma^2z^2}right) , dy ]

[ f_Z(z) -frac{1}{pi z^2} int_{y0}^infty frac{d}{dy} left(expleft(-frac{y^2}{2sigma^2z^2}right)right) , dy ]

As the integral of the exact derivative of a function, we simply evaluate this function at (infty) and (0) and take the difference.

[ f_Z(z) frac{1}{pi z^2} ]

Notice that the result is already an even function if we include non-positive (z) in its domain just as was required, so it is also valid for (z leq 0).

We recognize this as the density defining the Cauchy distribution.

Conclusion

In summary, if (X) and (Y) are independent (N(0, sigma^2)) random variables, then the ratio (frac{Y}{X}) indeed has a standard Cauchy distribution, confirming that (frac{Y}{X}) has a Cauchy density. This unique property highlights the intimate relationship between the normal and Cauchy distributions in the realm of probability and statistics.