Ratios of Independent Normals and the Cauchy Law

Why a Ratio of Normals Is Not Normal

Sums and averages of normal variables remain closely connected to the normal family, but a ratio can behave very differently. In particular, dividing two independent standard normal variables produces a distribution with substantial probability far from zero. The denominator can be arbitrarily close to zero, which allows the ratio to take very large positive or negative values.

Let $X$ and $Y$ be independent standard normal random variables. Since $P(Y=0)=0$, the ratio $R=X/Y$ is defined with probability one. For definiteness, assign any value to $R$ on the event $Y=0$; this choice cannot change its distribution. We will derive the density of $R$ directly from the joint density of $(X,Y)$.

Definition: The standard Cauchy distribution is the continuous distribution on the real line with density $$ f_C(t)=\frac{1}{\pi(1+t^2)},\qquad -\infty<t<\infty. $$ Its cumulative distribution function is $F_C(t)=\tfrac12+\tfrac{1}{\pi}\arctan(t)$.

Transforming the Joint Density

Independence and standard normality give the joint density $f_{X,Y}(x,y)=(2\pi)^{-1}\exp\{-(x^2+y^2)/2\}$. To isolate the ratio, transform from $(x,y)$ to $(t,s)$, where $t=x/y$ and $s=y$. On the set $y\ne0$, the inverse transformation is $x=ts,\ y=s$. It is one-to-one there, and its Jacobian determinant has absolute value $|s|$.

The excluded line $y=0$ has probability zero under the joint density. The change-of-variables formula therefore gives the joint density of $(T,S)=(X/Y,Y)$ for $s\ne0$. Integrating out $s$ gives the marginal density of the ratio.

Theorem: If $X$ and $Y$ are independent $N(0,1)$ random variables, then $R=X/Y$ has the standard Cauchy distribution.

Proof. Under the inverse transformation $x=ts,\ y=s$, the joint density of $(T,S)$ is

f_{T,S}(t,s) =\frac{1}{2\pi}\exp\left\{-\frac{(1+t^2)s^2}{2}\right\}|s|, \qquad t\in\mathbb R,\ s\ne0.

For each fixed $t$, set $a=1+t^2>0$. Using the substitution $u=as^2/2$ on $s>0$, and symmetry on $s<0$, we obtain

\int_{-\infty}^{\infty}|s|e^{-as^2/2}\,ds =2\int_0^\infty s e^{-as^2/2}\,ds =\frac{2}{a}\int_0^\infty e^{-u}\,du =\frac{2}{a}.

Consequently, the marginal density of $T=R$ is

f_R(t)=\int_{-\infty}^{\infty}f_{T,S}(t,s)\,ds =\frac{1}{2\pi}\frac{2}{1+t^2} =\frac{1}{\pi(1+t^2)}.

This is the standard Cauchy density. It integrates to one because $\int_{-\infty}^{\infty}(1+t^2)^{-1}\,dt =[\arctan(t)]_{-\infty}^{\infty}=\pi$. Thus the marginal calculation gives a valid density on the entire real line, and the theorem follows. $\square$

The CDF and Probability Calculations

Integrating the density gives the CDF. The antiderivative is the inverse tangent, and the limit of $\arctan(t)$ as $t$ tends to negative infinity is $-\pi/2$. Hence, for every real $t$,

F_R(t)=\int_{-\infty}^{t}\frac{1}{\pi(1+u^2)}\,du =\frac{1}{\pi}\left(\arctan(t)+\frac{\pi}{2}\right) =\frac12+\frac{\arctan(t)}{\pi}.

The density is even, so the distribution is symmetric about zero. The CDF makes probability calculations especially direct: for $a<b$, $P(a<R\le b)=F_R(b)-F_R(a)$. Because the distribution is continuous, including or excluding either endpoint does not change the probability.

Worked Example: Density at a Particular Ratio

For independent standard normal $X$ and $Y$, find the density of $R=X/Y$ at $t=1$, and interpret what that value means.

The derived density is $f_R(t)=1/[\pi(1+t^2)]$. Substitution gives

f_R(1)=\frac{1}{\pi(1+1^2)}=\frac{1}{2\pi}\approx0.1592.

This is a density value, not the probability that $R=1$. Since $R$ has a continuous distribution, $P(R=1)=0$. For a small interval of width $h$ around 1, its probability is approximately $f_R(1)h$, with the approximation improving as $h$ gets smaller.

Worked Example: Probability the Ratio Lies Between One and Two

Find $P(1<R<2)$. Apply the CDF and simplify the inverse-tangent expression:

\begin{aligned} P(1<R<2) &=F_R(2)-F_R(1)\\ &=\frac{\arctan(2)-\arctan(1)}{\pi}\\ &=\frac{1.107149-0.785398}{\pi}\\ &\approx 0.1024. \end{aligned}

Thus about $10.2\%$ of the ratio’s probability lies between 1 and 2. The subtraction of the two CDF values is essential: $F_R(2)$ alone is the probability of being at or below 2, not the probability of lying in this interval.

Worked Example: Ratios of Centered Normals With Different Variances

Suppose $X\sim N(0,4)$ and $Y\sim N(0,9)$ are independent. Find the density of $Q=X/Y$ and evaluate it at $q=2$.

The variables can be represented jointly as $X=2Z_1$ and $Y=3Z_2$, where $Z_1,Z_2$ are independent standard normals. Therefore $Q=(2/3)(Z_1/Z_2)$. If $C=Z_1/Z_2$, then $C$ is standard Cauchy, and scaling a continuous variable by $2/3$ gives

f_Q(q)=\frac{1}{2/3}f_C\left(\frac{q}{2/3}\right) =\frac{3}{2\pi\left(1+9q^2/4\right)} =\frac{6}{\pi(4+9q^2)}.

At $q=2$, this density is

f_Q(2)=\frac{6}{\pi(4+9\cdot 2^2)} =\frac{6}{40\pi} =\frac{3}{20\pi}\approx0.0477.

The density is rescaled because the numerator and denominator have different standard deviations. The ratio is not standard Cauchy in this example, but its distribution is a scaled standard Cauchy.

A Conditional-Density View

The same result can be checked by conditioning on the denominator. For a fixed $y\ne0$, $X/y$ is a scaled standard normal, with conditional density at $t$ equal to $|y|\phi(ty)$, where $\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}$. The factor $|y|$ is the absolute Jacobian for the scaling $x=ty$. Averaging this conditional density over $Y$ gives

\begin{aligned} f_R(t) &=\int_{-\infty}^{\infty}|y|\phi(ty)\phi(y)\,dy\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}|y| e^{-(1+t^2)y^2/2}\,dy\\ &=\frac{1}{\pi(1+t^2)}. \end{aligned}

The final integral is the same one evaluated in the change-of-variables proof. This conditioning calculation makes the source of the large ratios visible: when $|Y|$ is small, the conditional scaling factor $1/Y$ is large.

Why Symmetry Does Not Give a Mean

The Cauchy density is symmetric, and its median is zero. It is tempting to conclude that its mean is also zero. That conclusion is not valid: expectation requires integrability, not merely symmetry. Indeed,

E[|R|] =\int_{-\infty}^{\infty}\frac{|t|}{\pi(1+t^2)}\,dt =\frac{2}{\pi}\int_0^\infty\frac{t}{1+t^2}\,dt =\infty,

because the integral from 0 to $M$ equals $\tfrac{1}{2}\log(1+M^2)$, which tends to infinity as $M\to\infty$. In particular, both the positive and negative parts have infinite expectation, so $E[R]$ is undefined. A symmetric principal-value integral may cancel the two tails, but that cancellation does not define the expectation.

This heavy-tail behavior is important when interpreting averages of normal ratios. Even though both inputs are normal and have finite moments, their ratio has no finite mean or variance. The denominator’s possibility of being close to zero changes the distribution substantially.

Common Mistake

Wrong reasoning: “The ratio is symmetric around zero, so its expected value is zero.” Symmetry only implies a zero expectation when the expectation exists.
Correction: Check integrability. For the standard Cauchy density, $E[|R|]=\infty$, so the ordinary expectation does not exist. The symmetric principal value is not a substitute for a finite expectation.
Wrong reasoning: “The density at one is the probability that the ratio equals one.” A density is probability per unit of the variable’s scale, not point probability.
Correction: A continuous variable has zero probability at any single point. Use an integral or a CDF difference to find probabilities over intervals.

Key takeaway: For independent $N(0,1)$ variables $X$ and $Y$, $X/Y$ has density $1/[\pi(1+t^2)]$. The CDF uses the inverse tangent, and the ratio’s symmetry does not give it a finite mean.

Check Your Understanding

Try these questions using the density and CDF derived above.

Why is it legitimate to ignore the event $Y=0$ when deriving the distribution of $X/Y$?
Use the CDF to calculate $P(-1<R<1)$ for a standard Cauchy variable $R$.
Explain why $f_R(0)=1/\pi$ does not mean $P(R=0)=1/\pi$.
If $X\sim N(0,1)$ and $Y\sim N(0,4)$ are independent, express $X/Y$ as a scaled standard Cauchy variable.
What condition fails when one tries to define the standard Cauchy mean as zero by symmetry?