Ratios With Sign-Changing Denominators

Why the Denominator Near Zero Matters

A ratio can be extreme even when its numerator is ordinary: it is enough for the denominator to be close to zero. If the denominator can approach zero from either side, the ratio may become very large and positive or very large and negative. The resulting tails are not a small technical detail; they can make moments such as the mean fail to exist.

We study independent random variables $X$ and $Y$, with densities $f_X$ and $f_Y$, and the ratio $R=X/Y$. Assume $P(Y=0)=0$, so the ratio is defined with probability one. The key case is when $f_Y$ is continuous and strictly positive at zero: then $Y$ has positive density on both sides of zero near the origin.

Definition: The support of a random variable is the smallest closed set whose complement has probability zero. Saying that the support is unbounded in both directions means that it has no finite upper or lower bound. This does not, by itself, mean that the support is all of $\mathbb{R}$.

Density of a Ratio

For $y\ne 0$, the transformation from $(x,y)$ to $(r,y)$, where $r=x/y$, has inverse $x=ry$. Its Jacobian determinant has absolute value $|y|$. Independence therefore gives the ratio density

f_R(r)=\int_{-\infty}^{\infty}|y|f_X(ry)f_Y(y)\,dy,\qquad r\ne 0.

There is a second useful expression. For fixed $x\ne0$, the inverse transformation from $r=x/y$ is $y=x/r$, whose derivative in absolute value is $|x|/r^2$. Integrating over $X$ gives

f_R(r)=\frac{1}{r^2}\int_{-\infty}^{\infty}|x|f_X(x)f_Y(x/r)\,dx,\qquad r\ne 0.

These formulas apply wherever the integrals exist; for independent variables with densities, they give a density for the ratio away from zero. The second form is especially informative for large $|r|$: the argument $x/r$ of $f_Y$ moves toward zero. Thus the tail of the ratio is controlled by the denominator density near zero.

Worked Example: A Reciprocal of a Uniform Denominator

Let $X=1$ with probability one, and let $Y\sim\operatorname{Uniform}(-1,1)$, independently of $X$. Find the density and support of $R=X/Y$.

Here $R=1/Y$. The inverse is $y=1/r$, with derivative $-1/r^2$. Since $f_Y(y)=1/2$ for $-1<y<1$, the change-of-variables formula gives

f_R(r)=\frac{1}{2r^2},\qquad |r|>1.

For $|r|\le 1$, the inverse value $1/r$ is outside $(-1,1)$, so the density is zero. The support is therefore $(-\infty,-1]\cup[1,\infty)$. In particular, it is unbounded in both directions but does not contain the interval $(-1,1)$. The density also displays the inverse-square decay directly.

Two-Sided Unboundedness

A denominator with positive density on both sides of zero can send a nonzero numerator to either sign of infinity. Independence matters here: it lets a numerator event of positive probability occur together with denominator values arbitrarily close to zero on either side.

Theorem: Suppose $X$ and $Y$ are independent, $P(X\ne0)>0$, and $Y$ has a density $f_Y$ that is continuous and strictly positive at zero. Then, for every $t>0$, $$ P(X/Y>t)>0\quad\text{and}\quad P(X/Y<-t)>0. $$ Consequently, the support of $X/Y$ is unbounded in both directions.

Proof. Since $P(X\ne0)>0$, at least one of $P(X>0)$ and $P(X<0)$ is positive. First suppose $P(X>0)>0$. The event $\{X>0\}$ is the union over positive integers $k$ of $\{X\ge 1/k\}$. Hence there is some $a>0$ for which $P(X\ge a)>0$. Continuity and strict positivity of $f_Y$ at zero imply that $f_Y$ is positive throughout some interval $(-\delta,\delta)$. Choose $0<\epsilon<\min(\delta,a/t)$. Then $P(0<Y<\epsilon)>0$ and $P(-\epsilon<Y<0)>0$, and on the corresponding events,

X\ge a,\quad 0<Y<\epsilon \quad\Longrightarrow\quad X/Y>a/\epsilon>t,

X\ge a,\quad -\epsilon<Y<0 \quad\Longrightarrow\quad X/Y<-a/\epsilon<-t.

By independence, each joint event has positive probability. If instead $P(X<0)>0$, choose $a>0$ with $P(X\le -a)>0$; a positive denominator then gives a ratio below $-t$, and a negative denominator gives a ratio above $t$. This proves both claims for every $t>0$. A closed support that were bounded above or below would rule out one of these positive-probability events for some $t$, so it must be unbounded in both directions. $\square$

The theorem guarantees arbitrarily distant values of each sign, not every value between them. The actual support also depends on the numerator and on the full support of the denominator. For example, a positive numerator bounded away from zero divided by a denominator bounded in absolute value may leave a gap around zero.

Worked Example: Two Uniform Variables and a Gap in the Support

Let $X\sim\operatorname{Uniform}(1,2)$ and $Y\sim\operatorname{Uniform}(-1,1)$, independently. Find the support and density of $R=X/Y$.

Because $1<X<2$ and $|Y|<1$, the ratio has $|R|>1$. Both signs occur because $Y$ can be positive or negative. For $r>0$, the density formula using $x=ry$ is

f_R(r)=\frac12\int_{\{y:\,1<ry<2,\ 0<y<1\}}y\,dy.

The factor $1/2$ is $f_Y(y)$, while $f_X(ry)=1$ when $ry\in(1,2)$. If $1<r<2$, the allowed values of $y$ run from $1/r$ to $1$, so

f_R(r)=\frac12\int_{1/r}^{1}y\,dy =\frac14\left(1-\frac{1}{r^2}\right).

If $r\ge2$, the allowed interval is $1/r<y<2/r$, giving

f_R(r)=\frac12\int_{1/r}^{2/r}y\,dy =\frac{3}{4r^2}.

The symmetry of $Y$ makes the density even. Thus the complete density, up to immaterial endpoint values, is

f_R(r)= \begin{cases} \frac14(1-r^{-2}),&1<|r|<2,\\[2pt] \frac{3}{4r^2},&|r|\ge2,\\[2pt] 0,&|r|\le1. \end{cases}

As a check, the integral on the positive side is $\int_1^2(1-r^{-2})/4\,dr+\int_2^\infty 3/(4r^2)\,dr=1/8+3/8=1/2$. The negative side contributes the other $1/2$. The support has a gap around zero even though it extends without bound in both directions.

The Inverse-Square Tail

The density formula yields a precise tail result under mild additional conditions. Assume $X$ and $Y$ are independent with densities, $E[|X|]<\infty$, $f_Y$ is continuous at zero, and $f_Y$ is bounded on $\mathbb{R}$. The boundedness condition supplies an integrable dominating function in the argument below.

Theorem: Under these assumptions, as $r\to+\infty$ and as $r\to-\infty$, $$ r^2f_R(r)\longrightarrow f_Y(0)E[|X|]. $$ If $f_Y(0)>0$ and $E[|X|]>0$, then $$ P(R>t)\sim\frac{f_Y(0)E[|X|]}{t}, \qquad P(R<-t)\sim\frac{f_Y(0)E[|X|]}{t} \quad (t\to\infty). $$

Proof. Use the density formula expressed as an integral over $x$:

r^2 f_R(r)=\int_{-\infty}^{\infty}|x|f_X(x)f_Y(x/r)\,dx.

For each fixed $x$, $x/r\to0$ as $r\to+\infty$ or $r\to-\infty$. Continuity at zero gives $f_Y(x/r)\to f_Y(0)$ in either limit. If $f_Y(u)\le M$ for all $u$, the integrand is bounded by $M|x|f_X(x)$, whose integral is $M E[|X|]<\infty$. The dominated convergence theorem therefore gives the stated density limit for both signs of $r$.

Write $C=f_Y(0)E[|X|]>0$. For every $0<\eta<C$, the density limit ensures that for all sufficiently large positive $r$, $(C-\eta)/r^2\le f_R(r)\le(C+\eta)/r^2$. Integrating from $t$ to infinity gives

\frac{C-\eta}{t}\le P(R>t)\le\frac{C+\eta}{t}

for sufficiently large $t$. Letting $\eta$ decrease to zero proves $P(R>t)\sim C/t$. Applying the same bounds to $f_R(-u)$ for large $u>0$ proves the negative-tail result. $\square$

Worked Example: Exponential Numerator and Uniform Denominator

Let $X\sim\operatorname{Exponential}(1)$, with density $f_X(x)=e^{-x}$ for $x>0$, and let $Y\sim\operatorname{Uniform}(-1,1)$, independently. Here $f_Y(y)=1/2$ near zero, $E[|X|]=1$, and the theorem predicts a density tail constant of $1/2$ on each side.

Using the ratio-density formula over $x$, the condition $f_Y(x/r)>0$ requires $0<x<|r|$. Therefore, for $r\ne0$,

f_R(r)=\frac{1}{2r^2}\int_0^{|r|}xe^{-x}\,dx =\frac{1-(|r|+1)e^{-|r|}}{2r^2}.

The integral follows from $\int xe^{-x}\,dx=-(x+1)e^{-x}$. As $|r|\to\infty$, the exponential term tends to zero, so $r^2f_R(r)\to1/2$, as predicted. Consequently each one-sided tail is asymptotic to $1/(2t)$. In this example the support is all of $\mathbb{R}$: for every nonzero $r$, the interval $0<x<|r|$ has positive probability, and the density has a positive limiting value $1/4$ as $r$ approaches zero.

Why the Tail Rate Matters

An inverse-square density tail is heavy enough to prevent an absolute first moment. If $f_R(r)\sim C/r^2$ for $r\to+\infty$, with $C>0$, then for sufficiently large $r$, $f_R(r)\ge C/(2r^2)$. Hence

\int_A^\infty r f_R(r)\,dr \ \ge\ \frac{C}{2}\int_A^\infty\frac{1}{r}\,dr =\infty.

The negative tail gives the same divergence for the negative part. Thus $E[|R|]=\infty$, and the ordinary expectation $E[R]$ is undefined: both its positive and negative parts have infinite expectation. A symmetric principal-value cancellation, when available, is not a finite expectation. Since a finite second moment would imply a finite first absolute moment by the Cauchy–Schwarz inequality, the variance cannot be finite either.

Common Mistake

Wrong reasoning: “The numerator has a finite mean and the denominator has a finite mean, so their ratio should have a finite mean.”

This overlooks the denominator’s behavior near zero. A finite $E[|Y|]$ says nothing about whether $1/|Y|$ has an integrable tail. When $f_Y(0)>0$, values of $Y$ occur arbitrarily close to zero, and the ratio can have tails of order $1/t$ in probability. Check the density near zero and the resulting tail before asserting that a ratio has a finite moment.

Key takeaway: For an independent ratio with a denominator density that is positive and continuous at zero, nonzero numerator values create unbounded tails on both sides. Under a finite first absolute moment for the numerator, the ratio density typically decays like $1/r^2$, producing one-sided tail probabilities of order $1/t$.

Check Your Understanding

Use the assumptions and formulas in this tutorial to answer the following questions.

If $X>0$ with probability one and $Y$ has positive density on both sides of zero near zero, why can $X/Y$ take arbitrarily large values of either sign?
Does support unbounded in both directions imply that the support is all of $\mathbb{R}$? Use the two-uniform example to explain.
In the inverse-square density theorem, which integrable function dominates the integrand, and which assumption guarantees its integrability?
If $f_R(r)\sim C/r^2$ as $r\to\infty$, explain why the positive part of $R$ has infinite expectation.
For the exponential-over-uniform example, identify $f_Y(0)$ and $E[|X|]$, then give the asymptotic value of $P(R>t)$.