Ratios and Continuous Maps with Nonzero Denominators

Why Ratios Require Special Care

A ratio is a continuous function of its numerator and denominator wherever the denominator is nonzero. This makes the continuous mapping theorem a natural tool for studying limits of ratios. The qualification matters: division is not defined at zero, and the quotient map becomes discontinuous as the denominator approaches zero.

The key condition is not that the denominators $Y_n$ are always nonzero, nor that they stay a fixed distance from zero. The essential condition for a general weak-limit result is that the limiting denominator $Y$ is nonzero with probability one. Then any finite-$n$ zero denominators occur with probability tending to zero, and a measurable extension of the quotient can be used with the continuous mapping theorem.

A Precise Ratio Theorem

Definition: For real-valued random variables $X_n,Y_n$, define the extended ratio $R_n$ by $R_n=X_n/Y_n$ when $Y_n\ne 0$, and $R_n=0$ when $Y_n=0$. This convention makes $R_n$ defined everywhere. The chosen value on $Y_n=0$ is immaterial to the limit when the probability of that event tends to zero.

Write $(X_n,Y_n)\Rightarrow(X,Y)$ for weak convergence of the random vectors in $\mathbb R^2$. Joint convergence is the relevant input: it retains the dependence between numerator and denominator.

Theorem: Suppose $(X_n,Y_n)\Rightarrow(X,Y)$ and $\mathbb P(Y=0)=0$. With the extended ratios defined above, $$ R_n\Rightarrow \frac{X}{Y}. $$ The quotient on the right is defined almost surely, since $Y\ne 0$ almost surely.

Proof. Define $g:\mathbb R^2\to\mathbb R$ by $g(x,y)=x/y$ if $y\ne0$, and $g(x,0)=0$. This is a Borel-measurable function: it is continuous on the open set $\{(x,y):y\ne0\}$, and the remaining set $\{y=0\}$ is Borel. At every point with $y\ne0$, $g$ is continuous. Thus its set of discontinuities is contained in $\{(x,y):y=0\}$, which has probability $\mathbb P(Y=0)=0$ under the law of $(X,Y)$. The continuous mapping theorem for a map continuous almost surely under the limit law gives $g(X_n,Y_n)\Rightarrow g(X,Y)$. By definition, $g(X_n,Y_n)=R_n$, and $g(X,Y)=X/Y$ almost surely. Therefore $R_n\Rightarrow X/Y$. $\square$

The theorem is an application of the continuous mapping theorem on its almost-sure continuity set. The denominator condition is exactly what ensures that the limit vector lies in that set almost surely. No independence between $X$ and $Y$, or between $X_n$ and $Y_n$, is required.

Why Finite-Sample Zeros Become Negligible

The limiting condition also implies that denominators equal to zero at finite $n$ have vanishing probability. This explains why the particular value assigned to $R_n$ on $Y_n=0$ cannot affect the weak limit.

Proposition: If $Y_n\Rightarrow Y$ and $\mathbb P(Y=0)=0$, then $\mathbb P(Y_n=0)\to0$.

Proof. The set $\{0\}$ is closed in $\mathbb R$. The closed-set inequality in the Portmanteau theorem therefore gives $$ \limsup_{n\to\infty}\mathbb P(Y_n=0) \leq \mathbb P(Y=0)=0. $$ Probabilities are nonnegative, so their limit exists and equals zero. $\square$

In particular, if an application defines $X_n/Y_n$ only on the event $Y_n\ne0$, it can be completed by assigning any fixed finite value on the exceptional event. Under the proposition, different such completions differ only on events whose probabilities vanish.

Worked Example: A Denominator Converging to a Nonzero Constant

Suppose $Z_n\Rightarrow Z$, where the variables are real-valued, and $D_n\to_{\mathbb P}4$. By Slutsky’s theorem, $(Z_n,D_n)\Rightarrow(Z,4)$. Since the limiting denominator equals $4$, it is nonzero with probability one. The ratio theorem gives $$ \frac{Z_n}{D_n}\Rightarrow \frac{Z}{4}, $$ with any fixed finite convention used when $D_n=0$. In fact, $\mathbb P(D_n=0)\leq \mathbb P(|D_n-4|\geq4)\to0$, so the exceptional event has vanishing probability. The numerator need not converge in probability; weak convergence of $Z_n$ is sufficient because the denominator converges in probability to a constant.

Slutsky’s Theorem for Ratios

A frequent case follows directly from the joint-convergence form of Slutsky’s theorem. If $X_n\Rightarrow X$ and $Y_n\to_{\mathbb P}c$ for a fixed $c\ne0$, then $(X_n,Y_n)\Rightarrow(X,c)$. The ratio theorem then gives convergence to $X/c$. The condition $c\ne0$ is essential for this general conclusion.

Corollary: If $X_n\Rightarrow X$ and $Y_n\to_{\mathbb P}c\ne0$, then the ratios $X_n/Y_n$, extended by any fixed finite value on $Y_n=0$, satisfy $$ \frac{X_n}{Y_n}\Rightarrow\frac{X}{c}. $$

Worked Example: A Random Limiting Denominator

Let $G$ and $H$ be independent standard normal random variables, and set $X_n=G+1/n$ and $Y_n=2+H+1/n$. The pair converges almost surely, and hence weakly, to $(G,2+H)$. A standard normal variable has a density, so $\mathbb P(H=-2)=0$; consequently $\mathbb P(2+H=0)=0$. The ratio theorem yields $$ \frac{X_n}{Y_n}\Rightarrow\frac{G}{2+H}. $$ The limit can have substantial probability of being large when $2+H$ is close to zero. That does not invalidate the result: the denominator is nonzero almost surely, even though it is not bounded away from zero by a deterministic positive constant.

For a scalar denominator, an atomless distribution is sufficient for the nonzero condition, since it assigns probability zero to every singleton, including $\{0\}$. Having a continuous component alone is not sufficient: a law may combine such a component with a separate atom at zero. The condition to check is precisely $\mathbb P(Y=0)=0$.

What Goes Wrong at a Zero Limit

If the limiting denominator can equal zero with positive probability, the quotient map is discontinuous on a set that the limit may visit with positive probability. The continuous mapping theorem then provides no general ratio conclusion. Particular sequences can still have well-behaved ratios, but their behavior depends on details not captured by the limiting pair alone.

Worked Example: Divergence When the Denominator Tends to Zero

Take $X_n=1$ and $Y_n=1/n$. The pair converges deterministically to $(1,0)$, but the ratio is $X_n/Y_n=n$. It does not converge weakly to a real-valued random variable: for every finite $M$, the probability that $n\leq M$ is eventually zero. The proposed limiting expression $1/0$ is not a real-valued quotient. This example shows why a zero limiting denominator cannot simply be ignored.

Worked Example: A Zero Limit Does Not Determine the Ratio

Now take $X_n=Y_n=1/n$. The pair again converges to $(0,0)$, but $X_n/Y_n=1$ for every $n$. Thus a zero limiting denominator does not force divergence in every example. Rather, the limiting pair $(0,0)$ does not determine the ratio: the quotient $0/0$ is undefined, and different ways of approaching that point can produce different outcomes. For instance, $X_n=2/n$ and $Y_n=1/n$ have the same pair limit $(0,0)$ but ratio $2$ at every $n$.

Why Marginal Limits Are Not Enough

Knowing the separate weak limits of the numerator and denominator does not, in general, establish a ratio limit. Ratios depend on how the two coordinates vary together. The joint convergence hypothesis records that dependence and cannot be replaced by marginal convergence alone.

Worked Example: Identical Marginals, Different Ratios

Let $Z$ take values $1$ and $-1$, each with probability $1/2$. For every $n$, consider first the pair $(X_n,Y_n)=(Z,Z)$, and then the pair $(\widetilde X_n,\widetilde Y_n)=(Z,-Z)$. In both constructions, each numerator has the same marginal distribution as each denominator: both are symmetric on $\{-1,1\}$. Every denominator is nonzero. But the first ratio is always $1$, since $Z/Z=1$, while the second is always $-1$, since $Z/(-Z)=-1$. Separate marginal limits therefore cannot determine the ratio limit. The differing dependence is visible in the joint laws.

Practical Checks Before Applying the Result

Establish joint convergence.
Use joint weak convergence directly, or use Slutsky’s theorem when one coordinate converges in probability to a constant.

Check the limit denominator.
Verify that $\mathbb P(Y=0)=0$. It is not enough that most values of the denominator are nonzero.

Account for finite-sample zeros.
If ratios are undefined when $Y_n=0$, assign a finite value there; the zero-denominator probability tends to zero under the theorem’s assumptions.

Apply the quotient map.
Conclude that the ratio converges to $X/Y$, retaining the dependence encoded by the joint limit.

A common pitfall is to demand that the denominator stay uniformly away from zero. That stronger condition is unnecessary for weak convergence of the ratio. A random limiting denominator may come arbitrarily close to zero, as long as it equals zero with probability zero. Conversely, if it has positive probability of being zero, the general theorem does not apply, even if some specially structured ratios still converge.

Check Your Understanding

Use the joint ratio theorem and its conditions to answer these questions.

Why does the ratio theorem require joint convergence rather than only separate weak convergence of numerator and denominator?
If $Y_n\Rightarrow Y$ and $\mathbb P(Y=0)=0$, what does the closed-set Portmanteau inequality imply about $\mathbb P(Y_n=0)$?
Suppose $X_n\Rightarrow X$ and $Y_n\to_{\mathbb P}c$. Which condition on $c$ allows the ratio corollary to apply, and why?
Can a denominator approach zero with positive probability in arbitrarily small neighborhoods of zero while still satisfying the ratio theorem? Explain.
What does the pair $X_n=Y_n=1/n$ show about the role of a nonzero limiting denominator?