Ratios With Random Denominators

Why Random Denominators Need Care

Many econometric quantities are ratios: an average outcome per unit of exposure, a relative frequency, or one sample moment divided by another. Both parts of such a statistic are random in finite samples. It is tempting to take the limit of the numerator and denominator separately and divide the answers. That reasoning is valid only when the denominator’s limit is nonzero, or, in a more general distributional result, when the limiting random denominator is nonzero with probability one.

The issue is not merely algebraic. If a denominator can be close to zero, small fluctuations in it can produce large changes in the ratio. The convergence mode also matters: probability limits support a probability limit for the ratio, while distributional limits require joint information unless the denominator converges in probability to a constant.

Ratios With Constant Probability Limits

Theorem: Suppose $A_n\to_p a$ and $B_n\to_p b$, where $a$ and $b$ are constants and $b\ne 0$. Then, wherever the ratio is defined, $\frac{A_n}{B_n}\to_p\frac{a}{b}$. If $B_n$ can equal zero for some $n$, define the ratio arbitrarily on that event; under these assumptions, the probability of that event tends to zero, so this convention does not change the probability limit.

The condition $b\ne 0$ ensures that $B_n$ is bounded away from zero with probability approaching one. In particular, for any $\epsilon>0$, convergence in probability gives $P(|B_n-b|\ge |b|/2)\to 0$, and on the complementary event $|B_n|\ge |b|/2$. Thus an increasingly small error in the numerator and denominator cannot be magnified by a denominator approaching zero.

The algebra makes the role of this condition explicit:

\frac{A_n}{B_n}-\frac{a}{b} = \frac{b(A_n-a)-a(B_n-b)}{bB_n}.

The numerator on the right converges in probability to zero. With probability approaching one, the denominator has absolute value at least $b^2/2$, a positive constant. Consequently, the whole difference converges in probability to zero. This is the ratio result obtained by applying the continuous mapping theorem to the map $(x,y)\mapsto x/y$, which is continuous wherever $y\ne 0$.

Worked Example: A Ratio of Sample Averages

Suppose a hypothetical sample consists of independent, identically distributed pairs $(Y_i,X_i)$, where $Y_i$ is an outcome in dollars and $X_i$ is exposure measured in hours. Assume $E[Y_i]=120$, $E[X_i]=50$, both means are finite, and $E[X_i]\ne0$. Consider the ratio of sample averages $\hat r_n=\bar Y/\bar X$.

By the weak law of large numbers, $\bar Y\to_p120$ and $\bar X\to_p50$. Since the denominator limit is positive, the ratio theorem gives

\hat r_n=\frac{\bar Y}{\bar X} \ \xrightarrow{p}\ \frac{120}{50}=2.4.

For illustration, suppose one hypothetical sample gives $\bar Y=\$126$ and $\bar X=48$ hours. Its ratio is $\$126/48= \$2.625$ per hour. This finite-sample value need not equal the probability limit. The asymptotic interpretation is that the ratio of sample averages converges in probability to $\$2.40$ per hour as the sample size grows, under the stated sampling and moment conditions. No claim that the finite-sample ratio is exactly its limit is warranted.

Distributional Limits and Slutsky’s Theorem

A common case has a numerator with a non-degenerate distributional limit and a denominator that converges in probability to a constant. Write $A_n\to_d A$ and $B_n\to_p b$, where $b\ne0$. Slutsky’s theorem gives the joint convergence $(A_n,B_n)\to_d(A,b)$. Applying the continuous mapping theorem to division yields the following result.

Theorem: If $A_n\to_d A$ and $B_n\to_p b\ne0$, then $A_n/B_n\to_d A/b$, with the ratio defined arbitrarily on events where $B_n=0$. The numerator and denominator need not be independent.

The nonzero constant denominator is important. If both numerator and denominator have non-degenerate distributional limits, separate marginal convergence statements alone do not specify the limit of their ratio: the dependence between the two limits matters. When the random vectors converge jointly, a broader result applies.

Theorem: Suppose $(A_n,B_n)\to_d(A,B)$ and $P(B=0)=0$. Then $A_n/B_n\to_d A/B$, again defining the finite-sample ratio arbitrarily when $B_n=0$. The condition $P(B=0)=0$ makes division continuous at the limiting pair with probability one.

Worked Example: A Ratio With a Random Limiting Denominator

Let $Z_n\to_d Z$, where $Z\sim N(0,1)$, and define $A_n=Z_n$ and $B_n=1+Z_n^2$. The vector $(A_n,B_n)$ is a continuous function of $Z_n$, so the continuous mapping theorem gives $(A_n,B_n)\to_d(Z,1+Z^2)$. For every real $z$, $1+z^2\ge1$, and therefore $P(1+Z^2=0)=0$.

The joint ratio theorem applies:

\frac{A_n}{B_n} = \frac{Z_n}{1+Z_n^2} \ \xrightarrow{d}\ \frac{Z}{1+Z^2}.

For example, the limiting transformation gives $1/(1+1)=0.5$ at $Z=1$, and $-1/(1+1)=-0.5$ at $Z=-1$. These values illustrate that the numerator and denominator are linked through the same random variable. Replacing their joint behavior with two unrelated marginal limits would discard that dependence and need not give the correct ratio distribution.

Ratios at the Central Limit Scale

A particularly useful application is the asymptotic distribution of a ratio of sample averages. Let $(U_i,V_i)$ be independent, identically distributed random vectors with $E[U_i]=\mu_U$, $E[V_i]=\mu_V\ne0$, and finite second moments. Define $\bar U=n^{-1}\sum_{i=1}^n U_i$, $\bar V=n^{-1}\sum_{i=1}^n V_i$, and $\theta=\mu_U/\mu_V$. The multivariate central limit theorem gives a joint distributional limit for the centered, scaled averages. Also, the weak law gives $\bar V\to_p\mu_V$.

An exact identity isolates the fluctuation relevant to the ratio:

\sqrt n\left(\frac{\bar U}{\bar V}-\theta\right) = \frac{\sqrt n\left[(\bar U-\mu_U)-\theta(\bar V-\mu_V)\right]}{\bar V} = \frac{n^{-1/2}\sum_{i=1}^n(U_i-\theta V_i)}{\bar V}.

The second equality uses $\mu_U-\theta\mu_V=0$. The summands $U_i-\theta V_i$ have mean zero. If their variance $\sigma_R^2=\operatorname{Var}(U_i-\theta V_i)$ is finite, the central limit theorem makes the numerator converge in distribution to $N(0,\sigma_R^2)$. The denominator converges in probability to the nonzero constant $\mu_V$. Slutsky’s theorem therefore gives

\sqrt n\left(\frac{\bar U}{\bar V}-\theta\right) \ \xrightarrow{d}\ N\left(0,\frac{\operatorname{Var}(U_i-\theta V_i)}{\mu_V^2}\right).

This is a distributional limit for the centered ratio at scale $\sqrt n$. It is not a claim that the finite-sample ratio is normally distributed. The variance includes the covariance between $U_i$ and $V_i$, because the same observations contribute to both sample averages.

Worked Example: Asymptotic Distribution of a Sample Ratio

Suppose hypothetical independent, identically distributed pairs $(U_i,V_i)$ have means $\mu_U=3$, $\mu_V=2$, variances $\operatorname{Var}(U_i)=4$, $\operatorname{Var}(V_i)=1$, and covariance $\operatorname{Cov}(U_i,V_i)=1$. Assume finite second moments. The ratio of means is $\theta=3/2=1.5$, and the denominator limit $2$ is nonzero.

For one observation, the centered ratio contribution is $U_i-1.5V_i$. Its variance, calculated using the covariance term, is

\operatorname{Var}(U_i-1.5V_i) =4+(1.5)^2(1)-2(1.5)(1) =4+2.25-3 =3.25.

Thus the limiting variance for the scaled ratio is $3.25/2^2=0.8125$, and

\sqrt n\left(\frac{\bar U}{\bar V}-1.5\right) \ \xrightarrow{d}\ N(0,0.8125).

For $n=100$, this limit suggests an approximate standard error of $\sqrt{0.8125/100}=\sqrt{0.008125}\approx0.0901$ for the unscaled ratio. If a hypothetical sample has $\bar U=3.12$ and $\bar V=2.08$, its observed ratio is $3.12/2.08=1.50$. The approximate standard error describes sampling variation under the stated model; it does not establish that the sample ratio must be exactly 1.5 or that the approximation is exact at $n=100$.

When the Denominator Is Near Zero

A limit of zero in the denominator cannot be handled by the ratio theorem. For example, let $A_n=1/n$ and $B_n=1/n$. Both converge in probability to zero, but $A_n/B_n=1$, not an expression obtained by dividing the limits $0/0$. Other sequences with both limits zero can have ratios that diverge or fail to converge. The numerator and denominator rates of convergence then become essential.

Even a nonzero limiting denominator is useful only asymptotically: in small samples, a denominator may still be close to zero and the ratio may be highly variable. In distributional limits with a random denominator, $P(B=0)=0$ is enough for the mapping theorem, but it does not guarantee that $A/B$ has finite moments. A denominator can approach zero with positive density while equaling zero with probability zero, producing a ratio distribution with heavy tails.

Common Mistake

Wrong reasoning: “The numerator converges to $a$, and the denominator converges to $b$, so the ratio converges to $a/b$, regardless of $b$.”

Correction: The constant-limit ratio rule requires $b\ne0$. If $b=0$, the ratio’s behavior depends on relative rates and the joint behavior of numerator and denominator; dividing the two limits is not valid. For distributional limits with random numerator and denominator, establish joint convergence and verify that the limiting denominator is nonzero with probability one. Marginal limits alone do not generally determine a ratio limit.

Key takeaway: For probability limits, divide constant limits only when the denominator limit is nonzero. For distributional limits with a random denominator, use joint convergence and require the limiting denominator to be nonzero almost surely.

Check Your Understanding

Answer each question by stating the relevant limit condition and, where appropriate, the resulting ratio limit.

If $A_n\to_p 8$ and $B_n\to_p 4$, what is the probability limit of $A_n/B_n$, and which condition makes the result valid?
If $A_n\to_d A$ and $B_n\to_p 3$, what distributional limit follows for $A_n/B_n$? Is independence required?
Why are separate marginal distributional limits for $A_n$ and $B_n$ generally insufficient to determine the distributional limit of their ratio?
For a jointly convergent pair $(A_n,B_n)\to_d(A,B)$, state the condition on $B$ that permits the continuous mapping theorem for division.
In the sample-average ratio derivation, why does the variance involve $\operatorname{Cov}(U_i,V_i)$?