Combining Asymptotic Limits
Many asymptotic statistics are built by combining several random quantities: a sum of terms, a product with an estimated coefficient, or a ratio with an estimated scale. The key distinction is how each term converges. A term may have a nondegenerate distributional limit, while another settles near a constant in probability. Slutsky’s theorem lets us combine these limits without requiring independence.
The result is especially useful when a statistic has a random fluctuation whose limit is already known, but also contains a nuisance quantity that can be replaced asymptotically by a constant. Ratios require one additional check: the denominator’s probability limit must be nonzero.
The Combination Rules
The same idea applies to several terms and to continuous combinations such as \(X_n+Y_nZ_n\), provided the terms with nondegenerate limits converge jointly and the other terms converge in probability to constants. In particular, if \((X_n,Z_n)\xrightarrow{d}(X,Z)\) and \(Y_n\xrightarrow{p}c\), then \(X_n+Y_nZ_n\xrightarrow{d}X+cZ\).
The formal basis is the joint-convergence part of Slutsky’s theorem: \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{p}c\) imply \((X_n,Y_n)\xrightarrow{d}(X,c)\). Addition, multiplication, and division by a nonzero denominator are continuous functions at the relevant limiting values, so the continuous mapping theorem gives the displayed limits. For division, the map \((x,y)\mapsto x/y\) is continuous wherever \(y\neq0\).
Worked Applications
Worked Example: Adding and Multiplying by a Nearly Constant Term
Let \(Z\sim N(0,1)\) and \(E\sim N(0,1)\), and define \(X_n=Z\) and \(Y_n=2+E/n\). The variables \(Z\) and \(E\) may be dependent. Find the limits of \(X_n+Y_n\) and \(X_nY_n\).
For every \(\varepsilon>0\), \[ P(|Y_n-2|>\varepsilon)=P(|E|>n\varepsilon)\longrightarrow 0, \] because \(E\) is finite almost surely. Thus \(Y_n\xrightarrow{p}2\). Also \(X_n\) has the \(N(0,1)\) distribution for every \(n\), so \(X_n\xrightarrow{d}Z\).
For the sum, Slutsky’s theorem gives \(X_n+Y_n\xrightarrow{d}Z+2\). Hence the limiting distribution is \(N(2,1)\). For the product, it gives \(X_nY_n\xrightarrow{d}2Z\), whose distribution is \(N(0,4)\). The dependence between \(Z\) and \(E\) does not change either conclusion: the second term contributes only its probability limit.
Worked Example: Studentizing a Sample Mean
Let \(X_1,\ldots,X_n\) be IID observations with \(E[X_i]=\mu\) and \(0<\operatorname{Var}(X_i)=\sigma^2<\infty\). Define \[ \bar X_n=\frac{1}{n}\sum_{i=1}^nX_i,\qquad S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X_n)^2. \] Find the limiting distribution of \(\sqrt n(\bar X_n-\mu)/S_n\).
The IID central limit theorem gives \[ \sqrt n(\bar X_n-\mu)\xrightarrow{d}N(0,\sigma^2). \] The law of large numbers applied to \(X_i\) and \(X_i^2\) gives \(\bar X_n\xrightarrow{p}\mu\) and \(n^{-1}\sum_{i=1}^nX_i^2\xrightarrow{p}E[X_i^2]=\mu^2+\sigma^2\). Using \[ S_n^2=\frac{n}{n-1}\left(\frac{1}{n}\sum_{i=1}^nX_i^2-\bar X_n^2\right), \] we obtain \(S_n^2\xrightarrow{p}\sigma^2\), and therefore \(S_n\xrightarrow{p}\sigma\), since \(\sigma>0\).
The denominator has the nonzero probability limit \(\sigma\), so the ratio rule applies: \[ \frac{\sqrt n(\bar X_n-\mu)}{S_n} \xrightarrow{d}\frac{N(0,\sigma^2)}{\sigma} =N(0,1). \] Here the notation in the middle means scaling a random variable with distribution \(N(0,\sigma^2)\) by \(1/\sigma\). No independence between \(\bar X_n\) and \(S_n\) is assumed or needed.
Worked Example: A Sum of Jointly Convergent Random Terms
Suppose \((A_n,B_n)\) has, for every \(n\), a bivariate normal distribution with means zero, marginal variances one, and correlation \(1/2\). Let \(E\sim N(0,1)\) and \(C_n=1+E/n\). Find the limit of \(A_n+C_nB_n\).
The distribution of \((A_n,B_n)\) is the same for every \(n\), so \((A_n,B_n)\xrightarrow{d}(A,B)\), where \((A,B)\) has that bivariate normal distribution. As in the first example, \(C_n\xrightarrow{p}1\). Slutsky’s theorem for the joint vector and the continuous map \((a,b,c)\mapsto a+cb\) yield \[ A_n+C_nB_n\xrightarrow{d}A+B. \]
The sum \(A+B\) is normal. Its mean is zero, and its variance is \[ \operatorname{Var}(A+B)=\operatorname{Var}(A)+\operatorname{Var}(B)+2\operatorname{Cov}(A,B) =1+1+2\left(\frac12\right)=3. \] Thus \(A_n+C_nB_n\xrightarrow{d}N(0,3)\). The correlation matters because both terms retain random variation in the limit; it is not correct to treat them as independent.
Why Joint Convergence Matters
A single distributional limit for each of two terms does not, by itself, determine the limit of their sum. For example, let \(Z\sim N(0,1)\). If \(X_n=Z\) and \(Y_n=Z\), then each marginal has a standard normal distribution and \(X_n+Y_n\) has distribution \(N(0,4)\). If instead \(X_n=Z\) and \(Y_n=-Z\), the marginals are still standard normal, but \(X_n+Y_n=0\) with probability one. The difference is the dependence, which marginal limits alone do not specify.
When both terms have nondegenerate limits, use a joint limit—often supplied by a multivariate central limit theorem—and then apply the continuous mapping theorem to the desired sum or product. When one term converges in probability to a constant, Slutsky’s theorem supplies the needed joint convergence automatically. This is why a probability-convergent coefficient can be combined with a random term without first establishing independence.
Ratios and the Denominator Check
The nonzero-limit condition in the ratio rule is essential. If \(Y_n\xrightarrow{p}0\), then division by \(Y_n\) is not covered by the rule: the denominator is approaching the point where division is discontinuous. The ratio might diverge, remain bounded, or have a nontrivial limit, depending on the relative behavior of its numerator and denominator. It needs its own analysis rather than substitution of zero into the denominator.
Even when the denominator limit is nonzero, it is worth checking that the ratio is defined with probability approaching one. If \(Y_n\xrightarrow{p}c\neq0\), then \[ P(Y_n=0)\leq P(|Y_n-c|\geq |c|)\longrightarrow0. \] Thus any arbitrary definition of the ratio on the event \(Y_n=0\) cannot affect its asymptotic distribution.
A Reliable Calculation Sequence
Identify which terms converge in distribution and which converge in probability to constants.
A term converging in probability to a constant can be combined with a distribution-convergent term by Slutsky’s theorem. For multiple nondegenerate limits, establish their joint convergence.
Addition and multiplication are continuous everywhere; division requires a nonzero limiting denominator.
Apply the operation to the joint limit, retaining its dependence structure, and simplify the resulting distribution if possible.
These rules turn many asymptotic calculations into a disciplined substitution: keep the limiting randomness, replace terms converging in probability by their constants, and verify continuity at the resulting limit. The denominator check and the distinction between joint and marginal convergence prevent the most common errors.
Check Your Understanding
For each question, identify the convergence conditions needed before applying the relevant limit rule.
- If \(X_n\xrightarrow{d}X\) and \(Y_n\xrightarrow{p}4\), what are the limits of \(X_n+Y_n\) and \(X_n/Y_n\)?
- Why does the product rule not require \(X_n\) and \(Y_n\) to be independent when \(Y_n\xrightarrow{p}c\)?
- Can the limits of \(X_n\) and \(Y_n\) separately in distribution determine the limit of \(X_n+Y_n\)? Give a reason.
- What condition on the probability limit of \(Y_n\) is needed to apply Slutsky’s ratio rule to \(X_n/Y_n\)?
- If \((A_n,B_n)\xrightarrow{d}(A,B)\) and \(C_n\xrightarrow{p}c\), what is the candidate limit of \(A_n+C_nB_n\)?