Studentized Limits from Joint Convergence

Why Studentize an Asymptotic Limit?

A statistic may have a useful asymptotic distribution after centering and multiplying by a rate, but that distribution can still depend on an unknown variance. Studentization replaces the unknown scale with an estimate. The resulting statistic is often asymptotically pivotal: its limiting distribution no longer depends on the unknown scale parameter.

The key issue is not whether the numerator and estimated scale are independent. In many applications they are calculated from the same observations, so independence is absent. What matters for the usual studentized limit is that the numerator has a distributional limit and that the scale estimate converges in probability to a positive constant.

Definition: Let $a_n$ be a centering sequence, $r_n>0$ a sequence of rates, and $\hat s_n$ an estimate of a positive scale $s$. A statistic of the form $$ T_n=\frac{r_n(A_n-a_n)}{\hat s_n} $$ is a studentized statistic. Its numerator is centered and scaled by $r_n$; its denominator estimates the asymptotic scale of that numerator. The statistic can be assigned any value when $\hat s_n=0$, provided $P(\hat s_n=0)\to 0$.

The scale must match the numerator. For example, if $r_n(A_n-a_n)$ converges to a normal distribution with variance $\tau^2$, the denominator $\hat s_n$ should estimate the standard deviation $\tau$, not the variance $\tau^2$. In a sample-mean problem, this often means dividing by an estimated standard deviation and not by an estimated variance.

The Studentization Principle

The central result combines a distributional limit for the numerator with consistency of the estimated scale. The joint-convergence formulation makes clear why dependence between the two components causes no problem when the scale limit is constant.

Theorem: Suppose $U_n\xrightarrow{d}U$ and $S_n\xrightarrow{p}s$, where $s>0$ is a constant. Then $$ \frac{U_n}{S_n}\xrightarrow{d}\frac{U}{s}. $$ No independence assumption between $U_n$ and $S_n$ is required.

Here is a proof of the joint-convergence step used in the theorem. Let $g$ be any bounded continuous function on the product space of $(U_n,S_n)$. The sequence $U_n$ is tight because it converges in distribution. Given any $\varepsilon>0$, choose a compact set $K$ such that $P(U_n\notin K)<\varepsilon$ for all sufficiently large $n$. On a sufficiently small compact neighborhood of $K\times\{s\}$, continuity of $g$ is uniform. Since $S_n\xrightarrow{p}s$, it follows that $g(U_n,S_n)-g(U_n,s)\xrightarrow{p}0$. This difference is bounded, so its expected absolute value tends to zero. Also, $g(U_n,s)$ converges in expectation to $g(U,s)$, by $U_n\xrightarrow{d}U$. Thus $(U_n,S_n)\xrightarrow{d}(U,s)$. The continuous mapping theorem applied to $(u,v)\mapsto u/v$, which is continuous at every point with $v=s>0$, gives the claimed result. $\square$

In applications, a central limit theorem or another limit result supplies $U_n=r_n(A_n-a_n)$, while consistency supplies $S_n=\hat s_n$. If $U_n\xrightarrow{d}N(0,s^2)$, then the theorem and a linear rescaling give $U_n/\hat s_n\xrightarrow{d}N(0,1)$. The scale estimate may use the same data as $U_n$: the proof relies on convergence of the scale to a constant, not on independence.

Worked Example: Studentizing a Bernoulli Sample Proportion

Let $X_1,\ldots,X_n$ be independent Bernoulli random variables with success probability $p$, where $0<p<1$. Write $\hat p_n=n^{-1}\sum_{i=1}^n X_i$. Find the limit of the studentized statistic that uses the estimated standard error of $\hat p_n$.

A Bernoulli variable has mean $p$ and variance $p(1-p)$. The central limit theorem therefore gives $$ \sqrt n(\hat p_n-p)\xrightarrow{d}N\bigl(0,p(1-p)\bigr). $$ The sample proportion is consistent for $p$, so the plug-in scale $\hat s_n=\sqrt{\hat p_n(1-\hat p_n)}$ converges in probability to $s=\sqrt{p(1-p)}>0$. The estimated standard error of $\hat p_n$ is $\hat s_n/\sqrt n$; hence the studentized statistic is $$ T_n=\frac{\hat p_n-p}{\sqrt{\hat p_n(1-\hat p_n)/n}} =\frac{\sqrt n(\hat p_n-p)}{\hat s_n}. $$ The theorem yields $T_n\xrightarrow{d}N(0,1)$.

For $p=0.4$, the numerator has limiting variance $0.4(0.6)=0.24$, and the limiting denominator is $\sqrt{0.24}$, about $0.490$. Dividing a $N(0,0.24)$ limit by $\sqrt{0.24}$ gives a standard normal limit. The restriction $0<p<1$ ensures the limiting scale is positive; at either endpoint this argument does not apply.

Worked Example: Studentizing an Exponential Sample Mean

Suppose $X_1,\ldots,X_n$ are independent exponential observations with mean $\theta=2$. Let $\bar X_n$ be their sample mean and let $$ S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X_n)^2 $$ be the sample variance. Find the asymptotic limit of $\sqrt n(\bar X_n-2)/S_n$.

For an exponential distribution with mean $\theta$, the variance is $\theta^2$. At $\theta=2$, the variance is $4$, so the central limit theorem gives $$ \sqrt n(\bar X_n-2)\xrightarrow{d}N(0,4). $$ The distribution has finite second moment. The law of large numbers applied to $X_i$ and $X_i^2$ implies $S_n^2\xrightarrow{p}4$, and therefore $S_n\xrightarrow{p}2$, since the square-root function is continuous at $4$.

The scale $2$ is positive. Consequently, $$ \frac{\sqrt n(\bar X_n-2)}{S_n}\xrightarrow{d}N(0,1). $$ The observations used to form $S_n$ are the same observations used in $\bar X_n$, but this does not affect the conclusion. The denominator converges to the constant $2$, which is the condition needed for studentization here.

Joint Limits and Random Denominators

Consistency for a positive constant is a particularly convenient case of a more general ratio result. If the denominator has a nonconstant limit, the joint behavior matters: knowing the numerator and denominator limits separately does not generally determine the ratio limit.

Theorem: Suppose $(U_n,V_n)\xrightarrow{d}(U,V)$ and $P(V=0)=0$. Then $$ \frac{U_n}{V_n}\xrightarrow{d}\frac{U}{V}, $$ where the ratio may be defined arbitrarily on the event $V_n=0$. This is the continuous mapping theorem for division, since $(u,v)\mapsto u/v$ is continuous wherever $v\ne0$.

When $V_n\xrightarrow{p}s>0$, the joint convergence result above gives $(U_n,V_n)\xrightarrow{d}(U,s)$, and the random-denominator theorem reduces to the usual studentization principle. When $V$ is random, however, the ratio $U/V$ depends on how $U$ and $V$ vary together. Independence is neither automatic nor required; the joint limit must represent their actual dependence.

Worked Example: Studentizing a Ratio of Sample Means

Let $(X_i,Y_i)$, $i=1,\ldots,n$, be independent and identically distributed pairs with mean vector $(1,2)$ and covariance matrix $$ \Sigma= \begin{pmatrix} 4&1\\ 1&9 \end{pmatrix}. $$ Assume the observations have finite second moments. Consider $\hat R_n=\bar Y_n/\bar X_n$, which estimates $R=2$. Construct a studentized limit using a plug-in estimate of its asymptotic scale.

The gradient of $g(x,y)=y/x$ at $(1,2)$ is $c=(-2,1)^\mathsf{T}$. By the multivariate central limit theorem and the first-order delta method, $$ \sqrt n(\hat R_n-2)\xrightarrow{d}N(0,c^\mathsf{T}\Sigma c). $$ Calculating the variance gives $$ c^\mathsf{T}\Sigma c =(-2,1) \begin{pmatrix} 4&1\\ 1&9 \end{pmatrix} \begin{pmatrix} -2\\ 1 \end{pmatrix} =(-2,1) \begin{pmatrix} -7\\ 7 \end{pmatrix} =14+7=21. $$ Thus the asymptotic standard deviation of the numerator is $\sqrt{21}$.

Let $\hat\Sigma_n$ be the sample covariance matrix of the pairs, and define the estimated gradient $$ \hat c_n= \begin{pmatrix} -\bar Y_n/\bar X_n^2\\ 1/\bar X_n \end{pmatrix}. $$ Because $\bar X_n\xrightarrow{p}1$, $\bar Y_n\xrightarrow{p}2$, and $\hat\Sigma_n\xrightarrow{p}\Sigma$, we have $\hat c_n^\mathsf{T}\hat\Sigma_n\hat c_n\xrightarrow{p}21$. The estimated scale $\hat s_n=\sqrt{\hat c_n^\mathsf{T}\hat\Sigma_n\hat c_n}$ therefore converges to $\sqrt{21}>0$. It follows that $$ \frac{\sqrt n(\hat R_n-2)}{\sqrt{\hat c_n^\mathsf{T}\hat\Sigma_n\hat c_n}}\xrightarrow{d}N(0,1). $$ The scale estimate and ratio estimator are calculated from the same sample. Their possible dependence is allowed by the theorem.

Common Pitfall and Practical Meaning

A common mistake is to infer a ratio limit from marginal limits alone. For example, it is not enough to know separately that $U_n\xrightarrow{d}U$ and $V_n\xrightarrow{d}V$ when $V$ has a random limit. Different dependence structures between numerator and denominator can produce different ratio distributions. The correction is to establish joint convergence, or, in the usual studentization setting, to establish that the denominator converges in probability to a positive constant.

A second frequent error is to treat “positive scale” as merely “nonnegative.” If the limiting scale is zero, division can magnify fluctuations and the studentization result above does not apply; that situation requires a different normalization and analysis. If the scale converges to a positive constant, it is also enough that the denominator be positive with probability tending to one. Any exceptional cases where it is zero can be assigned an arbitrary value without changing the asymptotic distribution.

Studentized limits make asymptotic inference possible when a variance is unknown: they support approximate normal critical values and confidence intervals after replacing the scale by a consistent estimate. Their reliability rests on matching the scale to the numerator and verifying the limiting scale is strictly positive—not on treating the estimated scale as independent.

Check Your Understanding

Use the conditions on convergence and scale to answer each question.

If $U_n\xrightarrow{d}N(0,16)$ and $S_n\xrightarrow{p}4$, what is the limit of $U_n/S_n$?
Why does the studentized Bernoulli proportion example require $0<p<1$?
Does the studentization theorem require independence between a statistic and its estimated standard error? Explain.
What additional convergence information is needed to find the limit of $U_n/V_n$ when $V_n$ has a nonconstant random limit?
In the ratio-of-means example, why is the limiting scale $\sqrt{21}$, rather than $21$?

Studentized Limits from Joint Convergence

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