Anscombe Theorem for Random Sample Sizes

Why Random Sample Sizes Need a Transfer Argument

In an adaptive trial, the number of observations assigned to an arm is random: it depends on the allocation decisions made as the trial proceeds. A limit theorem for a response sequence at a fixed sample size therefore does not, by itself, establish the same limit when the sequence is stopped at an adaptive count. The count may depend on the observed responses, and the partial sum at that count may differ from the sum at a nearby deterministic index.

Anscombe’s theorem supplies a precise transfer argument. It combines concentration of the random index around a deterministic reference size with a condition ensuring that partial sums cannot change substantially over a short range of indices. Crucially, concentration does not require the random index to be independent of the response sequence.

Definition: Let $X_1,X_2,\ldots$ be centered random variables and let $S_m=\sum_{j=1}^{m}X_j$, with $S_0=0$. A sequence of integer-valued random indices $N_n$ is relatively concentrated around deterministic reference sizes $m_n$ if $m_n\to\infty$ and $N_n/m_n\to_p 1$. The partial sums satisfy the Anscombe condition at the scale $m_n$ if, for every $\epsilon>0$, $$ \lim_{\delta\downarrow 0}\limsup_{n\to\infty} P\left( \max_{\substack{j\geq 0\\|j-m_n|\leq \delta m_n}} \frac{|S_j-S_{m_n}|}{\sqrt{m_n}}>\epsilon \right)=0. $$

The condition says that, with high probability, moving the index by a sufficiently small fraction of $m_n$ changes the sum by less than a specified multiple of $\sqrt{m_n}$. It is a local regularity requirement on the partial sums, not an independence requirement on $N_n$. For a fixed number of response streams, the same condition can be imposed on each stream.

Anscombe’s Theorem and Its Justification

Theorem: Suppose $m_n\to\infty$, $N_n/m_n\to_p1$, the partial sums satisfy the Anscombe condition at $m_n$, and $$ \frac{S_{m_n}}{\sqrt{m_n}}\ \xrightarrow{d}\ Z. $$ Then $$ \frac{S_{N_n}}{\sqrt{N_n}}\ \xrightarrow{d}\ Z. $$ Here $Z$ is any proper limiting random variable for the deterministic-index statistic. The convergence result concerns distributions; it does not assert that $S_{N_n}$ and $S_{m_n}$ are close for every realized data set.

The result transfers a partial-sum limit when both the index and its normalization change to match the random sample size. The key step is to show that the difference between the random-index sum and the reference-index sum is negligible on the $\sqrt{m_n}$ scale.

Restrict the random index.
For any fixed $\delta>0$, the probability that $|N_n-m_n|>\delta m_n$ tends to zero as $n\to\infty$, by relative concentration.

Control the nearby sums.
On the event $|N_n-m_n|\leq\delta m_n$, the difference $|S_{N_n}-S_{m_n}|$ is bounded by the maximum in the Anscombe condition.

Make the difference negligible.
First take $n\to\infty$, then let $\delta\downarrow0$. The two preceding controls give $(S_{N_n}-S_{m_n})/\sqrt{m_n}\to_p0$.

Change the normalization.
Since $\sqrt{m_n/N_n}\to_p1$, the random and deterministic normalizations are asymptotically equivalent. Slutsky’s theorem and the deterministic-index limit give the stated result.

For independent, identically distributed centered $X_j$ with finite variance $\sigma^2$, the classical central limit theorem gives $S_m/\sqrt{m}\xrightarrow{d}N(0,\sigma^2)$, and the Anscombe condition follows from maximal inequalities. For example, when $m_n=m$, an interval of at most $\delta m$ increments has variance at most approximately $\delta m\sigma^2$. Maximal-inequality bounds make the probability of a fluctuation larger than $\epsilon\sqrt m$ of order $\delta\sigma^2/\epsilon^2$, which vanishes as $\delta\downarrow0$. The backward interval has the same type of bound. Thus this familiar independent, finite-variance setting meets the local regularity requirement. In adaptive applications, a suitable functional limit theorem or a direct maximal-inequality argument can provide the needed condition; it should not simply be assumed from a pointwise central limit theorem.

Worked Applications

Worked Example: A Random Index Near a Deterministic Size

Consider a centered response stream with partial sums $S_m$ such that $S_m/\sqrt m\xrightarrow{d}N(0,4)$. Assume the stream satisfies the Anscombe condition. For each deterministic reference size $m$, let $U$ be uniform on $[0,1]$ and set $N_m=m+\lfloor U\sqrt m\rfloor$. The index can depend on the trial’s randomization, so it need not be independent of the stream.

At $m=10{,}000$, the added count is between 0 and 99, so $10{,}000\leq N_m\leq10{,}099$. Consequently, $$ 1\leq\frac{N_m}{m}\leq\frac{10{,}099}{10{,}000}=1.0099. $$ More generally, $0\leq(N_m-m)/m\leq1/\sqrt m$, which tends to zero. Thus $N_m/m\to_p1$.

For an illustrative realized stream, suppose $U=0.63$, $S_{10{,}000}=18.0$, and the next 63 increments sum to $-2.5$. Then $N_m=10{,}063$, $S_{N_m}=18.0-2.5=15.5$, and the two scaled values are $$ \frac{S_{10{,}000}}{\sqrt{10{,}000}}=\frac{18.0}{100}=0.1800, \qquad \frac{S_{10{,}063}}{\sqrt{10{,}063}}\approx\frac{15.5}{100.315}=0.1545. $$ These realized numbers need not be close. The theorem instead says that, across repetitions, the random-index statistic has the same limiting distribution as the deterministic-index statistic: $N(0,4)$. Relative concentration and the assumed local oscillation condition justify that distributional conclusion.

Worked Example: Applying the Theorem to an Adaptive Arm Count

Suppose an adaptive two-arm design allocates a proportion tending to $\rho_A=0.40$ to arm A. Write $N_{A,n}$ for the number assigned to A by total trial size $n$. For the response stream on that arm, suppose the centered responses $X_{A,j}=Y_{A,j}-\mu_A$ have a deterministic-index limit $$ \frac{S_{A,m}}{\sqrt m}\xrightarrow{d}N(0,9), \qquad S_{A,m}=\sum_{j=1}^{m}X_{A,j}, $$ and satisfy the Anscombe condition. Assume the response stream represents the successive responses that would be observed on arm A, and that the model for those responses remains valid under the adaptive assignment mechanism.

The positive limiting allocation fraction gives $N_{A,n}/n\to_p0.40$. Choose the deterministic reference $m_n=\lfloor0.40n\rfloor$. Since $m_n/n\to0.40$, $$ \frac{N_{A,n}}{m_n} = \frac{N_{A,n}/n}{m_n/n} \xrightarrow{p}\frac{0.40}{0.40}=1. $$ For example, at $n=10{,}000$, the reference size is $m_n=4{,}000$. If a hypothetical realized trial assigns 3,976 participants to A, its observed ratio to the reference size is $3{,}976/4{,}000=0.9940$. This one ratio illustrates closeness; the theorem’s assumption is the asymptotic convergence in probability, not a single trial’s count.

Anscombe’s theorem now yields $$ \frac{S_{A,N_{A,n}}}{\sqrt{N_{A,n}}} \xrightarrow{d}N(0,9). $$ Thus, for the centered arm-A response sum, the asymptotic variance on the square-root-of-the-arm-count scale is 9 response-units squared. The corresponding asymptotic standard deviation on that scale is $\sqrt9=3$ response units. This conclusion transfers the stated stream limit to its adaptive index; it does not establish that the stream limit or allocation-fraction assumption holds in every adaptive design.

Worked Example: Why a Vanishing Allocation Fraction Needs a Different Reference

Suppose a hypothetical arm receives $N_{A,n}=\lfloor3\sqrt n\rfloor$ participants by total size $n$. Then $N_{A,n}/n\to0$. Consider an iid centered response stream with variance 1, so $S_m/\sqrt m\xrightarrow{d}N(0,1)$. If one informally uses $\sqrt n$ as the reference arm sample size, the relative concentration check fails: the proposed reference is $m_n=\lfloor\sqrt n\rfloor$, and $$ \frac{N_{A,n}}{m_n}\longrightarrow3,\quad\text{not }1. $$

On the scale $\sqrt{m_n}$, the sum at the actual index has limiting variance 3 rather than 1. In fact, $S_{N_{A,n}}/\sqrt{m_n}$ has variance $N_{A,n}/m_n\to3$ in this Gaussian example, whereas $S_{m_n}/\sqrt{m_n}$ has variance 1. The proposed transfer to the reference-index limit is therefore unjustified. This does not show that a central limit theorem using the actual count and normalization $\sqrt{N_{A,n}}$ must fail. It shows that a vanishing allocation fraction cannot be treated as concentration around a positive fraction of total trial size; a suitable arm-specific reference scale and its own concentration check are needed.

Diagnostics, Alternatives, and Limitations

In an adaptive design, a positive limiting allocation fraction $\rho_k$ for arm $k$ supplies a natural reference $m_n=\lfloor\rho_k n\rfloor$. If $N_{k,n}/n\to_p\rho_k>0$, then $N_{k,n}/m_n\to_p1$. A vanishing or unstable allocation fraction does not supply this conclusion. Check the ratio at the scale actually used in the deterministic-index limit, rather than checking only whether the arm count grows.

The other diagnostic is whether the partial sums are locally regular at that scale. A deterministic-index central limit theorem alone controls $S_{m_n}$; it does not automatically control the maximum change in $S_j$ over nearby indices. For iid finite-variance responses, standard maximal inequalities provide a route to this control. With adaptive or dependent response sequences, use assumptions and results appropriate to that dependence, such as a functional limit theorem or a suitable martingale argument. If the required bound cannot be established, the transfer step remains unproved.

Anscombe transfer complements, rather than replaces, other limit-theorem approaches. A deterministic-time central limit theorem is simpler when the sample size is fixed. A martingale central limit theorem may directly handle response noise indexed by successive trial assignments, given its own variance and Lindeberg conditions. Anscombe’s theorem is useful when a limit is already available along a response stream at deterministic indices and the statistic of interest stops at an adaptive count. These routes may use different assumptions; one should not treat them as interchangeable.

A common mistake is to argue, “The arm count is close to its target, so I can substitute the random count into any central limit theorem.” The correction is to check both relative concentration around the deterministic reference size and local control of partial-sum increments. Another mistake is to require independence between the count and the responses. That independence is not required by the theorem: the local maximum controls all indices in the neighborhood, including an index selected using the data. However, the response-stream model and its limit assumptions must still be justified under the adaptive design.

The theorem is a transfer result, not a guarantee of a normal limit. Its limit $Z$ can be nonnormal if that is the deterministic-index limit. It also does not automatically prove joint convergence for several arms or for allocation counts and response estimates, nor does it establish the consistency or central limit theorem for an arm-specific mean. Those conclusions require their own arguments.

Key takeaway: To transfer a deterministic-index partial-sum limit to an adaptive sample size, identify a deterministic reference size, verify that the random index is relatively concentrated around it, and establish local partial-sum regularity on the square-root scale.

Check Your Understanding

For each question, use the assumptions stated in the question; brief solutions follow.

If $N_n/n\to_p0.25$, what deterministic reference $m_n$ is natural when the total trial size is $n$, and what ratio must converge to 1?
Why does Anscombe’s theorem not require the random index $N_n$ to be independent of the response stream?
A deterministic-index central limit theorem holds, but no local oscillation condition or other control of nearby sums is available. Is the random-index transfer justified?
An arm count satisfies $N_n/n\to_p0$. Does this alone show that its count diverges or that it is concentrated around a usable deterministic reference? Explain.

Solutions. (1) Use $m_n=\lfloor0.25n\rfloor$; verify $N_n/m_n\to_p1$. (2) The Anscombe condition controls the maximum change over all indices in a shrinking relative neighborhood, so it also controls an index selected using the data. (3) No. A pointwise limit at the deterministic index does not by itself control changes in partial sums at nearby indices. (4) No. A ratio tending to zero relative to total size neither guarantees that the count diverges nor identifies a deterministic reference with which the count is relatively concentrated.

Anscombe Theorem for Random Sample Sizes

What You'll Learn