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Asymptotic Theory for Adaptive Allocation · Tutorial 227 of 501

Heavy-Tailed Responses and Non-Gaussian Limits

Learn when heavy-tailed responses lead to stable-law limits under adaptive allocation, and why each arm must receive enough observations for those limits to apply.

Advanced 12 min read

What You'll Learn

  • Define regularly varying response tails and normalize their positive and negative tail weights consistently.
  • Distinguish stable-limit centering for tail indices below, equal to, and above one.
  • State sufficient arm sample-size conditions for adaptive response sums to have stable limits.
  • Derive how limiting allocation fractions change the scale of arm means and contrasts.
  • Diagnose when rare-arm counts or outcome-dependent allocation invalidate a simple stable-limit argument.
  • Compare stable limits with Gaussian approximations and robust alternatives.

When Extreme Responses Determine the Limit

Gaussian limits describe many response sums when variance is finite and no small group of observations dominates. With sufficiently heavy tails, however, a few extreme responses can remain visible after normalization. The resulting limit is typically a stable law, whose scale and skewness reflect the extremes rather than a finite variance. In an adaptive trial, this response-level phenomenon must be combined with a second question: how quickly does each arm accumulate observations?

The focus here is the asymptotic law of arm-specific response sums when the number of observations in each arm is random. The conclusions require more than heavy-tailed marginal distributions: observations within each arm must have a stable tail structure, and the arm counts must grow at a suitable rate. The earlier material on moment conditions for Gaussian response limits treats the finite-moment route; here the relevant conditions are instead expressed through regularly varying tails and diverging arm sample sizes.

Definition: A response \(X\) has regularly varying two-sided tails with index \(\alpha>0\) if, for some slowly varying function \(L\) and constants \(c_+,c_-\geq 0\) with \(C=c_++c_->0\), \(P(X>x)\sim c_+x^{-\alpha}L(x)\) when \(c_+>0\), and \(P(X>x)=o(x^{-\alpha}L(x))\) when \(c_+=0\); likewise, \(P(X<-x)\sim c_-x^{-\alpha}L(x)\) when \(c_->0\), and \(P(X<-x)=o(x^{-\alpha}L(x))\) when \(c_-=0\), as \(x\to\infty\). Slowly varying means \(L(tx)/L(t)\to1\) for each fixed \(x>0\).

The index \(\alpha\) determines the scale of sums. When \(0<\alpha<2\), regularly varying tails are in the domain of attraction of an \(\alpha\)-stable law, subject to appropriate centering. The case \(\alpha=2\) is a boundary case and does not follow automatically from this stable-limit statement; its normalization may involve a slowly varying correction. The stable law for \(0<\alpha<2\) is generally non-Gaussian, except for the special index \(\alpha=2\).

Tail Normalization and Stable Limits

Choose a positive sequence \(b(m)\) so that \(mP(|X|>b(m))\to1\). Since \(P(|X|>t)\sim C t^{-\alpha}L(t)\), this choice entails \(mC b(m)^{-\alpha}L(b(m))\to1\). It follows, for every fixed \(x>0\), that the limiting positive and negative tail probabilities are

$$ mP(X>b(m)x)\to \frac{c_+}{C}x^{-\alpha}, \qquad mP(X<-b(m)x)\to \frac{c_-}{C}x^{-\alpha}. $$

Thus, with this normalization, the tail weights are fractions of the total tail mass, not the original constants \(c_+\) and \(c_-\). The corresponding limiting jump measure has density \(\alpha(c_+/C)x^{-\alpha-1}\) on \(x>0\) and density \(\alpha(c_-/C)|x|^{-\alpha-1}\) on \(x<0\). A zero tail weight means the corresponding normalized tail probability tends to zero. This normalization convention is important: using the unnormalized constants in the limit would contradict \(mP(|X|>b(m))\to1\).

Suppose \(X_1,X_2,\ldots\) are independent and identically distributed with these tails, and write \(S_m=\sum_{j=1}^m X_j\). The stable-sum limit uses different centerings in different index ranges:

  • If \(0<\alpha<1\), \(S_m/b(m)\) converges to an \(\alpha\)-stable random variable without centering. The population mean need not exist.
  • If \(1<\alpha<2\), the mean \(\mu=E[X]\) exists, and \((S_m-m\mu)/b(m)\) has an \(\alpha\)-stable limit.
  • If \(\alpha=1\), a standard centering is \(a_m=mE[X\,\mathbf{1}\{|X|\leq b(m)\}]\); then \((S_m-a_m)/b(m)\) has a stable limit under the stated regular-variation conditions. A finite uncentered mean is not generally available.

Here and below, an assertion of convergence to a stable law presumes the independent, identically distributed response sequence and the tail conditions just specified. One precise way to describe the limit is through its Lévy measure, whose positive- and negative-side densities were given above. For \(0<\alpha<1\), its characteristic exponent is \(\int (e^{itx}-1)\nu(dx)\); for \(1<\alpha<2\), after mean-centering it is \(\int(e^{itx}-1-itx)\nu(dx)\). These integrals converge in their respective ranges. At \(\alpha=1\), the truncation convention used for centering also determines the drift convention for the limit.

Worked Example: Checking the Tail Weights

Suppose a hypothetical response has \(P(X>x)\sim 2x^{-1.5}\) and \(P(X<-x)\sim x^{-1.5}\). Thus \(\alpha=1.5\), \(c_+=2\), \(c_-=1\), \(C=3\), and \(L(x)=1\). Find a normalization and the limiting positive-tail probability at \(x=2\).

The total tail satisfies \(P(|X|>t)\sim3t^{-1.5}\). Taking \(b(m)=(3m)^{1/1.5}=(3m)^{2/3}\) gives \(mP(|X|>b(m))\to m\cdot3(3m)^{-1}=1\).

With that normalization, \(mP(X>b(m)x)\to(2/3)x^{-1.5}\), not \(2x^{-1.5}\). At \(x=2\), the limit is \((2/3)2^{-1.5}=0.2357\). The negative-tail limit at the same \(x\) is \((1/3)2^{-1.5}=0.1179\); the two limits sum to \(2^{-1.5}=0.3536\), as required by the total-tail normalization.

The associated positive jump density is \(1.5(2/3)x^{-2.5}=x^{-2.5}\) for \(x>0\); the negative-side density is \(1.5(1/3)|x|^{-2.5}=0.5|x|^{-2.5}\) for \(x<0\). The factors \(2/3\) and \(1/3\) are essential.

Adaptive Arm Counts and a Stable-Limit Result

Consider a two-arm adaptive trial with \(n\) total assignments. For arm \(k\), let \(N_k(n)\) be its number of assignments and let \(X_{k,1},X_{k,2},\ldots\) denote its successive responses. A standard sufficient setup is that each arm has an independent, identically distributed response sequence with regularly varying tails of common index \(0<\alpha<2\). Assignments are made sequentially using information available before the next response, and the response sequence for an arm is not changed by the allocation rule. This representation allows the random arm count to be viewed as selecting a prefix of that arm's response sequence.

For a simple allocation-scale result, assume \(N_k(n)/n\to\rho_k\) in probability, where each \(\rho_k>0\) is deterministic and \(\rho_1+\rho_2=1\). In particular, both arm counts diverge and are of order \(n\). If counts are selected using observed outcomes, this ratio condition alone should not be treated as a universal random-index theorem: a suitable random-index argument is also needed. Under the stated independent response-sequence setup and deterministic positive limiting fractions, the asymptotic closeness to fixed indices supports the stable-sum conclusion below.

Theorem: Under the preceding tail, response-sequence, and arm-count conditions, the centered response sum in arm \(k\), divided by \(b_k(N_k(n))\), has the same stable limit as a fixed sample of \(N_k(n)\) independent responses from that arm. Here \(b_k(m)\) is the arm-specific tail normalization satisfying \(mP(|X_{k,1}|>b_k(m))\to1\). If the arm response sequences are independent, the two arm-specific limiting stable variables are independent. A random limiting allocation fraction, an arm count that fails to diverge, or allocation-dependent response distributions requires additional conditions and is not covered by this statement.

A short justification explains why the sample-size condition matters. For a fixed arm with \(N_k(n)/n\to\rho_k>0\), the random index differs from \(\lfloor \rho_k n\rfloor\) by \(o_p(n)\). Under the stable-domain-of-attraction conditions, sums over a window of \(o(n)\) consecutive observations are negligible in probability compared with \(b_k(n)\), provided the window endpoints are within \(o_p(n)\) of \(\lfloor\rho_k n\rfloor\). Therefore replacing the random count by that fixed index does not change the normalized limit. Regular variation gives \(b_k(\lfloor\rho_k n\rfloor)/b_k(n)\to\rho_k^{1/\alpha}\), up to the arm-specific tail scale encoded by \(b_k\). This yields the allocation fraction's effect on the limit's scale.

Worked Example: Scaling a Two-Arm Mean Contrast

Suppose two hypothetical arms have independent responses with a common tail index \(\alpha=1.5\), finite means \(\mu_1,\mu_2\), and normalizations \(b_k(m)\sim d_km^{2/3}\), where \(d_1,d_2>0\). Assume \(N_1(n)/n\to0.4\) and \(N_2(n)/n\to0.6\). Let \(\bar X_k\) be the observed mean in arm \(k\). Determine the allocation factors in the stable limit for the centered contrast.

For each arm, its sum centered at its mean is of order \(b_k(N_k)\), and its sample mean error is therefore of order \(b_k(N_k)/N_k\). Since \(b_k(m)\sim d_km^{2/3}\), this error has scale \(d_kN_k^{-1/3}\). Multiplying the contrast by \(n^{1/3}\) gives the factors

$$ n^{1/3}\frac{b_k(N_k)}{N_k} \sim d_k\left(\frac{N_k}{n}\right)^{-1/3}. $$

The arm-1 factor is \(d_1(0.4)^{-1/3}=1.357d_1\), and the arm-2 factor is \(d_2(0.6)^{-1/3}=1.186d_2\), to four significant figures. Thus, if \(Z_1,Z_2\) denote the arm-specific stable limits under their own normalizations, the centered contrast has limiting form \(1.357d_1Z_1-1.186d_2Z_2\). The minus sign comes from subtracting the arm-2 mean. This is a stable, generally non-Gaussian limit, not a normal approximation; the allocation fractions alter its scale even though they do not change either response distribution.

Because \(\alpha>1\), both population means exist and the contrast is meaningful. The assumed positive fractions ensure both arm counts grow proportionally to \(n\), and the independent arm sequences give independent \(Z_1,Z_2\). If either arm instead has \(\alpha\leq1\), this mean-contrast interpretation cannot be used without changing the estimand and centering.

What Happens When an Arm Is Rare?

The condition \(N_k(n)/n\to\rho_k>0\) gives a convenient common trial-size rate, but it is stronger than the basic requirement for an arm's own stable sum limit. An arm may still have a stable limit on its own sample-size scale if \(N_k(n)\to\infty\) and appropriate random-index conditions hold. What fails is the automatic use of a normalization based on \(n\), and the estimator may converge much more slowly.

Worked Example: A Sublinear Arm Count

For a hypothetical response tail index \(\alpha=1.5\), suppose \(b(m)=m^{2/3}\), but a rare arm receives \(N(n)\) of order \(n^{1/2}\), rather than a positive fraction of all \(n\) assignments. What is the scale of its centered sample-mean error?

The sum scale is \(b(N(n))\) and the mean error scale is \(b(N(n))/N(n)=N(n)^{-1/3}\). Since \(N(n)\) is of order \(n^{1/2}\), the error is of order \(n^{-1/6}\). With proportional allocation the corresponding scale would be \(n^{-1/3}\). The rare arm still has a diverging count, so an arm-specific stable limit may remain possible, but it is not valid to apply the proportional-allocation rate to its mean.

Diagnostics, Alternatives, and Limitations

A stable-limit argument is plausible only when its key conditions fit the design and response process. Check the following before using one:

  • Tail evidence and index: a few large responses can dominate a tail plot or estimate of \(\alpha\). Such diagnostics are uncertain in finite samples; an apparent heavy tail is not proof of regular variation.
  • Arm-specific growth: report or examine \(N_k(n)\), not just the total sample size. A small or nondiverging count cannot support a large-sample arm-specific stable approximation.
  • Response-sequence assumptions: dependence over time, changing response distributions, or allocation rules that alter how responses are generated may invalidate the independent, identically distributed sequence model.
  • Centering and estimand: for \(1<\alpha<2\), center at the population mean. For \(\alpha\leq1\), a finite mean need not exist, so calling the arm average an estimator of that mean can be wrong.

The main alternative is a Gaussian approximation under suitable finite-variance conditions, as in the earlier material on moment conditions for Gaussian response limits. It is not made valid merely by a large total sample if an arm has too few observations or the variance is infinite. Robust procedures such as trimming or winsorization can reduce the influence of extremes, but they generally change the target or require their own bias and variance analysis. A stable limit describes the unmodified heavy-tailed sum; it does not by itself provide a ready-made finite-sample confidence interval.

A frequent mistake is to combine an unnormalized tail constant with a normalization based on total tail probability. For example, writing \(mP(X>b(m)x)\to c_+x^{-\alpha}\) while also imposing \(mP(|X|>b(m))\to1\) is generally inconsistent when \(c_++c_-\neq1\). The correction is to use the fractions \(c_+/C\) and \(c_-/C\) in both limiting tails and jump densities, or choose a different normalization and keep its constants consistent throughout. Another mistake is to infer stable behavior solely from a heavy-tailed marginal distribution while ignoring the arm counts and the adaptive sampling mechanism.

Key takeaway: Regularly varying responses with \(0<\alpha<2\) can produce stable rather than Gaussian sum limits. In an adaptive trial, the correct limit also depends on centering, consistent tail normalization, and whether each arm's sample size grows sufficiently regularly.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. If \(c_+=3\), \(c_-=1\), and \(mP(|X|>b(m))\to1\), what positive-tail weight appears in the normalized limit?
  2. For which of the three ranges \(0<\alpha<1\), \(\alpha=1\), and \(1<\alpha<2\) is mean-centering by \(m\mu\) the standard choice?
  3. Why is \(N_k(n)/n\to0\) insufficient to use a proportional-allocation rate for the sample mean in arm \(k\)?
  4. In a two-arm trial with independent response sequences and deterministic positive limiting allocation fractions, what feature of the arm-specific stable limits permits their combination for a contrast?

Solutions. (1) Here \(C=3+1=4\), so the positive-tail weight is \(c_+/C=3/4\); the limiting tail is \((3/4)x^{-\alpha}\). (2) For \(1<\alpha<2\), the mean exists and the standard centering is \(m\mu\). For \(\alpha<1\), no centering is used for the sum limit; at \(\alpha=1\), truncated-mean centering is standard. (3) A vanishing allocation fraction may mean the arm count grows more slowly than \(n\), changing the sum and sample-mean rates; the count must be analyzed on its own scale. (4) Under the stated independent-sequence conditions, the arm-specific limits are independent, so their appropriately scaled and signed sum gives the contrast limit.

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