Why conditioning matters in adaptive trials
In an adaptive trial, the allocation history is random. Two trial paths can end with different proportions assigned to each arm, and those proportions can produce different asymptotic variances for the same estimator. Consequently, a statistic may be approximately normal conditional on the allocation history while having a non-normal distribution when the history is not conditioned upon.
This distinction is important for interpreting asymptotic results. A conditional normal approximation describes the sampling variation associated with a specified allocation environment. An unconditional approximation must also account for the fact that the environment itself varies across repetitions of the trial. Averaging over those environments generally produces a mixture of normal distributions, not one normal distribution with the average variance.
A random variance in this definition is not an ordinary fixed parameter. It represents variation across possible allocation histories. Conditional on a particular value of the limiting allocation environment, the normal component has a definite variance; unconditionally, the statistic averages over the distribution of those variances.
A conditional limit and its unconditional consequence
Here is a precise version of the averaging argument. Suppose \(V_n\) is a nonnegative quantity determined by the allocation history, \(V_n\) converges in distribution to \(V\), and, for each fixed real \(t\), the conditional characteristic function satisfies $$ E\left[e^{itT_n}\mid\mathcal A_n\right] - \exp\left(-\frac{t^2 V_n}{2}\right) \longrightarrow 0 $$ in probability. Assume also that \(V_n\) converges in distribution to a finite random variable \(V\). This conditional characteristic-function condition says that, given the history, \(T_n\) is asymptotically centered normal with variance \(V_n\). The normal component may change from one history to another.
The conditional error in the display is bounded in absolute value by 2. A bounded random variable that converges to zero in probability also converges to zero in mean absolute value. Taking expectations therefore shows that the unconditional characteristic function of \(T_n\) has the same limit as \(E[\exp(-t^2V_n/2)]\). The latter converges to \(E[\exp(-t^2V/2)]\), since \(v\mapsto\exp(-t^2v/2)\) is bounded and continuous on the nonnegative real line. Thus the unconditional limit has characteristic function \(E[\exp(-t^2V/2)]\), the characteristic function of the normal mixture.
This argument is a form of conditioning followed by averaging. It does not say that \(T_n\) is asymptotically normal unconditionally. It says that its limiting characteristic function is obtained by averaging the conditional normal characteristic functions over the limiting distribution of the variance. Equivalently, the limit can be represented as \(\sqrt V Z\), where \(Z\sim N(0,1)\) is independent of \(V\).
The conditions matter. In particular, the conditional approximation must hold for the histories relevant to the limit, and \(V_n\) must have a well-defined limiting distribution. If \(V\) can equal zero, the mixture includes a point mass at zero for those histories; many standard normal-limit results instead require \(V>0\) with probability one. If variances can grow without control or fail to converge in distribution, the stated averaging conclusion need not apply.
Worked Example: Random allocation fractions create a variance mixture
Worked Example: Random allocation fractions create a variance mixture
Consider a hypothetical two-arm trial with \(n\) total participants. Let \(\hat\mu_A\) and \(\hat\mu_B\) be the sample means, and consider \(T_n=\sqrt n\{(\hat\mu_A-\hat\mu_B)-(\mu_A-\mu_B)\}\). Suppose the response variances are \(\sigma_A^2=1\) and \(\sigma_B^2=4\). Assume the allocation fraction to arm A converges to a random \(Q\), which equals \(0.25\) or \(0.75\) with probability \(1/2\) each. For this example, assume each arm receives a positive limiting fraction and the conditional central limit approximation holds given the allocation history.
Given a limiting fraction \(Q=q\), the asymptotic variance of the scaled difference in sample means is $$ V(q)=\frac{\sigma_A^2}{q}+\frac{\sigma_B^2}{1-q}. $$ For \(q=0.25\), substitution gives \(V(0.25)=1/0.25+4/0.75=4+5.333=9.333\). For \(q=0.75\), it gives \(V(0.75)=1/0.75+4/0.25=1.333+16=17.333\).
Thus, conditional on the limiting allocation environment, the limit is \(N(0,9.333)\) for one type of history and \(N(0,17.333)\) for the other. Unconditionally, the limit is the equally weighted mixture $$ \frac12N(0,9.333)+\frac12N(0,17.333). $$ Its variance is \(E[V]=\frac12(9.333)+\frac12(17.333)=13.333\). That variance summarizes the mixture's second moment, but does not make the mixture \(N(0,13.333)\). For example, its characteristic function is \(\frac12 e^{-9.333t^2/2}+\frac12 e^{-17.333t^2/2}\), not \(e^{-13.333t^2/2}\) in general.
The assumptions used here are explicit: both possible limiting allocation fractions lie strictly between 0 and 1, and the conditional Gaussian approximation is assumed to hold in each environment. The calculation would not justify a Gaussian limit if an arm's limiting fraction were zero or if the conditional approximation failed.
Moments and tail probabilities of a mixture
For a centered normal mixture with finite \(E[V]\), the law of total expectation gives \(E[\sqrt V Z]=0\). Independence of \(Z\) and \(V\), along with \(E[Z^2]=1\), gives $$ \operatorname{Var}(\sqrt V Z)=E[VZ^2]=E[V]. $$ This is the variance decomposition for this centered mixture: the conditional mean is zero, so there is no additional between-history variance of conditional means. If conditional means are random and nonzero, the unconditional variance also includes the variance of those means, and centering only by an overall target may not produce the centered mixture above.
Although the variance averages simply, other distributional features do not. For any \(c>0\), the unconditional two-sided tail probability is $$ P(|\sqrt V Z|>c)=E\left[2\left\{1-\Phi\left(\frac{c}{\sqrt V}\right)\right\}\right], $$ with the zero-variance case interpreted as a statistic equal to zero. This expression averages tail probabilities, not standard deviations or quantiles. A history with a large variance can contribute substantial tail probability even if it occurs infrequently.
Worked Example: Compare a mixture tail with an averaged-variance normal
Let \(V=1\) or \(V=4\), each with probability \(1/2\), and let the conditional limit be \(N(0,V)\). Calculate the unconditional probability that the absolute value of the limit exceeds 2, then compare it with the probability from a normal distribution having the same unconditional variance.
Conditional on \(V=1\), the tail probability is \(P(|Z|>2)=2\{1-\Phi(2)\}\approx 0.0455\). Conditional on \(V=4\), the limit is \(2Z\), so the tail probability is \(P(|Z|>1)=2\{1-\Phi(1)\}\approx 0.3173\). Averaging gives $$ P(|\sqrt V Z|>2)=\tfrac12(0.0455)+\tfrac12(0.3173)=0.1814. $$
The unconditional variance is \(E[V]=\tfrac12(1)+\tfrac12(4)=2.5\). A single normal distribution with that variance would give \(P(|N(0,2.5)|>2)=2\{1-\Phi(2/\sqrt{2.5})\}\approx 2\{1-\Phi(1.265)\}\approx 0.206\). It has the correct variance but a different tail probability. The mixture has less probability beyond 2 in this particular calculation; the direction and size of differences depend on the threshold and the mixing distribution.
Why a normal distribution with average variance can mislead
A mixture of centered normals is itself a single centered normal only in the special case that its variance is constant with probability one. To see why, suppose its characteristic function equals that of \(N(0,c)\). Then \(E[e^{-sV}]=e^{-sc}\) for all \(s\geq 0\), taking \(s=t^2/2\). Differentiating at \(s=0\), when the relevant moments are finite, gives \(E[V]=c\) and \(E[V^2]=c^2\). Hence \(\operatorname{Var}(V)=0\), so \(V=c\) with probability one. Thus a genuinely varying conditional variance generally produces a genuinely non-normal unconditional limit.
This distinction can affect uncertainty statements. A confidence interval justified by a fixed-variance conditional approximation may have a coverage probability that depends on the allocation history. Averaging those conditional coverages gives unconditional coverage, but it need not equal the nominal level. Using \(E[V]\) in a single normal interval is not automatically a remedy.
Worked Example: Coverage of a normal interval using the average variance
Suppose, hypothetically, that a centered limiting statistic has conditional variance 1 with probability \(1/2\) and 9 with probability \(1/2\). A common shortcut is to use the unconditional average variance \(E[V]=5\) and the nominal 95% normal cutoff \(1.96\sqrt 5\approx 4.383\). Calculate the actual limiting coverage of the interval \([-4.383,4.383]\).
When \(V=1\), coverage is \(P(|Z|\leq 4.383)\), which is effectively 1 to three significant figures. When \(V=9\), the conditional standard deviation is 3, so coverage is \(P(|Z|\leq 4.383/3)=P(|Z|\leq 1.461)\approx 0.856\). Therefore the unconditional coverage is $$ \tfrac12(1.000)+\tfrac12(0.856)=0.928. $$
The interval uses the mixture's variance but has limiting coverage of about 92.8%, not 95%. This calculation does not establish that every average-variance interval undercovers; it shows why a nominal level cannot be inferred from the average variance alone. In practice, the relevant limiting distribution or a suitable conditional calibration must be justified.
Diagnostics, alternatives, and limitations
Several checks help determine whether a conditional-to-unconditional calculation is credible:
- Check the conditioning information. State exactly what is included in \(\mathcal A_n\), and verify that the proposed conditional variance is determined by that information.
- Check the limiting allocation environment. Confirm that the allocation fractions have a limiting distribution and that the variance formula is valid in the environments receiving positive probability.
- Check centering and conditional means. If conditional means have nonzero history-dependent limits, a centered normal mixture may be inappropriate; the unconditional limit can instead mix normal distributions with different means as well as different variances.
- Check variance degeneracy or instability. If an arm's allocation fraction approaches zero, the usual arm-specific variance expression may diverge or the standard conditional normal approximation may fail. The mixture result does not repair a failed conditional limit.
- Compare distributions, not just variances. Examine tail probabilities or quantiles under the mixture, not only \(E[V]\). Simulations across repeated trial histories can reveal whether the allocation environments materially change uncertainty.
The main alternative descriptions answer different questions. A conditional normal approximation is useful when inference is explicitly calibrated to a realized allocation environment and its variance can be estimated appropriately. An unconditional mixture describes repeated trials in which both outcomes and allocation histories vary. A single normal approximation with averaged variance is simpler, but it needs separate justification for the distributional feature of interest; matching the second moment alone is insufficient.
The result also has limits. It relies on a valid conditional limit and suitable convergence of the random variance; it does not prove either. It describes asymptotic behavior, not necessarily the finite-sample distribution. Further, if the target itself shifts with the trial size, centering must match the relevant local target before applying a conditional limit. Finally, these assumptions are pointwise for the stated setting; they do not establish uniform validity over trial parameters.
Check Your Understanding
Answer each question, then compare with the solutions below.
- If the conditional limit is \(N(0,V)\), with \(V=2\) or \(V=8\) equally likely, what is the unconditional variance? Is the unconditional limit necessarily \(N(0,5)\)?
- For a centered normal mixture, write the unconditional two-sided tail probability at a positive threshold \(c\) in terms of \(V\) and \(\Phi\).
- Suppose an allocation history determines the conditional variance, but the conditional approximation has a nonzero history-dependent mean. Why might a centered variance mixture be inadequate?
- A proposed variance formula contains \(1/Q\), where \(Q\) is the limiting fraction assigned to an arm. What issue arises if \(Q=0\) with positive probability?
Solutions. 1. The variance is \(E[V]=\tfrac12(2)+\tfrac12(8)=5\). The limit is generally a mixture of \(N(0,2)\) and \(N(0,8)\), not \(N(0,5)\), because the conditional variances differ. 2. It is \(E[2\{1-\Phi(c/\sqrt V)\}]\), with a zero-variance component treated as a point mass at zero. 3. The unconditional distribution must average both the history-dependent means and variances; centering by one overall target does not necessarily leave zero conditional means. 4. The variance expression is undefined or divergent at \(Q=0\), and the usual conditional Gaussian approximation may fail because that arm is not represented by a positive limiting fraction. The standard mixture conclusion cannot be applied without a different argument.