Deriving the Conditional Distribution from Independent Increments

Why Work With Score Increments?

At an interim analysis, the observed statistic summarizes information collected so far. To describe what remains, it is useful to separate the score accumulated by that time from the score that will be added later. Under the canonical joint normal model, this separation is especially simple: the future score increment is independent of the interim score. Its distribution can therefore be written down directly, and the conditional distribution of the final statistic follows by adding the observed interim score.

The distinction between the unstandardized score increment and the final standardized statistic matters. The increment has a conditional distribution that does not depend on the observed interim score. The final statistic does depend on that score, because it includes the score already accumulated.

Model and Notation

Consider an interim analysis at information $I_1$ and a later analysis at information $I_2$, where $0<I_1<I_2$. Write $Z_1$ and $Z_2$ for the standardized statistics at those analyses. In the canonical model, under a fixed standardized effect $\theta$,

E(Z_j)=\theta\sqrt{I_j}, \qquad \operatorname{Var}(Z_j)=1, \qquad \operatorname{Cov}(Z_1,Z_2)=\sqrt{\frac{I_1}{I_2}}.

The pair $(Z_1,Z_2)$ is jointly normal. The parameter $\theta$ is the effect per square root of information; positive $\theta$ is taken to favor the direction of larger $Z$ statistics. Information is assumed known, and the effect is treated as fixed over the two analyses. These assumptions describe the canonical approximation; they need not be exact for every endpoint or analysis method.

Define the unstandardized score at information $I_j$ as $S_j=\sqrt{I_j}Z_j$. The model implies $E(S_j)=\theta I_j$ and $\operatorname{Var}(S_j)=I_j$. The score added between analyses is

\Delta S=S_2-S_1=\sqrt{I_2}Z_2-\sqrt{I_1}Z_1.

Here $\Delta I=I_2-I_1$ denotes the information increment. The result below concerns the distribution of $\Delta S$ given the observed interim score, or equivalently the observed interim statistic.

Deriving the Conditional Distribution

The score increment is a linear combination of jointly normal variables, so it is normal jointly with $Z_1$. Its mean and variance follow from the model:

E(\Delta S)=\theta(I_2-I_1)=\theta\Delta I, \qquad \operatorname{Var}(\Delta S) =I_2+I_1-2\sqrt{I_1I_2}\sqrt{\frac{I_1}{I_2}} =I_2-I_1=\Delta I.

To determine whether conditioning on $Z_1$ changes this distribution, calculate the covariance:

\operatorname{Cov}(\Delta S,Z_1) =\sqrt{I_2}\operatorname{Cov}(Z_2,Z_1) -\sqrt{I_1}\operatorname{Var}(Z_1) =\sqrt{I_1}-\sqrt{I_1}=0.

Because $\Delta S$ and $Z_1$ are jointly normal, zero covariance implies independence. This implication relies on joint normality; zero covariance alone would not establish independence for arbitrary distributions. Independence means that learning $Z_1=z_1$ does not alter the distribution of $\Delta S$.

Theorem: Under the canonical joint normal model, for any observed interim value $Z_1=z_1$, the future score increment has conditional distribution $$ \Delta S\mid Z_1=z_1\sim N(\theta\Delta I,\Delta I). $$ Thus its conditional mean is $\theta\Delta I$, and its conditional variance is $\Delta I$. These are also its unconditional mean and variance.

The final standardized statistic is $Z_2=(S_1+\Delta S)/\sqrt{I_2}$. Once $Z_1=z_1$ is observed, $S_1=\sqrt{I_1}z_1$ is fixed. Adding this fixed value to the random increment and rescaling gives

Z_2\mid Z_1=z_1 \sim N\left( \frac{\sqrt{I_1}z_1+\theta\Delta I}{\sqrt{I_2}}, \frac{\Delta I}{I_2} \right).

The conditional mean can also be written as $\theta\sqrt{I_2}+\sqrt{I_1/I_2}(z_1-\theta\sqrt{I_1})$. This form makes clear that the final mean starts at its model mean and adjusts according to how far the interim observation lies from its model mean. The conditional variance is $1-I_1/I_2$, which is positive because $I_2>I_1$.

Equivalently, the standardized future increment $U=\Delta S/\sqrt{\Delta I}$ satisfies $U\mid Z_1=z_1\sim N(\theta\sqrt{\Delta I},1)$. This is the new, standardized information supplied after the interim analysis. It should not be confused with $Z_2$, which combines that new information with the already observed score.

Worked Example: A Future Score Increment and Final Statistic

Suppose a hypothetical trial has $I_1=40$, $I_2=100$, and observes $Z_1=1.20$. For illustration, take the fixed effect to be $\theta=0.25$. Find the conditional distribution of the future score increment and of $Z_2$.

The information increment is $\Delta I=100-40=60$. Therefore

\Delta S\mid Z_1=1.20 \sim N(0.25(60),60)=N(15,60).

So the conditional mean of the score added after the interim analysis is 15 score units and its conditional standard deviation is $\sqrt{60}=7.746$ score units. The interim value does not enter this increment distribution.

For the final statistic, the observed score already accumulated is $\sqrt{40}(1.20)=7.589$. Thus

E(Z_2\mid Z_1=1.20) =\frac{7.589+15}{10}=2.259, \qquad \operatorname{Var}(Z_2\mid Z_1=1.20) =\frac{60}{100}=0.600.

Consequently, $Z_2\mid Z_1=1.20\sim N(2.259,0.600)$, with conditional standard deviation $\sqrt{0.600}=0.7746$. The mean incorporates both the observed interim score and the expected future score.

Worked Example: Conditioning Under the Null Effect

Consider a hypothetical trial with $I_1=25$, $I_2=64$, observed $Z_1=-0.40$, and $\theta=0$. Calculate the increment distribution and the conditional final distribution.

Here $\Delta I=64-25=39$, so under the null effect,

\Delta S\mid Z_1=-0.40\sim N(0(39),39)=N(0,39).

The interim score is $\sqrt{25}(-0.40)=-2.00$. Since $\sqrt{64}=8$, the conditional final mean and variance are

E(Z_2\mid Z_1=-0.40)=\frac{-2.00+0}{8}=-0.250, \qquad \operatorname{Var}(Z_2\mid Z_1=-0.40)=\frac{39}{64}=0.6094.

Thus $Z_2\mid Z_1=-0.40\sim N(-0.250,0.6094)$. The final mean is not zero, despite setting $\theta=0$, because it includes the observed negative interim score. The future increment itself remains centered at zero under this assumed effect.

Worked Example: Standardizing the Remaining Increment

In a hypothetical trial, let $I_1=60$, $I_2=90$, $Z_1=2.00$, and $\theta=0.20$. Find the distribution of the standardized future increment $U=\Delta S/\sqrt{\Delta I}$, then recover the conditional distribution of $Z_2$.

The remaining information is $\Delta I=90-60=30$. Therefore

U\mid Z_1=2.00 \sim N(0.20\sqrt{30},1)=N(1.095,1).

In score units, $\Delta S\mid Z_1=2.00\sim N(0.20(30),30)=N(6,30)$. The observed interim score is $\sqrt{60}(2.00)=15.492$, so

E(Z_2\mid Z_1=2.00) =\frac{15.492+6}{\sqrt{90}} =\frac{21.492}{9.487}=2.265, \qquad \operatorname{Var}(Z_2\mid Z_1=2.00) =\frac{30}{90}=0.3333.

The conditional standard deviation of $Z_2$ is $\sqrt{0.3333}=0.5774$. The observed interim evidence contributes to the conditional mean, while the standardized future increment has unit variance by construction.

How to Check the Result

A useful way to verify a calculation is to derive the final conditional law directly from the bivariate normal formula. For jointly normal variables, the conditional mean of $Z_2$ given $Z_1=z_1$ is $E(Z_2)+\operatorname{Cov}(Z_2,Z_1)\{z_1-E(Z_1)\}/\operatorname{Var}(Z_1)$, and its conditional variance is $\operatorname{Var}(Z_2)-\operatorname{Cov}(Z_2,Z_1)^2/\operatorname{Var}(Z_1)$. Substituting the canonical model gives

E(Z_2\mid Z_1=z_1) =\theta\sqrt{I_2} +\sqrt{\frac{I_1}{I_2}}(z_1-\theta\sqrt{I_1}) =\frac{\sqrt{I_1}z_1+\theta\Delta I}{\sqrt{I_2}}, \qquad \operatorname{Var}(Z_2\mid Z_1=z_1) =1-\frac{I_1}{I_2} =\frac{\Delta I}{I_2}.

This agrees with the increment derivation. The direct bivariate formula is a compact check; the increment representation explains why the result has this form and identifies the new information separately. Both approaches require the stated joint normal model and the same information covariance.

Several diagnostics follow from the formulas. If $I_2$ is only slightly larger than $I_1$, then $\Delta I/I_2$ is small, so the conditional final statistic has little remaining uncertainty. If $I_2$ is much larger, that conditional variance is larger, though it remains below 1. A computed variance outside the range $0<\Delta I/I_2<1$ signals an information-ordering or arithmetic error. The increment variance should equal $\Delta I$, not $I_2$: it describes only information collected after the interim.

Assumptions, Limitations, and a Common Mistake

The result depends on a fixed-effect canonical normal approximation, correctly specified information levels, and the canonical covariance structure. It is appropriate when the standardized statistics have the stated joint normal law and information is treated as known. If the variance is estimated, the endpoint is discrete, information is misspecified, or the effect changes over time, the exact conditional law may differ. In such settings, the model may need adjustment or the conditional distribution may require a different approximation.

If continuation is determined by the interim observation, conditioning on a particular observed $Z_1$ still leaves the future increment independent of $Z_1$ under this model. This statement does not mean that every post-stopping analysis has the same distribution as an analysis planned in advance; it concerns the future increment in the canonical model when future data are in fact collected. Nor does it remove the need to account for the design when interpreting estimates or intervals after a stopping decision.

Common mistake: “The future increment has mean $\theta\sqrt{I_2}$, and its distribution depends on the interim Z statistic.” This confuses the mean of the final standardized statistic with the mean of the unstandardized increment.
Correction: The increment has mean $\theta\Delta I$, variance $\Delta I$, and is independent of $Z_1$. The final statistic adds the observed score $\sqrt{I_1}z_1$ and divides by $\sqrt{I_2}$.
Another warning: Zero covariance implies independence here because the variables are jointly normal. Do not use that conclusion from covariance alone in a nonnormal model.

Key takeaway: Under the canonical joint normal model, $\Delta S=\sqrt{I_2}Z_2-\sqrt{I_1}Z_1$ is independent of the interim statistic and has distribution $N(\theta\Delta I,\Delta I)$. The conditional law of the final statistic follows by adding the observed interim score and rescaling.

Check Your Understanding

Use the canonical model. In each question, information is known and the effect is fixed. Solutions are included below the questions.

For $I_1=16$, $I_2=49$, and $\theta=0.10$, what are the conditional mean and variance of the score increment, given any observed $Z_1$?
With $I_1=36$, $I_2=100$, $Z_1=1.00$, and $\theta=0$, calculate the conditional mean and variance of $Z_2$.
Why does the conditional distribution of $\Delta S$ not depend on the observed interim statistic?
If $I_1=50$ and $I_2=45$, can the stated two-look model be used as written? Explain.

Solutions. (1) $\Delta I=49-16=33$, so $\Delta S\mid Z_1\sim N(0.10(33),33)=N(3.30,33)$; its conditional mean is 3.30 score units and its variance is 33 score units squared. (2) $\Delta I=64$, and the interim score is $\sqrt{36}(1.00)=6$. Thus the conditional mean is $(6+0)/\sqrt{100}=0.600$, and the variance is $64/100=0.640$. (3) The increment and $Z_1$ are jointly normal with covariance zero, so they are independent; conditioning on $Z_1$ therefore does not change the increment distribution. (4) No. The later information must exceed the interim information, but $45<50$, giving a negative information increment and violating the assumed ordering.