Gaussian Transition Kernels Between Looks

Why a Transition Kernel Is Needed

A group sequential design observes a test statistic at ordered information times. To calculate how the statistic can move from one look to the next, it is not enough to know the marginal distribution at each look: the looks are dependent. The Gaussian transition kernel describes the distribution at a later look conditional on the statistic’s value at an earlier look. It is the local probability model needed to carry a numerical state forward.

The earlier material on representing interim looks as numerical states treats a state as a value of the canonical statistic at a specified look. Here, the state at the earlier time is denoted $x$, and the state at the later time is denoted $y$. The kernel is a conditional density in $y$, not a probability of crossing a boundary by itself.

Canonical Model and Assumptions

Let $0<t_1<\cdots<t_K=1$ be the planned information times, expressed as fractions of the information at the planned final analysis. Write $Z_i$ for the canonical test statistic at time $t_i$. The parameter $\theta$ is the standardized effect at the planned final information; under the null hypothesis, $\theta=0$. The canonical model can be represented as

Z_i=\frac{\theta t_i+B(t_i)}{\sqrt{t_i}},

where $B(t)$ is standard Brownian motion: it has mean zero, variance $t$, and independent normal increments over disjoint time intervals. Thus $Z_i\sim N(\theta\sqrt{t_i},1)$. For $t_i<t_j$, the correlation between $Z_i$ and $Z_j$ is $\sqrt{t_i/t_j}$, so the two statistics are generally not independent.

Definition: For two planned information times $0<t_i<t_j\leq1$, the Gaussian transition kernel $k_{ij}(y\mid x;\theta)$ is the conditional density of $Z_j=y$, given $Z_i=x$, under the canonical model with effect parameter $\theta$. It has conditional mean $$ m_{ij}(x;\theta)=\sqrt{\frac{t_i}{t_j}}\,x+\theta\frac{t_j-t_i}{\sqrt{t_j}} $$ and conditional variance $$ v_{ij}=\frac{t_j-t_i}{t_j}. $$

These formulas assume ordered, deterministic information times and the canonical joint-normal model. In applications, that model may be exact or a normal approximation to the standardized score process. The same effect parameter $\theta$ is used across the interval. If the sampling plan or data-generating process changes in a way that invalidates the canonical joint distribution, the kernel may no longer describe the conditional distribution correctly.

Deriving the Conditional Distribution

The Brownian representation separates the information accumulated before $t_i$ from the increment between $t_i$ and $t_j$. In particular, $B(t_j)=B(t_i)+\{B(t_j)-B(t_i)\}$, and the increment in braces is independent of $B(t_i)$, with distribution $N(0,t_j-t_i)$. Substituting this decomposition into $Z_j$ gives

\begin{aligned} Z_j &=\frac{\theta t_j+B(t_i)+\{B(t_j)-B(t_i)\}}{\sqrt{t_j}}\\ &=\sqrt{\frac{t_i}{t_j}}Z_i+ \frac{\theta(t_j-t_i)+\{B(t_j)-B(t_i)\}}{\sqrt{t_j}}. \end{aligned}

Conditional on $Z_i=x$, the first term is fixed. The remaining Brownian increment is independent of $Z_i$. Its contribution has mean zero and variance $(t_j-t_i)/t_j$; the effect contribution is fixed. Therefore,

Z_j\mid Z_i=x\sim N\left( \sqrt{\frac{t_i}{t_j}}\,x+ \theta\frac{t_j-t_i}{\sqrt{t_j}}, \frac{t_j-t_i}{t_j} \right).

The second argument of $N$ is the variance. A normal density with mean $m$ and variance $v$ is $(2\pi v)^{-1/2}\exp\{-(y-m)^2/(2v)\}$. Applying this formula yields the transition kernel:

k_{ij}(y\mid x;\theta) = \frac{1}{\sqrt{2\pi v_{ij}}} \exp\left\{ -\frac{[y-m_{ij}(x;\theta)]^2}{2v_{ij}} \right\}, \qquad v_{ij}=\frac{t_j-t_i}{t_j}.

Equivalently, the conditional standard deviation is $\sqrt{(t_j-t_i)/t_j}$. The kernel is a density with respect to the later statistic $y$: its units are inverse units of $Z_j$, and its integral over the real line is one. To obtain a conditional probability, integrate the density over the relevant set of later states.

Calculating a Transition Density

Worked Example: Density Under the Null

Suppose two hypothetical looks occur at $t_i=0.36$ and $t_j=0.64$. At the first look, the observed state is $x=1.20$. Calculate the density of the later statistic at $y=1.50$ under $H_0$.

Under $H_0$, $\theta=0$. The conditional mean and variance are

m_{ij}= \sqrt{\frac{0.36}{0.64}}(1.20)=0.75(1.20)=0.900, \qquad v_{ij}=\frac{0.64-0.36}{0.64}=0.4375.

Substitute these values into the normal density:

\begin{aligned} k_{ij}(1.50\mid1.20;0) &=\frac{1}{\sqrt{2\pi(0.4375)}} \exp\left\{-\frac{(1.50-0.900)^2}{2(0.4375)}\right\}\\ &=\frac{1}{\sqrt{2.7489}}\exp(-0.4114)\\ &\approx 0.3998. \end{aligned}

This is a density value, not the probability that $Z_j$ equals 1.50; for a continuous variable, the probability of one exact value is zero. A probability would require an interval, such as integrating the kernel over a range of $y$-values.

Using the Kernel for a Conditional Probability

The kernel can also answer a boundary-relevant question once a boundary is specified. The conditional probability of a later statistic exceeding an upper threshold $u$, given the current state $x$, is the upper tail of its conditional normal distribution. This probability is conditional on the stated earlier state; it is not the unconditional probability of crossing over the whole trial.

Worked Example: Conditional Probability of Exceeding a Threshold

In a hypothetical design, $t_i=0.25$, $t_j=0.75$, and the current state is $x=0.80$. Suppose the effect parameter is $\theta=0.50$, and the later upper threshold is $u=1.50$. Calculate $P(Z_j\geq1.50\mid Z_i=0.80)$.

The conditional mean and variance are

\begin{aligned} m_{ij} &=\sqrt{\frac{0.25}{0.75}}(0.80) +0.50\frac{0.75-0.25}{\sqrt{0.75}}\\ &\approx 0.4619+0.2887=0.7506,\\ v_{ij}&=\frac{0.75-0.25}{0.75}=0.6667. \end{aligned}

The conditional standard deviation is $\sqrt{0.6667}\approx0.8165$. Standardizing the threshold gives

z=\frac{1.50-0.7506}{0.8165}\approx0.9178.

Consequently,

P(Z_j\geq1.50\mid Z_i=0.80) =1-\Phi(0.9178)\approx0.179,

where $\Phi$ is the standard normal cumulative distribution function. Under this model and effect, conditional on the specified current state, the later statistic has about a 17.9% probability of being at least 1.50. This is not the trial-wide probability of crossing, which would also depend on the distribution of the current state and on the stopping rule.

Effect Shifts and Direction

The effect parameter shifts the conditional mean but does not change the conditional variance under this canonical model. A positive $\theta$ adds $\theta(t_j-t_i)/\sqrt{t_j}$ to the null conditional mean; a negative $\theta$ shifts it downward. The variance depends only on the information increment relative to the later information time. Thus, holding the times fixed, different values of $\theta$ move the kernel’s center without changing its spread.

Worked Example: A Negative Effect and a Later-State Density

Consider hypothetical looks at $t_i=0.40$ and $t_j=0.90$. Let the observed state be $x=-1.10$, the effect parameter be $\theta=-0.30$, and evaluate the density at $y=-0.50$.

The conditional mean is

\begin{aligned} m_{ij} &=\sqrt{\frac{0.40}{0.90}}(-1.10) +(-0.30)\frac{0.90-0.40}{\sqrt{0.90}}\\ &\approx -0.7333-0.1581=-0.8914. \end{aligned}

The conditional variance is

v_{ij}=\frac{0.90-0.40}{0.90}=0.5556.

Using the mean and variance in the normal density,

\begin{aligned} k_{ij}(-0.50\mid-1.10;-0.30) &=\frac{1}{\sqrt{2\pi(0.5556)}} \exp\left\{-\frac{[-0.50-(-0.8914)]^2}{2(0.5556)}\right\}\\ &\approx \frac{1}{1.868}\exp(-0.1379)\\ &\approx 0.466. \end{aligned}

The negative effect shifts the conditional mean below the contribution from the earlier state alone. The computed value is the density at $-0.50$; it does not indicate a 46.6% probability at that exact statistic value.

Checks, Alternatives, and Limitations

Before using a transition kernel, check the order and scale of the information times, the specified effect, and the role of the result. The following relationships provide useful diagnostics:

For $t_i<t_j$, the variance $(t_j-t_i)/t_j$ must be positive and less than one.
As $t_j$ approaches $t_i$, the conditional variance approaches zero and the conditional mean approaches $x$ when the time interval vanishes.
Under the null, the effect-shift term disappears. Under a nonzero effect, its sign must agree with the sign of $\theta$.
For fixed $x$, integrating $k_{ij}(y\mid x;\theta)$ over all real $y$ must give one.

A common mistake is to use $t_i/t_j$ as the conditional variance because it appears in the correlation. That quantity is the square of the correlation, not the conditional variance. The correct calculation is $1-t_i/t_j=(t_j-t_i)/t_j$. Another mistake is to treat the two looks as independent and use the marginal distribution of $Z_j$ after observing $Z_i=x$. Conditioning matters: the earlier state contributes $\sqrt{t_i/t_j}\,x$ to the later conditional mean.

The kernel is one way to represent dependence between looks. A joint multivariate normal calculation can instead represent all planned statistics together; it is useful when working directly with the joint distribution, but it does not remove the need to account for dependence. Simulation can approximate transition probabilities and accommodate more complicated data-generating mechanisms, though its numerical accuracy depends on the simulation design and number of replicates. The canonical kernel is especially convenient when the normal independent-increment assumptions are appropriate and a calculation needs the conditional distribution for a specified pair of looks.

The model’s limitations matter. The formula relies on the canonical normal process and the information times used in its derivation. Material departures from normality, changes to the sampling process, or information times that are not fixed as assumed can make the stated conditional law inaccurate. Also, the kernel alone does not specify which states continue or stop: boundary rules determine that. It supplies the transition density between states, rather than the complete sequential decision calculation.

Key takeaway: For ordered information times, condition on the earlier state, shift the later mean by the earlier state and the effect increment, and use variance $(t_j-t_i)/t_j$. Keep the distinction between a density, a conditional probability, and an overall crossing probability explicit.

Check Your Understanding

Use the canonical model and show how the conditional mean, variance, or probability follows from it.

Under $H_0$, let $t_i=0.20$, $t_j=0.80$, and $Z_i=1.00$. Find the conditional mean and variance of $Z_j$.
For the same times, suppose $\theta=0.40$. What changes in the conditional distribution?
Why is $k_{ij}(y\mid x;\theta)$ not the probability that $Z_j=y$?
If a calculation gives a negative conditional variance for $t_i<t_j$, name a likely error.

Solutions. (1) The mean is $\sqrt{0.20/0.80}(1.00)=0.500$, and the variance is $(0.80-0.20)/0.80=0.750$. (2) The effect adds $0.40(0.60)/\sqrt{0.80}\approx0.2683$ to the mean, giving approximately $0.7683$; the variance remains $0.750$. (3) It is a density over a continuous variable, so the probability of one exact value is zero; probabilities require integration over a set. (4) Check that the times are correctly ordered and that the variance is calculated as $(t_j-t_i)/t_j$, not with the difference reversed.