Deriving the Survivor-Density Recursion

Why Propagate a Survivor Density?

At an interim analysis, the distribution of the statistic among paths that have continued is generally not its ordinary marginal distribution. Paths that crossed an earlier boundary have already stopped, so they must not contribute to later looks. A survivor-density recursion keeps track of this restriction as each new look is reached.

The Gaussian transition kernel describes how the statistic moves from one look to the next, conditional on its current value. The recursion combines that kernel with the density of paths that survived so far, integrating only over states that were allowed to continue. This gives the density of surviving paths at the new look, without first having to calculate a conditional density given survival.

Definition: Let \(Z_i\) be the test statistic at look \(i\), and let \(A_i\) be the continuation region at that look. A path survives through look \(i\) if \(Z_j\in A_j\) for every \(j=1,\ldots,i\). The survivor density \(f_i(y)\) is the subdensity of \(Z_i=y\) jointly with survival through look \(i\). Thus, for a measurable set \(B\), \(\int_B f_i(y)\,dy=P(Z_i\in B,\ Z_j\in A_j\text{ for }j=1,\ldots,i)\). Its total integral is the probability of survival through look \(i\), not 1 in general.

For two-sided stopping, a common continuation region is \(A_i=(l_i,u_i)\), where \(l_i\) and \(u_i\) are the lower and upper boundaries at look \(i\). Asymmetric designs allow unequal distances from zero or otherwise different lower and upper limits; the recursion does not require symmetry. A one-sided design can be represented by taking one endpoint to infinity. With continuous Gaussian distributions, including or excluding an endpoint has probability zero.

Assumptions and the Recursion

Suppose monitoring occurs at discrete looks \(1,\ldots,K\), and the process \(Z_1,\ldots,Z_K\) is Markov at those looks: conditional on \(Z_{i-1}\), the distribution of \(Z_i\) does not depend on the earlier states. Assume that, under the parameter value being evaluated, the transition density is Gaussian:

Z_i\mid Z_{i-1}=x\sim N(m_i(x),v_i),\qquad v_i>0,

where \(m_i(x)\) is the conditional mean and \(v_i\) is the conditional variance. Write its density as \(K_i(y\mid x)\). For this Gaussian model,

K_i(y\mid x)=\frac{1}{\sqrt{2\pi v_i}} \exp\left\{-\frac{(y-m_i(x))^2}{2v_i}\right\}.

The initial look has marginal density \(g_1(y)\). The starting survivor density and the recursion for later looks are

f_1(y)=\mathbf{1}_{A_1}(y)g_1(y), \qquad f_i(y)=\mathbf{1}_{A_i}(y)\int_{A_{i-1}} f_{i-1}(x)K_i(y\mid x)\,dx,\quad i=2,\ldots,K.

Here \(\mathbf{1}_{A_i}(y)\) equals 1 when \(y\in A_i\) and 0 otherwise. The previous survivor density already accounts for all restrictions through look \(i-1\); the integral is restricted to \(A_{i-1}\) because only states in that region can proceed. The indicator at the new look sets the density to zero at states where the process would stop.

Why the Recursion Holds

Let \(S_{i-1}\) denote survival through look \(i-1\). For a small interval around \(y\), paths that survive through look \(i\) must have both survived through look \(i-1\) and landed in \(A_i\) at the new look. Partition those paths according to their value \(x\) at look \(i-1\). The joint density of prior survival and \(Z_{i-1}=x\) is \(f_{i-1}(x)\), while the Markov assumption gives the conditional density \(K_i(y\mid x)\) for the next state. Multiplying and integrating over eligible prior states gives the new joint density:

\begin{aligned} P(S_{i-1},Z_i\in dy) &=\int_{A_{i-1}}P(S_{i-1},Z_{i-1}\in dx)\, P(Z_i\in dy\mid Z_{i-1}=x,S_{i-1})\\ &=\left[\int_{A_{i-1}}f_{i-1}(x)K_i(y\mid x)\,dx\right]dy. \end{aligned}

The second equality uses the Markov property: given \(Z_{i-1}=x\), survival history before that look adds no further information about the transition to \(Z_i\). Restricting the resulting density to \(y\in A_i\) yields the stated recursion. At the first look, survival is simply the event \(Z_1\in A_1\), which explains the initialization \(f_1(y)=\mathbf{1}_{A_1}(y)g_1(y)\).

Theorem: Under the stated Markov, transition-density, and discrete-monitoring assumptions, the recursion produces the joint density of the current state and survival through that look. In particular, \(f_i(y)\geq 0\), it is zero outside \(A_i\), and \(\int_{A_i} f_i(y)\,dy=P(S_i)\), where \(S_i\) is survival through look \(i\).

Worked Examples

Worked Example: Setting the Integration Region

Suppose the first two continuation regions are \(A_1=(-1,1)\) and \(A_2=(-1.5,1)\). Let \(Z_1\sim N(0,1)\), and suppose \(Z_2\mid Z_1=x\sim N(0.6x,0.64)\). Write the survivor density at look 2.

The look-1 density is the standard normal density \(\phi(x)\), restricted to the states that continued: \(f_1(x)=\mathbf{1}_{(-1,1)}(x)\phi(x)\). The transition variance is \(0.64\), so the transition standard deviation is \(\sqrt{0.64}=0.8\), and \(K_2(y\mid x)=\frac{1}{0.8}\phi((y-0.6x)/0.8)\).

The recursion first integrates over \(x\in(-1,1)\), then restricts the destination \(y\) to \(A_2\):

f_2(y)=\mathbf{1}_{(-1.5,1)}(y) \int_{-1}^{1}\phi(x)\frac{1}{0.8} \phi\left(\frac{y-0.6x}{0.8}\right)\,dx.

For example, at \(y=1.2\), the expression is zero because \(1.2\notin A_2\), regardless of the value of the integral. At \(y=0\), it is the integral over prior states from \(-1\) to \(1\); states outside that interval have already stopped and contribute nothing.

Worked Example: A Three-Look Density With a Survival Restriction

Consider standard-normal states with \(Z_2\sim N(0,1)\) and transition \(Z_3\mid Z_2=x\sim N(x/\sqrt{3},2/3)\). Suppose the process is monitored at look 2 with continuation region \(A_2=(-1,1)\), and at look 3 the state \(y=0\) is within its continuation region. Find \(f_3(0)\).

The density at look 2 after applying its boundary is \(f_2(x)=\mathbf{1}_{(-1,1)}(x)\phi(x)\). Therefore,

f_3(0)=\int_{-1}^{1}\phi(x) \frac{1}{\sqrt{2\pi(2/3)}}\exp\left\{-\frac{(0-x/\sqrt{3})^2}{2(2/3)}\right\}dx.

The joint distribution of \(Z_2,Z_3\) is standard bivariate normal with correlation \(1/\sqrt{3}\). Since both marginal variances are 1, \(Z_2\mid Z_3=0\sim N(0,1-1/3)=N(0,2/3)\), and the marginal density of \(Z_3\) at zero is \(\phi(0)\). Thus the integrand can be expressed as \(\phi(0)\) times the conditional density of \(Z_2\) given \(Z_3=0\). Consequently,

\begin{aligned} f_3(0) &=\phi(0)P(-1<Z_2<1\mid Z_3=0)\\ &=\phi(0)\left[ \Phi\left(\frac{1}{\sqrt{2/3}}\right) -\Phi\left(\frac{-1}{\sqrt{2/3}}\right)\right]\\ &\approx 0.39894(0.7793) \approx 0.3109. \end{aligned}

The density value \(0.3109\) is a subdensity at look 3, not the probability of surviving to that look. It incorporates the restriction at look 2 and, by assumption, the restriction at look 3. Its units are density per unit of the standardized statistic.

Worked Example: A Shifted Model and an Asymmetric Boundary

Suppose \(Z_1\sim N(0.2,1)\), and \(Z_2\) is independent of \(Z_1\) with \(Z_2\sim N(0.2,1)\). The continuation region at look 1 is \((-\infty,0.5)\), and at look 2 it is \((-1,1.5)\). Calculate the survivor density at \(y=0\).

Independence means the transition density is \(K_2(y\mid x)=\phi(y-0.2)\), regardless of \(x\). Since 0 lies inside the look-2 continuation region, the destination indicator is 1. The look-1 survivor mass is \(P(Z_1<0.5)=\Phi(0.5-0.2)=\Phi(0.3)\approx0.6179\). The transition density at zero is \(\phi(0-0.2)=\phi(0.2)\approx0.3910\) per unit of \(Z_2\). Hence

f_2(0)=\int_{-\infty}^{0.5} \frac{1}{\sqrt{2\pi}}e^{-(x-0.2)^2/2} \frac{1}{\sqrt{2\pi}}e^{-(0-0.2)^2/2}\,dx =\Phi(0.3)\phi(0.2) \approx 0.6179(0.3910) \approx 0.2416.

This is a subdensity at an allowed state, not a conditional density among survivors. The unequal look-2 limits cause no special change to the integral: the region is used to decide whether the destination is retained, while the earlier limit determines which prior states contribute.

Checks, Failure Modes, and Alternatives

Several checks follow directly from the definition. The propagated density must be nonnegative and supported only on the current continuation region. Its integral is the probability of surviving through the current look, so it cannot exceed the preceding survivor mass. To see the last fact, integrate the recursion over \(A_i\). Since a transition density integrates to 1 over the whole real line,

\int_{A_i} f_i(y)\,dy \leq \int_{A_{i-1}} f_{i-1}(x) \left[\int_{-\infty}^{\infty}K_i(y\mid x)\,dy\right]dx =\int_{A_{i-1}} f_{i-1}(x)\,dx.

The inequality holds because \(A_i\) is only part of the full destination state space. It provides a useful diagnostic: if computed survivor mass increases from one look to the next, check the limits, kernel normalization, and boundary restrictions.

Integrating over all previous states. Using \(\int_{-\infty}^{\infty} f_{i-1}(x)K_i(y\mid x)\,dx\) without restricting the integration to \(A_{i-1}\) can reintroduce paths that already stopped. Use \(A_{i-1}\) as the integration region.
Applying the boundary to the wrong variable. The integration restriction concerns the previous state \(x\); the indicator \(\mathbf{1}_{A_i}(y)\) concerns the new state \(y\). Both restrictions are needed.
Renormalizing after each look. Dividing \(f_i\) by its integral creates a density conditional on survival. That is a different quantity and no longer retains the probability mass lost to stopping.
Using an incorrect transition variance or mean. A kernel with the wrong variance can yield a plausible-looking but incorrect recursion. Verify that \(K_i(\cdot\mid x)\) integrates to 1 and matches the transition model under the parameter value being considered.
Interpreting discrete monitoring as continuous monitoring. This recursion accounts for stopping at scheduled looks only. A path can cross a threshold between looks and return before the next scheduled look; this method does not count that as a stop.

A direct calculation from the joint multivariate Gaussian distribution can be an alternative when the number of looks is small and the relevant joint probability or density is manageable. The recursive method instead carries forward the surviving state distribution one look at a time, which is useful when the monitoring sequence is longer. Simulation is another alternative: it approximates the same survival behavior by generating paths, but its results have sampling error. The recursion itself is exact for the specified Markov transition model; numerical integration used to evaluate it can introduce additional approximation error.

The key limitation is that the recursion depends on the model being correctly specified. If the state process is not Markov at the chosen looks, or if the transition kernel does not describe its conditional distribution, the factorization in the derivation need not hold. It also assumes that the continuation regions are defined at the monitored looks and that the Gaussian transition densities are appropriate under the parameter value of interest. These assumptions should be checked before treating a computed survivor density as a design result.

Key takeaway: At each look, multiply the previous survivor subdensity by the Gaussian transition kernel, integrate only over the previous continuation region, and set the new density to zero outside the current continuation region. Do not normalize away the probability of paths that have stopped.

Check Your Understanding

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

If \(f_{i-1}\) is already a survivor density, why is the recursion’s integral restricted to \(A_{i-1}\)?
Solution: The restriction explicitly excludes previous states that would have stopped at look \(i-1\). The survivor density is zero outside that region, but writing the domain as \(A_{i-1}\) makes the stopping rule explicit.
Suppose \(K_i(y\mid x)\) integrates to 1 over all \(y\). Why can the total survivor mass still decrease at look \(i\)?
Solution: The kernel’s total mass includes destinations both inside and outside \(A_i\). The recursion retains only destinations in \(A_i\), so \(\int_{A_i}f_i(y)\,dy\leq\int_{A_{i-1}}f_{i-1}(x)\,dx\).
What goes wrong if the computed \(f_i(y)\) is nonzero for a value \(y\notin A_i\)?
Solution: The current-look continuation restriction has not been applied correctly. The factor \(\mathbf{1}_{A_i}(y)\) must make the survivor density zero outside the current region.
Does this recursion count a boundary crossing that occurs between scheduled looks if the statistic returns inside the boundaries before the next look? Explain.
Solution: No. The recursion checks continuation at the specified looks only. Between-look crossing requires a monitoring model that explicitly accounts for behavior between those looks.