Gauss-Legendre Quadrature for Boundary Integrals

Why Integrate a Transition Density Over an Interval?

In a group sequential calculation, a state at one analysis determines a distribution for the state at the next analysis. To find the probability that the next state lies between two finite limits, we integrate the transition density over that interval. The integral is usually simple to define but may need numerical evaluation when a closed-form expression is unavailable or when the integrand is part of a larger calculation.

Gauss-Legendre quadrature approximates a definite integral by evaluating its integrand at a small, carefully chosen set of points. Its finite-interval form is especially convenient when the state limits are already known. The method applies directly to a one-dimensional transition probability; it does not, by itself, calculate a complete sequential crossing probability.

Definition: Suppose the next state $Y$, conditional on the current state $X=x$, has a continuous density $f(y\mid x)$. For finite limits $a<b$, the transition probability of landing in the interval $[a,b]$ is $$ P(a\leq Y\leq b\mid X=x)=\int_a^b f(y\mid x)\,dy. $$ If the transition is Gaussian, write $Y\mid X=x\sim N(m(x),v)$, where $m(x)$ is the conditional mean and $v>0$ is the conditional variance. Its density is $$ f(y\mid x)=\frac{1}{\sqrt{2\pi v}}\exp\!\left(-\frac{(y-m(x))^2}{2v}\right). $$

The calculation assumes that the stated conditional distribution is an appropriate transition model, that $v$ is positive, and that $a$ and $b$ correctly describe the event of interest. If the process is represented as Markovian, the transition distribution must also depend on the past only through the current state (and any other variables explicitly included in the state). These are model assumptions, not consequences of the quadrature rule.

Transforming the Interval and Applying the Rule

The standard Gauss-Legendre rule integrates on $[-1,1]$. To use it on $[a,b]$, set $y=c+ht$, where $c=(a+b)/2$ is the interval midpoint and $h=(b-a)/2$ is its half-width. Then $dy=h\,dt$, and

\int_a^b f(y\mid x)\,dy = h\int_{-1}^{1} f(c+ht\mid x)\,dt \approx h\sum_{i=1}^{n} w_i f(c+ht_i\mid x).

Here $t_i$ are the standard nodes and $w_i$ their weights. The half-width factor $h$ is essential: it accounts for the change in scale from the standard interval to the actual state interval. For two nodes, $t_i=\pm1/\sqrt{3}$ and each weight is 1. For three nodes, the nodes are $0$ and $\pm\sqrt{3/5}$, with weight $8/9$ at 0 and $5/9$ at each outer node.

Key result: An $n$-node Gauss-Legendre rule is exact when the transformed integrand is a polynomial of degree at most $2n-1$. A Gaussian density is not a polynomial, so its integral is generally approximate; smoothness makes the rule useful, but does not guarantee a specified accuracy for every interval or node count.

This exactness result follows from the construction of the nodes: they are the roots of the $n$th Legendre polynomial, and the corresponding weights make the rule integrate polynomials through degree $2n-1$ exactly. Changing variables from $[a,b]$ to $[-1,1]$ preserves polynomial degree, so the same exactness applies to the transformed integrand. For example, with two nodes the rule integrates any cubic or lower-degree polynomial exactly. A Gaussian density has nonzero derivatives of arbitrarily high order, so that polynomial guarantee does not make its integral exact.

Worked Transition-Probability Calculations

Worked Example: A Three-Node Probability on a Finite State Interval

Suppose, hypothetically, that at a fixed current state $x$, the next state satisfies $Y\mid X=x\sim N(0.4,1.44)$. Estimate the probability that $Y$ lies in $[0,1.2]$ using three-node Gauss-Legendre quadrature. The assumed transition is a valid Gaussian density because its variance is $1.44>0$; the limits are finite and ordered.

The midpoint is $c=(0+1.2)/2=0.6$, and the half-width is $h=0.6$. The three state-scale nodes are $y_i=0.6+0.6t_i$, giving approximately $0.135242,\ 0.600000,\ 1.064758$. The density is $f(y)=\{1/(1.2\sqrt{2\pi})\}\exp\{-(y-0.4)^2/(2(1.44))\}$.

\begin{aligned} P(0\leq Y\leq1.2\mid X=x) &\approx 0.6\left[ \frac{5}{9}f(0.135242)+\frac{8}{9}f(0.600000)+\frac{5}{9}f(1.064758) \right]\\ &\approx 0.6\left[ \frac{5}{9}(0.32446)+\frac{8}{9}(0.32787)+\frac{5}{9}(0.28519) \right]\\ &\approx 0.37808. \end{aligned}

As a check, the exact normal probability is $\Phi((1.2-0.4)/1.2)-\Phi((0-0.4)/1.2) =\Phi(0.6667)-\Phi(-0.3333)\approx0.37807$. The quadrature estimate agrees to the displayed precision. It estimates the conditional probability of ending in the stated interval, which is dimensionless; the state itself is measured in the units used for $Y$.

Worked Example: A Transition Centered Within the Interval

Consider a hypothetical transition $Y\mid X=x\sim N(1.0,0.25)$ and the interval $[0.5,1.5]$. The positive variance $0.25$ gives standard deviation $0.5$. The conditional mean lies at the interval midpoint, but the calculation still requires integrating the density across the full interval.

Here $c=1.0$ and $h=0.5$. The three state-scale nodes are $1\pm0.5\sqrt{3/5}$ and $1$, or approximately $0.612702,\ 1.000000,\ 1.387298$. The density at the center is $f(1)=1/(0.5\sqrt{2\pi})\approx0.79788$. At either outer node, the standardized distance from the mean is $\sqrt{3/5}$, so the density is approximately $0.79788\exp(-0.3)=0.59109$.

\begin{aligned} P(0.5\leq Y\leq1.5\mid X=x) &\approx 0.5\left[\frac{5}{9}(0.59109)+\frac{8}{9}(0.79788)+\frac{5}{9}(0.59109)\right]\\ &\approx 0.68300. \end{aligned}

The exact value is $\Phi(1)-\Phi(-1)\approx0.68269$. This close agreement is a check for this particular density and interval, not a guarantee that three nodes will always be sufficient.

Worked Example: Comparing Two- and Three-Node Estimates

Suppose hypothetically that $Y\mid X=x\sim N(0.5,0.09)$, and the target interval is $[0.2,0.8]$. The transition has standard deviation $0.3$, and the interval is centered at the conditional mean. The model has positive variance and the limits are finite.

The midpoint is $0.5$ and the half-width is $0.3$. With two nodes, the standardized locations relative to the mean are $\pm1/\sqrt{3}$. The density at either node is $f(0.5\pm0.3/\sqrt{3})=1/(0.3\sqrt{2\pi})\exp(-1/6)\approx1.12566$. Thus the two-node estimate is

0.3\{f(0.5-0.3/\sqrt{3})+f(0.5+0.3/\sqrt{3})\} =0.3(2)(1.12566)\approx0.67540.

With three nodes, the outer standardized locations are $\pm\sqrt{3/5}$, where the density is approximately $1.32981\exp(-0.3)=0.98515$; the center density is $1.32981$. Therefore,

0.3\left[\frac{5}{9}(0.98515)+\frac{8}{9}(1.32981)+\frac{5}{9}(0.98515)\right]\approx0.68300.

The exact probability is $\Phi(1)-\Phi(-1)\approx0.68269$. Here the three-node estimate is closer than the two-node estimate. This example illustrates why comparing node counts can be informative; it does not establish that estimates must improve monotonically as nodes are added.

Checks, Alternatives, and Limitations

A reliable calculation begins by confirming that the integral represents the intended event. In a survivor-density recursion, for example, the integration limits must be the survivor interval at the next look; integrating over a crossing region answers a different question. At a fixed current state, the transition density must be normalized over the full real line, even though the desired probability integrates it only over $[a,b]$.

Several practical checks help identify errors:

Check that $a<b$, $h=(b-a)/2$, and every transformed node $c+ht_i$ lies inside the interval.
Check that the density uses the variance in the denominator as $2v$, and uses $\sqrt{v}$ in its normalizing constant.
Check that the weights sum to 2 on $[-1,1]$, then retain the multiplier $h$ after transforming.
Confirm that the resulting probability is nonnegative. If the event is an interval, its probability cannot exceed 1, although a numerical approximation can occasionally violate probability bounds slightly.
When a normal transition is specified, compare with the exact difference of normal CDF values. Otherwise, compare estimates using different node counts or split the interval into smaller pieces if the density changes rapidly.

The main alternative for a single Gaussian transition is direct evaluation of the normal CDF difference. That is preferable when the transition is Gaussian and reliable CDF computation is available. Gauss-Legendre quadrature is useful when the integrand is a smooth transition density whose integral is not available in closed form, and it can also serve as a numerical check on a CDF calculation. Rectangular-cell approximations instead evaluate or represent mass over cells; Gauss-Legendre uses weighted point evaluations and can be more efficient for a smooth integrand on a finite interval.

The rule has limits. A fixed, small number of nodes may be inadequate when the density is sharply concentrated relative to the interval, when the interval is wide, or when the integrand is not smooth. A discontinuity inside the interval is particularly problematic because the polynomial exactness guarantee concerns smooth polynomial behavior, not jumps. Also, quadrature error is distinct from model error: increasing numerical accuracy cannot repair an incorrect transition distribution, an incorrectly specified state, or mistaken limits.

Common mistake: Treating the weighted sum of density values as the probability without multiplying by the half-width. The weights are defined for $[-1,1]$; on $[a,b]$, the correct estimate is $h\sum_i w_i f(c+ht_i\mid x)$. Omitting $h$ makes the result depend incorrectly on the interval’s scale and generally gives the wrong probability.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

For the interval $[2,6]$, what are the midpoint and half-width, and how do you map a standard node $t_i$ to the state scale?
What is the two-node Gauss-Legendre estimate of $\int_0^2 1\,dy$?
For a Gaussian transition, why is a Gauss-Legendre estimate not guaranteed to be exact even with three nodes?
A transition density has mean $m$ and variance $v$. Which quantity appears in the exponent’s denominator: $v$ or $\sqrt{v}$?

Solutions. (1) $c=4$, $h=2$, and $y_i=4+2t_i$. (2) The nodes are $1\pm1/\sqrt{3}$, the weights are both 1, and the estimate is $1+1=2$; it is exact because the integrand is constant and the transformed interval contributes the half-width 1. (3) A Gaussian density is not a polynomial of degree at most 5, the degree through which a three-node rule is guaranteed exact. (4) The exponent denominator is $2v$; $\sqrt{v}$ appears in the density’s normalizing constant.)

Gauss-Legendre Quadrature for Boundary Integrals

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