Gauss-Hermite Quadrature for Gaussian Transitions

Why Use Gauss-Hermite Quadrature for a Gaussian Transition?

A survivor-density update integrates the previous look's density against a Gaussian transition kernel. The kernel is smooth, but its Gaussian shape is part of the integrand; a quadrature rule designed for Gaussian-weighted integrals can use that structure directly. Gauss-Hermite quadrature does this by converting the kernel into the weight $e^{-u^2}$ and evaluating the remaining function at a finite set of transformed nodes.

The calculation is useful only if the transition model, quadrature scaling, and survivor region are all handled correctly. In particular, Gauss-Hermite nodes can fall outside the region in which the previous density is nonzero. The density must then be treated as zero there, rather than evaluated as though the state had survived.

Model and notation: Let $X$ be the standardized statistic at information time $t_{j-1}$, and $Y$ the statistic at the next look, time $t_j>t_{j-1}$. Under the canonical Gaussian-increment model, conditional on $X=x$, $$ Y\mid X=x \sim N(rx+d,q^2), \qquad r=\sqrt{\frac{t_{j-1}}{t_j}}, \qquad q=\sqrt{1-r^2}. $$ Here $d$ is a fixed drift increment; under the null, $d=0$. Thus $0<r<1$, $q>0$, and the transition density is $\phi((y-rx-d)/q)/q$, where $\phi$ is the standard normal density. The survivor density is an unnormalized density: its integral is the probability of having survived the looks already conducted.

Completing the Square and Mapping the Nodes

Suppose the surviving states at look $j-1$ lie in the interval $(a_{j-1},b_{j-1})$, and the current survivor interval is $(a_j,b_j)$. Write $f_{j-1}(x)$ for the previous unnormalized survivor density, extended as zero outside its survivor interval. Before imposing current-look survival, the transitioned density at $y$ is

g_j(y)=\int_{a_{j-1}}^{b_{j-1}} f_{j-1}(x)\frac{1}{q}\phi\!\left(\frac{y-rx-d}{q}\right)\,dx.

To expose a Gaussian weight in the integration variable $x$, define

m(y)=\frac{y-d}{r}, \qquad s_x=\frac{q}{r}.

Completing the square gives

\frac{1}{q}\phi\!\left(\frac{y-rx-d}{q}\right) =\frac{1}{r}\frac{1}{s_x}\phi\!\left(\frac{x-m(y)}{s_x}\right).

The factor $1/r$ is essential: viewed as a function of $x$, the transition kernel integrates to $1/r$, not 1. Substituting $x=m(y)+\sqrt{2}s_xu$ converts the normal density in $x$ to the physicists' Gauss-Hermite weight:

g_j(y)=\frac{1}{r\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^2}f_{j-1}\!\left(m(y)+\sqrt{2}s_xu\right)\,du.

For order $K$, let $u_k,w_k$, $k=1,\ldots,K$, be nodes and weights satisfying $\int_{-\infty}^{\infty}e^{-u^2}h(u)\,du\approx\sum_{k=1}^K w_kh(u_k)$. Then the quadrature approximation is

g_j(y)\approx \frac{1}{r\sqrt{\pi}}\sum_{k=1}^{K}w_k f_{j-1}\!\left(m(y)+\sqrt{2}s_xu_k\right).

The density after applying the current-look stopping rule is $f_j(y)=\mathbf{1}\{a_j<y<b_j\}g_j(y)$, where $\mathbf{1}\{\cdot\}$ equals 1 when its condition holds and 0 otherwise. Because $f_{j-1}$ is zero outside its survivor region, nodes mapped outside $(a_{j-1},b_{j-1})$ contribute zero.

Justification of the transformation: With $m=(y-d)/r$, the numerator $y-rx-d$ equals $-r(x-m)$. Hence its squared standardized value is $(x-m)^2/s_x^2$. The normal-density prefactor on the right is $r/q$; multiplying by $1/r$ gives $1/q$, the prefactor on the left. Under $x=m+\sqrt{2}s_xu$, the normal density times $dx$ becomes $e^{-u^2}du/\sqrt{\pi}$, giving the stated integral and scaling.

Worked Calculations

Worked Example: Mapping Two Hermite Nodes

Suppose $r=0.8$, $q=0.6$, $d=0$, and the target value is $y=0$. Use the two-node physicists' Gauss-Hermite rule, with nodes $u=\pm 1/\sqrt{2}$ and weights $w=\sqrt{\pi}/2$. Find the state values and coefficients in the approximation.

Here $m=0/0.8=0$ and $s_x=0.6/0.8=0.75$. The mapped states are $x=m+\sqrt{2}s_xu$, so the two nodes map to $x=-0.75$ and $x=0.75$. Each coefficient is $$ \frac{1}{r\sqrt{\pi}}w =\frac{1}{0.8\sqrt{\pi}}\frac{\sqrt{\pi}}{2} =0.625. $$ Thus $g_j(0)\approx0.625[f_{j-1}(-0.75)+f_{j-1}(0.75)]$. For example, if the previous survivor density has values $0.20$ and $0.40$ at those states, the approximation is $0.625(0.20+0.40)=0.375$. This is a density value at $y=0$, not a crossing probability.

Worked Example: Masking Nodes Outside the Survivor Interval

Let $r=0.6$, $q=0.8$, $d=0$, and $y=0.3$. Use the three-node standard-normal version of the rule, with nodes $v=0,\pm\sqrt{3}$ and weights $2/3,1/6,1/6$. These nodes integrate expectations under a standard normal distribution. The previous survivor interval is $(-1,1)$, and the density values needed below are $f_{j-1}(0.5)=0.30$, with density zero outside that interval. Find the three mapped states and the quadrature result.

The mapping in this version is $x=m+s_xv$, with the same $m=(y-d)/r$ and $s_x=q/r$. Therefore $m=0.3/0.6=0.5$ and $s_x=0.8/0.6=1.3333$. The nodes map to $x=0.5$, $x=0.5+1.3333\sqrt{3}\approx2.809$, and $x=0.5-1.3333\sqrt{3}\approx-1.809$. Only $x=0.5$ is inside the previous survivor interval. Since the weights in this rule sum to 1, the update is $$ g_j(0.3)\approx \frac{1}{0.6}\left[\frac{2}{3}(0.30)+\frac{1}{6}(0)+\frac{1}{6}(0)\right]=0.3333. $$ This is the rule's approximation, not a guarantee of high accuracy. Here two of the three nodes miss the support entirely, a warning that a low-order rule may represent a restricted integral poorly.

Worked Example: Updating at Two Current-Look States

Suppose $r=0.8$, $q=0.6$, $d=0$, and the previous survivor density is $f_{j-1}(x)=0.2(1-x^2)$ for $-1<x<1$, and zero elsewhere. Use the two-node standard-normal rule with nodes $v=\pm1$ and weights $1/2,1/2$. Let the current survivor interval be $(-1,1)$. Calculate the approximate updated density at $y=0$ and $y=0.4$.

For this rule, $x=m+s_xv$, where $s_x=0.6/0.8=0.75$, and each weight is multiplied by $1/r=1.25$. At $y=0$, $m=0$, so the mapped states are $-0.75$ and $0.75$. At either state, $f_{j-1}(x)=0.2(1-0.75^2)=0.0875$. Therefore $$ g_j(0)\approx1.25\left[\frac{1}{2}(0.0875)+\frac{1}{2}(0.0875)\right]=0.1094. $$ At $y=0.4$, $m=0.4/0.8=0.5$, giving mapped states $-0.25$ and $1.25$. The first density value is $0.2(1-0.25^2)=0.1875$; the second is zero because $1.25$ is outside the previous survivor interval. Hence $$ g_j(0.4)\approx1.25\left[\frac{1}{2}(0.1875)+\frac{1}{2}(0)\right]=0.1172. $$ Both target values are inside the current survivor interval, so these are also the updated survivor-density values there. If a target $y$ were outside that interval, the updated density would instead be zero.

Using the Rule Reliably

For each target state $y$, the transformed nodes depend on $y$. The survivor-density update therefore evaluates the previous density at a different set of states for each target. In a numerical implementation, those evaluations must use the density representation available at the previous look; values outside its survivor region must be zero. After the transition, the current survivor restriction is applied as well. This is the same survivor-mass accounting used in the earlier material on survivor-density recursion, with the integral evaluated by a different quadrature rule.

Set the transition parameters.
Calculate $r$, $q$, and, if applicable, $d$ from the specified Gaussian-increment model. Check that $0<r<1$ and $q=\sqrt{1-r^2}>0$.

Map the nodes for each target.
Use $m(y)=(y-d)/r$, $s_x=q/r$, and $x_k=m(y)+\sqrt{2}s_xu_k$ for physicists' nodes, or $x_k=m(y)+s_xv_k$ for standard-normal nodes.

Evaluate and weight the previous density.
Apply the matching weight convention, include the factor $1/r$, and assign zero to mapped states outside the previous survivor region.

Apply current survival.
Keep the transitioned density only for target states inside the current survivor region; set it to zero outside.

Useful diagnostics include checking the weight convention, confirming the mapped nodes lie where the density is defined, and inspecting the resulting density for unexpected jumps or negative values. The Gaussian kernel and a nonnegative prior density imply a nonnegative transitioned density. Also, if the previous density is constant over the whole real line, the integral over $x$ is that constant times $1/r$; this checks the scale factor, though a constant over the whole line is not a probability density.

The most important numerical warning is node masking. The full-line Gauss-Hermite rule is exact for polynomials of sufficiently low degree times its Gaussian weight, but multiplying the evaluated function by a sharp survivor-region indicator creates a discontinuity. A few nodes can miss a narrow surviving interval, as in the second example. Increasing quadrature order, checking the result at higher orders, and comparing with a suitable alternative integration are practical diagnostics—not proofs that every implementation is accurate. The earlier material on Gauss-Legendre quadrature for boundary integrals offers a useful contrast: Gauss-Legendre rules are naturally applied over finite intervals, whereas Gauss-Hermite rules exploit Gaussian weighting over the real line. The best choice depends on the integration domain and integrand, and a boundary discontinuity can challenge either method if it is not handled appropriately.

Other failure modes include using physicists' nodes with standard-normal weights, omitting $1/r$, using $s_x=q$ instead of $q/r$, and forgetting the current survivor restriction. The method also relies on the stated Gaussian transition model. If increments are not Gaussian, information times are mishandled, or the density values at mapped nodes are inaccurate, the quadrature sum does not repair those problems. A coarse quadrature may smooth or distort the density, especially near a boundary; accuracy should therefore be judged by convergence checks and agreement with the model's mass accounting, not by the appearance of a single numerical output.

Common Mistake

Common mistake: “The transition density integrates to 1, so after changing variables the transformed integral needs no scale factor.” This confuses integration in $y$ with integration in $x$. For fixed $y$, the kernel as a function of $x$ integrates to $1/r$, so the completed-square form includes $1/r$. Omitting it changes the survivor-density scale and can distort subsequent mass calculations.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $r=0.5$, $q=\sqrt{0.75}$, and $d=0$, write $m(y)$ and $s_x$.
Why must a mapped node outside the previous survivor interval contribute zero?
In the physicists' Gauss-Hermite formula, what factor multiplies the weighted sum?
Name one reason a low-order Gauss-Hermite rule can be inaccurate for a survivor update.

Solutions. (1) $m(y)=y/0.5=2y$, and $s_x=\sqrt{0.75}/0.5=\sqrt{3}$. (2) The previous survivor density is defined to be zero outside the region in which the process has survived; those states have no surviving mass to transition forward. (3) The factor is $1/(r\sqrt{\pi})$, because the physicists' weights approximate an integral weighted by $e^{-u^2}$. (4) A low-order rule may place all or most mapped nodes outside a restricted survivor interval, or may represent the sharp zero/nonzero cutoff poorly. Higher-order and alternative-rule comparisons are useful diagnostics.