Composite quadrature

For a quant, this turns an integral for an expected payoff, probability or accumulated cash flow into a finite set of function evaluations. Refining the partition can reduce integration error without changing the financial model. It cannot correct a misspecified density or missing payoff feature.

What gets added

Take $a<b$, choose $n$ equal subintervals and set

The composite trapezoid rule integrates the straight line connecting each pair of neighbouring values:

Interior nodes have full weight because they belong to two neighbouring trapezoids; each endpoint belongs to only one. The composite midpoint rule instead samples once in the middle of each subinterval:

Here $n$ counts subintervals, so Simpson uses $n/2$ quadratic pieces and $n+1$ sampled nodes. The three formulas require different samples and weights. Endpoint data alone do not supply the midpoint values, and these equal-spacing weights should not be copied onto an irregular grid.

Four subintervals, three answers

Use the dimensionless benchmark $f(x)=x^2$ on $[0,1]$. Its exact integral is $I=1/3$. One financial reading is the second moment $\mathbb{E}[X^2]$ of a synthetic uniform risk factor $X$ on $[0,1]$, whose density is one on that interval. This is a mathematical benchmark, not a fitted return distribution.

For $n=4$, $h=1/4$ and the endpoint samples are $0,1/16,1/4,9/16,1$. The midpoint samples are taken at $1/8,3/8,5/8,7/8$.

Simpson's exact answer here does not mean that every composite approximation is exact, or that Simpson is exact for arbitrary smooth functions. On an equally spaced grid with even $n$, it integrates every polynomial through degree three exactly. Trapezoid and midpoint are exact for affine functions.

More pieces improve resolution, not the rule's order

For this quadratic, a single trapezoid of width $h$ overestimates its local integral by $h^3/6$. Summing $n$ such errors gives $nh^3/6=1/(6n^2)$ on $[0,1]$. Thus the local cubic error becomes a global quadratic error as the number of pieces grows.

Doubling $n$ from four to eight reduces the trapezoid error from $1/96$ to $1/384$, and the midpoint error magnitude from $1/192$ to $1/768$. Both fall by a factor of four. Composition has reduced the size of the errors while retaining second-order convergence.

More generally, on a fixed finite interval with a continuous second derivative, trapezoid and midpoint errors are $O(h^2)$. Simpson's error is $O(h^4)$ when the fourth derivative is continuous. In a regime where a nonzero leading error term dominates, halving $h$ gives error ratios approaching four or sixteen respectively. Exactness, cancellation, kinks and numerical rounding can change those observed ratios. A midpoint estimate is not guaranteed to beat a trapezoid estimate for every smooth function.

When pricing a payoff with a known kink, placing a partition boundary at the kink can be more useful than assuming a smooth-function error formula applies across it. For an unbounded integral, error from truncating the domain is another calculation: adding panels inside a fixed window does not account for omitted tails.

Finally, preserve units. If $f(t)$ is a discounted cash-flow rate in USD/year and $h$ is in years, every weighted contribution and the final sum are in USD. Signed rates must retain their signs; taking absolute areas changes the integral.

Sources

• James Stewart, Single Variable Calculus: Early Transcendentals, 6th edition, Thomson Brooks/Cole, 2008, §7.7, “Approximate Integration”: composite formulas and derivative-based error bounds.
• Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §5.6, “Numerical integration”: weighted samples, piecewise interpolation and global trapezoid convergence.
• SciPy, scipy.integrate.simpson: composite Simpson integration and polynomial exactness under equal and unequal spacing.