5.6 Mean-value theorem and high-D Monte Carlo

Source: Ideas from Computational Physics (Landau, Páez, Bordeianu), §5.6. Rewritten in original wording. Continues 5.5 Monte Carlo integration. Labs: 5.6b Mean-value and high-D labs.

Mean-value form

For a continuous $f$ on $[a,b]$ there is some $c\in[a,b]$ with $\int_a^b f=(b-a)f(c)$. Monte Carlo replaces the unknown $f(c)$ by a sample mean. Walk the mean-value identity, the estimator $(b-a)\langle f\rangle$, equal weights $w_i=(b-a)/N$, $\sigma_I$, and the 2D volume form — pick the matching equation at each step.

Doubling accuracy costs about $4\times$ more samples — slow but predictable.

Why high dimensions favor Monte Carlo

A conventional product grid with $64$ points per axis in $D=36$ would need $64^{36}$ evaluations — not a serious plan. Monte Carlo still needs only $N$ integrand calls, and the leading error remains $\sim 1/\sqrt{N}$ independent of $D$ (for plain independent sampling). Rough crossover lore: around $D\approx 3$–$4$, Monte Carlo starts competing with simple deterministic rules; for larger $D$ it usually wins on cost-per-digit.

A 10-dimensional check

A textbook calibration (smooth, known exact value) is

$$I=\int_{[0,1]^{10}}\Biggl(\sum_{i=1}^{10} x_i\Biggr)^{2}d^{10}x=\frac{155}{6}.$$

One Monte Carlo trial: draw $N$ points in the unit hypercube, average $f(\mathbf{x})=(\sum x_i)^{2}$. Averaging several independent seeds reduces run-to-run jitter. Relative error $|I_{\mathrm{MC}}-155/6|/(155/6)$ should shrink like $1/\sqrt{N}$ as you raise $N$ or average trials.

Two-dimensional generalization

On a rectangle (or hyper-rectangle of volume $V$), the same pattern is $I\simeq V\langle f\rangle$ — the guide’s last step. Averaging several independent seeds reduces run-to-run jitter. Relative error should shrink like $1/\sqrt{N}$ as you raise $N$ or average trials.

Quiz

Mean value and high-D

1 / 3

Mean-value MC writes I ≈ …

Pick an answer to see if you are right.

Next: 5.6b Mean-value and high-D labs.