Book 6D

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Course 6Book 6D: Diffusions, Finance and ComputationChapter 10

Monte Carlo, Fourier and Finite Element Methods

Monte Carlo error and variance reduction, Euler–Maruyama, Fourier pricing, and finite element error estimates.

34 min read · Updated Oct 23, 2026

Read with Glasserman, Monte Carlo Methods in Financial Engineering (chapters 1, 4 and 6: foundations, variance reduction and discretisation of SDEs); Kloeden and Platen, Numerical Solution of Stochastic Differential Equations, for the convergence theory of Euler–Maruyama; Brenner and Scott, The Mathematical Theory of Finite Element Methods (chapters 0–2 and 5); and Jost, Partial Differential Equations, section 10.5 (the finite element method as a Galerkin method). Carr and Madan, "Option valuation using the fast Fourier transform" (J. Comput. Finance 2, 1999), is the source for Fourier pricing. This chapter proves the Monte Carlo error estimate, the variance formulas for antithetic and control variates, the strong order 12\frac12 of Euler–Maruyama, Céa's lemma, the one-dimensional interpolation estimate and the Aubin–Nitsche duality argument; weak order, multilevel Monte Carlo and the Carr–Madan formula are stated. The numerical results quoted were computed for this chapter.

In this chapter · 7 sections
  1. 10.1Solitaire and the bomb
  2. 10.2Monte Carlo
  3. 10.3Simulating SDEs: Euler–Maruyama
  4. 10.4Fourier pricing
  5. 10.5Finite elements
  6. 10.6History
  7. 10.7Exercises
  8. ↗Free resources

6D.9 Finite Difference Methods solved PDE on grids. This chapter covers the two other great families of numerical methods, which rest on the two other pillars of Course 6. Monte Carlo methods use the probabilistic representation: by Feynman–Kac (6D.3 Stochastic Differential Equations and Feynman–Kac), a solution of a parabolic equation is an expectation over random paths, and an expectation can be estimated by simulating many paths and averaging. The error decreases like N−1/2N^{-1/2} with the number of paths, slowly, but independently of the dimension, which makes Monte Carlo the method of choice for options on many assets and for path-dependent contracts. Finite element methods use the weak formulation: by Lax–Milgram (4A.5 Hilbert Spaces and Lax–Milgram, 6A.6 Weak Solutions and Elliptic Regularity), a solution of an elliptic equation is the solution of a variational problem in a Hilbert space, and restricting that problem to a finite-dimensional subspace of piecewise polynomials gives a linear system whose solution is, up to a constant, the best approximation available in the subspace. In between sits Fourier pricing, which uses the explicit characteristic functions of many models and the FFT.

By the end of this chapter you will be able to:

  • estimate an expectation by Monte Carlo with a confidence interval, and reduce its variance with antithetic and control variates;
  • simulate SDEs by the Euler–Maruyama scheme, prove its strong order 12\frac12, and explain weak order 11 and multilevel Monte Carlo;
  • describe how option prices are computed from characteristic functions by the FFT;
  • set up the finite element method for a second-order elliptic problem, and prove Céa's lemma, the interpolation estimate and the Aubin–Nitsche L2L^2 estimate in one dimension;
  • compute finite element stiffness matrices, including the cotangent formula on triangles.

Solitaire and the bomb

In the world Data The origin of Monte Carlo

In 1946, recovering from an illness, Stanisław Ulam wondered what the chance was that a game of Canfield solitaire would come out, gave up on combinatorics, and realised it would be easier to play many games and count. At Los Alamos he and John von Neumann turned the idea into a method for the neutron transport problems of weapons design: simulate the random histories of many neutrons on the ENIAC and average. Nicholas Metropolis named it after the casino in Monaco, and he and Ulam published "The Monte Carlo method" in 1949. Four years later Metropolis, the Rosenbluths and the Tellers introduced the Markov chain sampling algorithm that bears Metropolis's name, the ancestor of the Langevin Monte Carlo of 6D.3 Stochastic Differential Equations and Feynman–Kac.

Phelim Boyle brought the method to option pricing in 1977, and today a large share of the computing power of banks goes into Monte Carlo simulation of portfolios of derivatives under many correlated risk factors, where grids are hopeless: a finite difference grid with 100100 points per dimension needs 102010^{20} points in 1010 dimensions, while Monte Carlo's error σ/N\sigma/\sqrt N does not care about dimension. The finite element method has a parallel history: Richard Courant proposed piecewise linear approximation on triangles in 1943, aerospace engineers at Boeing and elsewhere developed it in the 1950s for aircraft structures, and Ray Clough named it the "finite element method" in 1960.

Monte Carlo

To estimate m=EXm = \mathbb{E}X for an integrable random variable XX that we can simulate, draw independent copies X1,…,XNX_1, \ldots, X_N and form the sample mean XˉN=1N∑iXi\bar X_N = \frac1N\sum_iX_i.

Proposition 10.1 The Monte Carlo error

If σ2=Var⁡X<∞\sigma^2 = \operatorname{Var}X < \infty, then EXˉN=m\mathbb{E}\bar X_N = m, Var⁡XˉN=σ2N\operatorname{Var}\bar X_N = \frac{\sigma^2}{N}, and XˉN→m\bar X_N \to m almost surely. By the central limit theorem, N(XˉN−m)/σ→N(0,1)\sqrt N(\bar X_N - m)/\sigma \to N(0, 1) in distribution, so XˉN±1.96σ^N\bar X_N \pm 1.96\frac{\hat\sigma}{\sqrt N}, with σ^\hat\sigma the sample standard deviation, is an approximate 95%95\% confidence interval for mm.

Proof. Linearity of expectation and independence give the mean and variance (Var⁡∑Xi=∑Var⁡Xi\operatorname{Var}\sum X_i = \sum\operatorname{Var}X_i). Almost sure convergence is the strong law of large numbers; the interval comes from the central limit theorem with σ^→σ\hat\sigma \to \sigma (statements; 3A.4 Measures, Probability and Weights).

The error is O(N−1/2)O(N^{-1/2}) whatever the dimension: to gain one decimal digit, multiply the work by 100100. Every improvement therefore aims at the constant σ\sigma.

Example 10.2 Pricing a call by simulation

For the call of 6D.6 Black–Scholes (S0=K=100S_0 = K = 100, r=0.05r = 0.05, σ=0.2\sigma = 0.2, T=1T = 1), simulate ST=S0exp⁡((r−12σ2)T+σTZ)S_T = S_0\exp\big((r - \frac12\sigma^2)T + \sigma\sqrt TZ\big) under the risk-neutral measure and average the discounted payoffs e−rT(ST−K)+e^{-rT}(S_T - K)^+. With N=100,000N = 100{,}000 samples, one run gave 10.49710.497 with a standard error of 0.0470.047, consistent with the exact 10.450610.4506 (the difference is one standard error). The discounted payoff has standard deviation about 14.714.7, so a standard error of one cent would need about (14.7/0.01)2≈2×106(14.7/0.01)^2 \approx 2\times10^6 samples.

Antithetic variates. If X=f(Z)X = f(Z) with ZZ standard normal, −Z-Z has the same law, and Y=12(f(Z)+f(−Z))Y = \frac12\big(f(Z) + f(-Z)\big) also has mean mm, with

Var⁡Y=12(Var⁡f(Z)+Cov⁡(f(Z),f(−Z))).\operatorname{Var}Y = \frac12\big(\operatorname{Var}f(Z) + \operatorname{Cov}(f(Z), f(-Z))\big).

When ff is monotone, f(Z)f(Z) and f(−Z)f(-Z) are negatively correlated (Exercise 10.8), and averaging the pair costs two evaluations but reduces variance by more than a factor 22. For the call above the reduction per evaluation was about a factor 2.02.0: modest, because the payoff is far from linear.

Control variates. If WW is a simulated quantity with known mean EW=w\mathbb{E}W = w, then X−b(W−w)X - b(W - w) has mean mm for every bb, and variance Var⁡X−2bCov⁡(X,W)+b2Var⁡W\operatorname{Var}X - 2b\operatorname{Cov}(X, W) + b^2\operatorname{Var}W, minimised at

b∗=Cov⁡(X,W)Var⁡W,Var⁡(X−b∗(W−w))=(1−ρ2)Var⁡X,b^* = \frac{\operatorname{Cov}(X, W)}{\operatorname{Var}W}, \qquad \operatorname{Var}\big(X - b^*(W - w)\big) = (1 - \rho^2)\operatorname{Var}X,

where ρ\rho is the correlation of XX and WW. For the call, W=STW = S_T with EQST=S0erT\mathbb{E}^{\mathbb{Q}}S_T = S_0e^{rT} is a natural control: the estimated correlation gave ρ2≈0.85\rho^2 \approx 0.85 and a variance reduction by a factor of about 77, equivalent to seven times as many samples. The best controls are prices of similar contracts with known formulas, for example a geometric-average Asian option as a control for an arithmetic-average one.

Simulating SDEs: Euler–Maruyama

For dX=b(X) dt+σ(X) dBdX = b(X)\,dt + \sigma(X)\,dB, with step Δt=T/n\Delta t = T/n and grid times tk=kΔtt_k = k\Delta t, the Euler–Maruyama scheme is

Yk+1=Yk+b(Yk) Δt+σ(Yk) ΔBk,ΔBk=Btk+1−Btk∼N(0,Δt),Y_{k+1} = Y_k + b(Y_k)\,\Delta t + \sigma(Y_k)\,\Delta B_k, \qquad \Delta B_k = B_{t_{k+1}} - B_{t_k} \sim N(0, \Delta t),

the stochastic analogue of Euler's method, with the noise increments sampled exactly. Two kinds of error matter. The strong error (E∣XT−Yn∣2)1/2\big(\mathbb{E}|X_T - Y_n|^2\big)^{1/2} measures how close individual paths are (needed when the paths themselves matter, or for multilevel methods); the weak error ∣Ef(XT)−Ef(Yn)∣|\mathbb{E}f(X_T) - \mathbb{E}f(Y_n)| measures how close expectations are (all that matters for pricing).

Theorem 10.3 Strong order 12\frac12

If bb and σ\sigma are Lipschitz, there is CC depending on TT, the Lipschitz constants and xx such that Esup⁡k≤n∣Xtk−Yk∣2≤CΔt\mathbb{E}\sup_{k\leq n}|X_{t_k} - Y_k|^2 \leq C\Delta t.

Proof. Extend the scheme to continuous time: with η(s)=tk\eta(s) = t_k for tk≤s<tk+1t_k \leq s < t_{k+1}, let Yt=x+∫0tb(Yη(s)) ds+∫0tσ(Yη(s)) dBsY_t = x + \int_0^tb(Y_{\eta(s)})\,ds + \int_0^t\sigma(Y_{\eta(s)})\,dB_s, which agrees with YkY_k at the grid times. Then

Xt−Yt=∫0t(b(Xs)−b(Yη(s)))ds+∫0t(σ(Xs)−σ(Yη(s)))dBs,X_t - Y_t = \int_0^t\big(b(X_s) - b(Y_{\eta(s)})\big)ds + \int_0^t\big(\sigma(X_s) - \sigma(Y_{\eta(s)})\big)dB_s,

and ∣b(Xs)−b(Yη(s))∣≤K(∣Xs−Xη(s)∣+∣Xη(s)−Yη(s)∣)|b(X_s) - b(Y_{\eta(s)})| \leq K\big(|X_s - X_{\eta(s)}| + |X_{\eta(s)} - Y_{\eta(s)}|\big), likewise for σ\sigma. Two facts about the exact solution are needed: Esup⁡t≤T∣Xt∣2≤C\mathbb{E}\sup_{t\leq T}|X_t|^2 \leq C (6D.3 Stochastic Differential Equations and Feynman–Kac), and hence, by the same Cauchy–Schwarz and isometry estimates, E∣Xs−Xη(s)∣2≤C∣s−η(s)∣≤CΔt\mathbb{E}|X_s - X_{\eta(s)}|^2 \leq C|s - \eta(s)| \leq C\Delta t. Let f(t)=Esup⁡s≤t∣Xs−Ys∣2f(t) = \mathbb{E}\sup_{s\leq t}|X_s - Y_s|^2. The two inequalities of the existence proof in 6D.3 Stochastic Differential Equations and Feynman–Kac (Cauchy–Schwarz for the dsds integral, Doob and the isometry for the dBdB integral) give

f(t)≤C∫0t(E∣Xs−Xη(s)∣2+f(s))ds≤C TΔt+C∫0tf(s) ds,f(t) \leq C\int_0^t\big(\mathbb{E}|X_s - X_{\eta(s)}|^2 + f(s)\big)ds \leq C\,T\Delta t + C\int_0^tf(s)\,ds,

and Gronwall's inequality (6D.3 Stochastic Differential Equations and Feynman–Kac) gives f(T)≤CTΔteCTf(T) \leq CT\Delta te^{CT}.

The order 12\frac12 comes from the term E∣Xs−Xη(s)∣2≈Δt\mathbb{E}|X_s - X_{\eta(s)}|^2 \approx \Delta t: Brownian motion moves a distance Δt\sqrt{\Delta t} in time Δt\Delta t (6D.1 Brownian Motion). Including the next term of the Itô–Taylor expansion (the Milstein scheme, which adds 12σσ′((ΔB)2−Δt)\frac12\sigma\sigma'\big((\Delta B)^2 - \Delta t\big) in one dimension) raises the strong order to 11.

The weak order of Euler–Maruyama is 11: for smooth ff and coefficients, Ef(XT)−Ef(Yn)=CfΔt+O(Δt2)\mathbb{E}f(X_T) - \mathbb{E}f(Y_n) = C_f\Delta t + O(\Delta t^2) (statement; Denis Talay and Luciano Tubaro, 1990, who also proved the expansion that justifies Richardson extrapolation). For geometric Brownian motion it can be seen directly: EYn=x(1+μΔt)n\mathbb{E}Y_n = x(1 + \mu\Delta t)^n while EXT=xeμT\mathbb{E}X_T = xe^{\mu T}, and the difference is about −12xμ2TΔteμT-\frac12x\mu^2T\Delta te^{\mu T}, halving when Δt\Delta t halves (Exercise 10.10).

Multilevel Monte Carlo. To reach accuracy ε\varepsilon with Euler–Maruyama and plain Monte Carlo one needs Δt∼ε\Delta t \sim \varepsilon (weak error) and N∼ε−2N \sim \varepsilon^{-2} paths (statistical error), a cost of order ε−3\varepsilon^{-3}. Michael Giles (2008) observed that the expectation on the finest grid can be written as a telescoping sum EPL=EP0+∑ℓ=1LE(Pℓ−Pℓ−1)\mathbb{E}P_L = \mathbb{E}P_0 + \sum_{\ell=1}^L\mathbb{E}(P_\ell - P_{\ell-1}) over grids with Δtℓ=2−ℓT\Delta t_\ell = 2^{-\ell}T, and that the corrections Pℓ−Pℓ−1P_\ell - P_{\ell-1}, computed on the same Brownian path at two resolutions, have variance of order Δtℓ\Delta t_\ell by the strong error estimate. So few samples are needed on the expensive fine levels, and the total cost falls to order ε−2(log⁡ε)2\varepsilon^{-2}(\log\varepsilon)^2 (statement): strong convergence, useless on its own for pricing, buys a speedup for weak problems.

Fourier pricing

Many models used in practice are not solvable for prices but have explicit characteristic functions ϕT(v)=EQ[eivlog⁡ST]\phi_T(v) = \mathbb{E}^{\mathbb{Q}}\big[e^{iv\log S_T}\big]: Lévy (jump) models, and Steven Heston's stochastic volatility model (1993). Prices can then be computed by Fourier inversion. A call price, as a function of the log-strike k=log⁡Kk = \log K, is not integrable (it tends to S0S_0 as k→−∞k \to -\infty), so Peter Carr and Dilip Madan (1999) damp it: for α>0\alpha > 0 with EST1+α<∞\mathbb{E}S_T^{1+\alpha} < \infty, the function eαkC(k)e^{\alpha k}C(k) is integrable, and its Fourier transform is explicit in terms of ϕT\phi_T,

ψ(v)=∫eivkeαkC(k) dk=e−rTϕT(v−(α+1)i)α2+α−v2+i(2α+1)v\psi(v) = \int e^{ivk}e^{\alpha k}C(k)\,dk = \frac{e^{-rT}\phi_T\big(v - (\alpha + 1)i\big)}{\alpha^2 + \alpha - v^2 + i(2\alpha + 1)v}

(statement; derived by Fubini from C(k)=e−rTE(ST−ek)+C(k) = e^{-rT}\mathbb{E}(S_T - e^k)^+, Exercise 10.11). Inverting, C(k)=e−αkπ∫0∞Re⁡(e−ivkψ(v)) dvC(k) = \frac{e^{-\alpha k}}{\pi}\int_0^\infty\operatorname{Re}\big(e^{-ivk}\psi(v)\big)\,dv, and discretising the integral on a grid of NN frequencies gives prices on a grid of NN strikes by a single FFT (6D.9 Finite Difference Methods), in O(Nlog⁡N)O(N\log N) operations. This is how the parameters of stochastic volatility models are calibrated to hundreds of quoted options in seconds.

Finite elements

Consider the model problem −(pu′)′+qu=f-(pu')' + qu = f on (0,1)(0, 1) with u(0)=u(1)=0u(0) = u(1) = 0, p≥p0>0p \geq p_0 > 0, q≥0q \geq 0 (the method works the same way in any dimension). Its weak form (6A.6 Weak Solutions and Elliptic Regularity) is: find u∈V=H01(0,1)u \in V = H^1_0(0, 1) with

a(u,v)=∫01(pu′v′+quv) dx=∫01fv dx=ℓ(v)for all v∈V.a(u, v) = \int_0^1\big(pu'v' + quv\big)\,dx = \int_0^1fv\,dx = \ell(v) \quad\text{for all } v \in V.

The form aa is bounded, ∣a(u,v)∣≤M∥u∥H1∥v∥H1|a(u, v)| \leq M\|u\|_{H^1}\|v\|_{H^1}, and coercive, a(v,v)≥α∥v∥H12a(v, v) \geq \alpha\|v\|_{H^1}^2 (by Poincaré's inequality), so Lax–Milgram gives a unique solution (4A.5 Hilbert Spaces and Lax–Milgram).

The Galerkin method chooses a finite-dimensional subspace Vh⊂VV_h \subset V and asks for uh∈Vhu_h \in V_h with a(uh,vh)=ℓ(vh)a(u_h, v_h) = \ell(v_h) for all vh∈Vhv_h \in V_h. In a basis φ1,…,φN\varphi_1, \ldots, \varphi_N of VhV_h, with uh=∑jcjφju_h = \sum_jc_j\varphi_j, this is the linear system

∑ja(φj,φi) cj=ℓ(φi),i=1,…,N,\sum_ja(\varphi_j, \varphi_i)\,c_j = \ell(\varphi_i), \qquad i = 1, \ldots, N,

with the stiffness matrix Aij=a(φj,φi)A_{ij} = a(\varphi_j, \varphi_i), which is positive definite by coercivity. The finite element method chooses VhV_h to be piecewise polynomials on a mesh, with a local basis, so that AA is sparse. The simplest choice in one dimension: nodes xj=jhx_j = jh, h=1N+1h = \frac{1}{N+1}, and the hat functions φj\varphi_j, piecewise linear, equal to 11 at xjx_j and 00 at the other nodes. For p=1p = 1, q=0q = 0, the stiffness matrix is 1htridiag⁡(−1,2,−1)\frac1h\operatorname{tridiag}(-1, 2, -1), the finite difference matrix of 6D.9 Finite Difference Methods times hh (Exercise 10.12).

Lemma 10.4 Céa's lemma

Let uu and uhu_h be the exact and Galerkin solutions. Then a(u−uh,vh)=0a(u - u_h, v_h) = 0 for all vh∈Vhv_h \in V_h (Galerkin orthogonality), and

∥u−uh∥H1≤Mαinf⁡vh∈Vh∥u−vh∥H1.\|u - u_h\|_{H^1} \leq \frac M\alpha\inf_{v_h\in V_h}\|u - v_h\|_{H^1}.

Proof. Subtracting a(uh,vh)=ℓ(vh)a(u_h, v_h) = \ell(v_h) from a(u,vh)=ℓ(vh)a(u, v_h) = \ell(v_h) gives orthogonality. For any vh∈Vhv_h \in V_h, uh−vh∈Vhu_h - v_h \in V_h, so

α∥u−uh∥2≤a(u−uh,u−uh)=a(u−uh,u−vh)+a(u−uh,vh−uh)=a(u−uh,u−vh)≤M∥u−uh∥∥u−vh∥.\alpha\|u - u_h\|^2 \leq a(u - u_h, u - u_h) = a(u - u_h, u - v_h) + a(u - u_h, v_h - u_h) = a(u - u_h, u - v_h) \leq M\|u - u_h\|\|u - v_h\|.

Divide by α∥u−uh∥\alpha\|u - u_h\|.

So the Galerkin solution is quasi-optimal: up to the constant Mα\frac M\alpha, it is as good as the best approximation of uu from VhV_h. (For symmetric aa it is exactly the best approximation in the energy norm a(v,v)1/2a(v, v)^{1/2}: the Galerkin solution is the orthogonal projection of uu onto VhV_h for the inner product aa.) Error estimates therefore reduce to approximation theory.

Proposition 10.5 Interpolation error for hat functions

Let Ihu∈VhI_hu \in V_h be the piecewise linear interpolant of u∈H2(0,1)∩H01u \in H^2(0, 1)\cap H^1_0 at the nodes. Then

∥(u−Ihu)′∥L2≤h∥u′′∥L2,∥u−Ihu∥L2≤h2∥u′′∥L2.\|(u - I_hu)'\|_{L^2} \leq h\|u''\|_{L^2}, \qquad \|u - I_hu\|_{L^2} \leq h^2\|u''\|_{L^2}.

Proof. On an element [xj,xj+1][x_j, x_{j+1}], e=u−Ihue = u - I_hu vanishes at both ends, so by Rolle's theorem e′(ξ)=0e'(\xi) = 0 for some ξ\xi in the element. Since (Ihu)′′=0(I_hu)'' = 0 there, for xx in the element ∣e′(x)∣=∣∫ξxu′′∣≤h(∫xjxj+1∣u′′∣2)1/2|e'(x)| = \big|\int_\xi^xu''\big| \leq \sqrt h\big(\int_{x_j}^{x_{j+1}}|u''|^2\big)^{1/2} by Cauchy–Schwarz. Squaring and integrating over the element gives ∫∣e′∣2≤h2∫∣u′′∣2\int|e'|^2 \leq h^2\int|u''|^2 on each element; sum. Similarly ∣e(x)∣=∣∫xjxe′∣≤h∥e′∥L2(element)|e(x)| = \big|\int_{x_j}^xe'\big| \leq \sqrt h\|e'\|_{L^2(\text{element})}, so ∫∣e∣2≤h2∫∣e′∣2≤h4∫∣u′′∣2\int|e|^2 \leq h^2\int|e'|^2 \leq h^4\int|u''|^2 on each element.

With Céa's lemma: ∥u−uh∥H1≤Mα∥u−Ihu∥H1≤Ch∥u′′∥L2\|u - u_h\|_{H^1} \leq \frac M\alpha\|u - I_hu\|_{H^1} \leq Ch\|u''\|_{L^2}, first order in the energy norm. But the L2L^2 error of the Galerkin solution is better than the H1H^1 error by a full power of hh, by a duality argument of Jean-Pierre Aubin and Joachim Nitsche (1967–68).

Theorem 10.6 The Aubin–Nitsche estimate

For the model problem with p,qp, q smooth, ∥u−uh∥L2≤Ch2∥u′′∥L2\|u - u_h\|_{L^2} \leq Ch^2\|u''\|_{L^2}.

Proof. Let e=u−uhe = u - u_h, and let z∈H01z \in H^1_0 solve the dual problem a(v,z)=∫eva(v, z) = \int ev for all v∈Vv \in V, that is, −(pz′)′+qz=e-(pz')' + qz = e. By elliptic regularity in one dimension (6A.6 Weak Solutions and Elliptic Regularity), z∈H2z \in H^2 with ∥z′′∥L2≤C∥e∥L2\|z''\|_{L^2} \leq C\|e\|_{L^2}. Take v=ev = e and use Galerkin orthogonality with vh=Ihzv_h = I_hz:

∥e∥L22=a(e,z)=a(e,z−Ihz)≤M∥e∥H1∥z−Ihz∥H1≤M⋅Ch∥u′′∥L2⋅Ch∥z′′∥L2≤Ch2∥u′′∥L2∥e∥L2.\|e\|_{L^2}^2 = a(e, z) = a(e, z - I_hz) \leq M\|e\|_{H^1}\|z - I_hz\|_{H^1} \leq M\cdot Ch\|u''\|_{L^2}\cdot Ch\|z''\|_{L^2} \leq Ch^2\|u''\|_{L^2}\|e\|_{L^2}.

Divide by ∥e∥L2\|e\|_{L^2}.

The same structure works in higher dimensions on meshes of triangles or tetrahedra: Céa's lemma unchanged; interpolation estimates ∥u−Ihu∥H1≤Ch∣u∣H2\|u - I_hu\|_{H^1} \leq Ch|u|_{H^2} for piecewise linear elements on meshes whose triangles are not too thin (statement; Brenner and Scott, chapter 4); and Aubin–Nitsche on convex polygonal domains, where H2H^2 regularity of the dual problem holds. Polynomials of degree pp give O(hp)O(h^p) and O(hp+1)O(h^{p+1}). For parabolic problems, finite elements in space are combined with the time-stepping schemes of 6D.9 Finite Difference Methods (the method of lines). Unlike finite differences, finite elements handle curved boundaries, unstructured meshes and variable coefficients naturally, and their error theory is the Hilbert space theory of 4A.5 Hilbert Spaces and Lax–Milgram, which is why they dominate engineering simulation.

Where this goes Discrete geometry

On a triangulated surface, the stiffness matrix of piecewise linear elements is the cotangent Laplacian (Exercise 10.14), the discrete Laplace–Beltrami operator of discrete differential geometry. Gerhard Dziuk (1990) computed mean curvature flow of surfaces with exactly these elements, and Bennett Chow and Feng Luo (2003) introduced a combinatorial Ricci flow on triangulated surfaces whose convergence reproves discrete versions of uniformization (11A.12 Ricci Flow on Surfaces); discrete conformal maps built on the same ideas are used in computer graphics and in computational anatomy to flatten the cortex of the brain.

History

Ulam, von Neumann and Metropolis created the Monte Carlo method at Los Alamos in 1946–49, and Boyle applied it to options in 1977. Gisiro Maruyama analysed the Euler scheme for SDEs in 1955 and Grigori Milstein his higher-order scheme in 1974; Talay and Tubaro's weak error expansion dates from 1990 and Giles's multilevel method from 2008. Richard Courant proposed piecewise linear elements in 1943, engineers developed the method in the 1950s, Clough named it in 1960, and the mathematical theory (Céa's lemma, 1964; Aubin and Nitsche, 1967–68; Ciarlet's book, 1978) followed. Carr and Madan's FFT method dates from 1999 and Heston's model from 1993.

Recall Where we stand

Monte Carlo estimates EX\mathbb{E}X by a sample mean with error σ/N\sigma/\sqrt N independent of dimension; antithetic variates and control variates (variance (1−ρ2)Var⁡X(1 - \rho^2)\operatorname{Var}X) reduce σ\sigma. Euler–Maruyama converges strongly with order 12\frac12 (proved by the SDE estimates and Gronwall) and weakly with order 11; multilevel Monte Carlo exploits strong convergence to cut the cost from ε−3\varepsilon^{-3} to about ε−2\varepsilon^{-2}. Fourier methods price from characteristic functions with one FFT. Finite elements solve the weak formulation on a subspace of piecewise polynomials; Galerkin orthogonality gives Céa's quasi-optimality, interpolation gives O(h)O(h) in H1H^1, and the Aubin–Nitsche duality argument gives O(h2)O(h^2) in L2L^2. 6D.11 Computing Geometric Flows applies these methods, and discrete geometry, to the geometric flows of the guide, and closes Course 6.

Exercises

Exercise 10.7 How many samples?

For the call of Example 10.2, the discounted payoff has standard deviation about 14.714.7. (a) How many samples give a 95%95\% confidence interval of half-width 0.050.05? (b) With the control variate of the text (ρ2≈0.85\rho^2 \approx 0.85), how many?

Solution

(a) 1.96×14.7/N=0.051.96\times14.7/\sqrt N = 0.05 gives N≈(576)2≈3.3×105N \approx (576)^2 \approx 3.3\times10^5. (b) The standard deviation becomes 14.70.15≈5.714.7\sqrt{0.15} \approx 5.7, so N≈(1.96×5.7/0.05)2≈5.0×104N \approx (1.96\times5.7/0.05)^2 \approx 5.0\times10^4, about seven times fewer.

Exercise 10.8 Antithetic variates for monotone functions

Let ff be non-decreasing and ZZ symmetric. Show that Cov⁡(f(Z),f(−Z))≤0\operatorname{Cov}(f(Z), f(-Z)) \leq 0. (Use g(z)=−f(−z)g(z) = -f(-z), also non-decreasing, and the fact that two non-decreasing functions of the same random variable are non-negatively correlated: E[(f(Z)−f(Z′))(g(Z)−g(Z′))]≥0\mathbb{E}[(f(Z) - f(Z'))(g(Z) - g(Z'))] \geq 0 for an independent copy Z′Z'.)

Solution

(f(Z)−f(Z′))(g(Z)−g(Z′))≥0(f(Z) - f(Z'))(g(Z) - g(Z')) \geq 0 pointwise, since both factors have the sign of Z−Z′Z - Z'. Taking expectations, 2Cov⁡(f(Z),g(Z))≥02\operatorname{Cov}(f(Z), g(Z)) \geq 0. With g(z)=−f(−z)g(z) = -f(-z): Cov⁡(f(Z),−f(−Z))≥0\operatorname{Cov}(f(Z), -f(-Z)) \geq 0, i.e. Cov⁡(f(Z),f(−Z))≤0\operatorname{Cov}(f(Z), f(-Z)) \leq 0.

Exercise 10.9 The optimal control coefficient

Prove the formulas for b∗b^* and the minimal variance in the text, and show that the variance reduction factor is 11−ρ2\frac{1}{1 - \rho^2}.

Solution

V(b)=Var⁡X−2bCov⁡(X,W)+b2Var⁡WV(b) = \operatorname{Var}X - 2b\operatorname{Cov}(X, W) + b^2\operatorname{Var}W is a quadratic in bb minimised at b∗=Cov⁡(X,W)/Var⁡Wb^* = \operatorname{Cov}(X, W)/\operatorname{Var}W, with V(b∗)=Var⁡X−Cov⁡(X,W)2Var⁡W=(1−ρ2)Var⁡XV(b^*) = \operatorname{Var}X - \frac{\operatorname{Cov}(X, W)^2}{\operatorname{Var}W} = (1 - \rho^2)\operatorname{Var}X. The factor is Var⁡X/V(b∗)=(1−ρ2)−1\operatorname{Var}X/V(b^*) = (1 - \rho^2)^{-1}; for ρ2=0.85\rho^2 = 0.85, about 6.76.7.

Exercise 10.10 Weak error for geometric Brownian motion

For dX=μX dt+σX dBdX = \mu X\,dt + \sigma X\,dB, show that the Euler–Maruyama approximation satisfies EYn=x(1+μΔt)n\mathbb{E}Y_n = x(1 + \mu\Delta t)^n. With x=100x = 100, μ=0.05\mu = 0.05, T=1T = 1, compute the weak error EYn−EXT\mathbb{E}Y_n - \mathbb{E}X_T for n=10n = 10, 2020, 4040, and confirm weak order 11.

Solution

E[Yk+1∣Yk]=Yk(1+μΔt)\mathbb{E}[Y_{k+1}\mid Y_k] = Y_k(1 + \mu\Delta t) since ΔBk\Delta B_k has mean zero and is independent of YkY_k; iterate. The errors are 100(1.005)10−100e0.05≈−0.0131100(1.005)^{10} - 100e^{0.05} \approx -0.0131, then −0.0066-0.0066 and −0.0033-0.0033: halving with Δt\Delta t. They match −12xμ2TΔteμT=−0.0131-\frac12x\mu^2T\Delta te^{\mu T} = -0.0131 for Δt=0.1\Delta t = 0.1.

Exercise 10.11 The damped call transform

Assume STS_T has a density and EST1+α<∞\mathbb{E}S_T^{1+\alpha} < \infty. Writing C(k)=e−rT∫k∞(es−ek)q(s) dsC(k) = e^{-rT}\int_k^\infty(e^s - e^k)q(s)\,ds with qq the density of s=log⁡STs = \log S_T, compute ∫eivkeαkC(k) dk\int e^{ivk}e^{\alpha k}C(k)\,dk by Fubini (integrate over k<sk < s first) and obtain the formula for ψ(v)\psi(v).

Solution

∫−∞se(α+iv)k(es−ek) dk=e(α+1+iv)sα+iv−e(α+1+iv)sα+1+iv=e(α+1+iv)s(α+iv)(α+1+iv)\int_{-\infty}^se^{(\alpha + iv)k}(e^s - e^k)\,dk = \frac{e^{(\alpha + 1 + iv)s}}{\alpha + iv} - \frac{e^{(\alpha + 1 + iv)s}}{\alpha + 1 + iv} = \frac{e^{(\alpha + 1 + iv)s}}{(\alpha + iv)(\alpha + 1 + iv)}. Integrating against e−rTq(s) dse^{-rT}q(s)\,ds gives e−rTEei(v−(α+1)i)s(α+iv)(α+1+iv)\frac{e^{-rT}\mathbb{E}e^{i(v - (\alpha + 1)i)s}}{(\alpha + iv)(\alpha + 1 + iv)}, and (α+iv)(α+1+iv)=α2+α−v2+i(2α+1)v(\alpha + iv)(\alpha + 1 + iv) = \alpha^2 + \alpha - v^2 + i(2\alpha + 1)v.

Exercise 10.12 The one-dimensional stiffness matrix

For −u′′=f-u'' = f on (0,1)(0, 1) with hat functions on a uniform mesh, compute a(φj,φi)=∫φj′φi′a(\varphi_j, \varphi_i) = \int\varphi_j'\varphi_i' and show that the stiffness matrix is 1htridiag⁡(−1,2,−1)\frac1h\operatorname{tridiag}(-1, 2, -1). If the load is approximated by ℓ(φi)≈hf(xi)\ell(\varphi_i) \approx hf(x_i), show that the finite element equations are the finite difference equations −ui+1−2ui+ui−1h2=f(xi)-\frac{u_{i+1} - 2u_i + u_{i-1}}{h^2} = f(x_i).

Solution

φi′=1h\varphi_i' = \frac1h on (xi−1,xi)(x_{i-1}, x_i) and −1h-\frac1h on (xi,xi+1)(x_i, x_{i+1}). So ∫(φi′)2=2h\int(\varphi_i')^2 = \frac2h, ∫φi′φi+1′=−1h\int\varphi_i'\varphi_{i+1}' = -\frac1h (overlap on one element), and 00 otherwise. Dividing the equations 1h(−ci−1+2ci−ci+1)=hf(xi)\frac1h(-c_{i-1} + 2c_i - c_{i+1}) = hf(x_i) by hh gives the difference scheme, with ci=uh(xi)c_i = u_h(x_i).

Exercise 10.13 Constants in Céa's lemma

For a(u,v)=∫01(u′v′+uv)a(u, v) = \int_0^1(u'v' + uv) on H01(0,1)H^1_0(0, 1) with the norm ∥v∥H12=∫(v′2+v2)\|v\|_{H^1}^2 = \int(v'^2 + v^2), show that M=α=1M = \alpha = 1, so the Galerkin solution is the best approximation in H1H^1. For a(u,v)=∫01u′v′a(u, v) = \int_0^1u'v', find α\alpha using the Poincaré inequality ∫v2≤1π2∫v′2\int v^2 \leq \frac{1}{\pi^2}\int v'^2.

Solution

In the first case a(v,v)=∥v∥H12a(v, v) = \|v\|_{H^1}^2 and ∣a(u,v)∣≤∥u∥∥v∥|a(u, v)| \leq \|u\|\|v\|, so M=α=1M = \alpha = 1. In the second, ∥v∥H12≤(1+1π2)∫v′2\|v\|_{H^1}^2 \leq (1 + \frac{1}{\pi^2})\int v'^2, so a(v,v)≥π2π2+1∥v∥H12a(v, v) \geq \frac{\pi^2}{\pi^2 + 1}\|v\|_{H^1}^2, α=π21+π2\alpha = \frac{\pi^2}{1 + \pi^2} and M=1M = 1.

Exercise 10.14 Rehearsal: the cotangent Laplacian

Let TT be a triangle with vertices ii, jj, kk, and φi\varphi_i, φj\varphi_j the linear functions on TT equal to 11 at one vertex and 00 at the other two. (a) Show that ∇φi\nabla\varphi_i is perpendicular to the edge jkjk, points towards ii, and has length 1hi\frac{1}{h_i}, where hih_i is the height from ii. (b) Show that ∫T∇φi⋅∇φj=−12cot⁡θk\int_T\nabla\varphi_i\cdot\nabla\varphi_j = -\frac12\cot\theta_k, where θk\theta_k is the angle at kk. (Use hi=∣ik∣sin⁡θkh_i = |ik|\sin\theta_k, hj=∣jk∣sin⁡θkh_j = |jk|\sin\theta_k and ∣T∣=12∣ik∣∣jk∣sin⁡θk|T| = \frac12|ik||jk|\sin\theta_k, and the angle between the two gradients.) (c) Conclude that on a triangulated surface the stiffness matrix entry for an interior edge ijij is −12(cot⁡αij+cot⁡βij)-\frac12(\cot\alpha_{ij} + \cot\beta_{ij}), with αij\alpha_{ij}, βij\beta_{ij} the angles opposite the edge. This is the discrete Laplace–Beltrami operator used in discrete mean curvature flow, discrete conformal geometry and combinatorial Ricci flow on surfaces (11A.12 Ricci Flow on Surfaces).

Solution

(a) φi\varphi_i is constant (00) along the edge jkjk, so its gradient is perpendicular to it; it increases from 00 to 11 over the distance hih_i towards ii. (b) The gradients are inward normals to the edges jkjk and ikik, and the angle between them is π−θk\pi - \theta_k, so ∇φi⋅∇φj=−cos⁡θkhihj\nabla\varphi_i\cdot\nabla\varphi_j = -\frac{\cos\theta_k}{h_ih_j}. Multiplying by the area, ∣T∣⋅−cos⁡θkhihj=12∣ik∣∣jk∣sin⁡θk(−cos⁡θk)∣ik∣∣jk∣sin⁡2θk=−12cot⁡θk|T|\cdot\frac{-\cos\theta_k}{h_ih_j} = \frac{\frac12|ik||jk|\sin\theta_k(-\cos\theta_k)}{|ik||jk|\sin^2\theta_k} = -\frac12\cot\theta_k. (c) The edge ijij belongs to two triangles, with opposite angles αij\alpha_{ij} and βij\beta_{ij}, and their contributions add.

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