A Fourier Transform Method for Solving Backward Stochastic Differential Equations

Abstract

We propose a method based on the Fourier transform for numerically solving backward stochastic differential equations. Time discretization is applied to the forward equation of the state variable as well as the backward equation to yield a recursive system with terminal conditions. By assuming the integrability of the functions in the terminal conditions and applying truncation, the solutions of the system are shown to be integrable and we derive recursions in the Fourier space. The fractional FFT algorithm is applied to compute the Fourier and inverse Fourier transforms. We showcase the efficiency of our method through various numerical examples.

Appendices

Appendix A: Proofs

Proof Proposition 1

Consider the j th component of $\mathbb {E}_{m}^{\boldsymbol {x}}\left [v\left (t_{m+1}, \boldsymbol {X}_{m+1}^{\Delta }\right ) {\Delta } \boldsymbol {B}_{m+1}\right ]$, which can be calculated as follows:

$$ \begin{array}{@{}rcl@{}} &&\mathbb{E}_{m}^{\boldsymbol{x}}\left[v\left( t_{m+1},\boldsymbol{x}+\boldsymbol{\mu}{\Delta} t+\boldsymbol{\Sigma}{\Delta} \boldsymbol{B}_{m+1}\right) {\Delta} {B}_{m+1,j}\right]\\ \!&=&\!\frac{1}{(2\pi{\Delta} t)^{d/2}}\!{\int}_{\mathbb{R}^{d-1}}\!\exp\!\left( \!\!-\frac{{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}^{\intercal}{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}}{2{\Delta} t}\!\right)\!d{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}\!{\int}_{\mathbb{R}}v\!\left( t_{m+1},\boldsymbol{x} + \boldsymbol{\mu}{\Delta} t + \boldsymbol{\Sigma}\boldsymbol{\zeta}\right)\zeta_{j}\exp\!\left( \!-\frac{{\zeta_{j}^{2}}}{2{\Delta} t}\!\right)d\zeta_{j}\\ &=&\frac{\Delta t}{(2\pi{\Delta} t)^{d/2}}{\int}_{\mathbb{R}^{d-1}}\exp(-\frac{{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}^{\intercal}{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}}{2{\Delta} t})d{\boldsymbol{\zeta}_{[\boldsymbol{j}]}}\\ &&{\int}_{\mathbb{R}}\left( {\Sigma}_{1,j}\frac{\partial}{\partial x_{1}}+\cdots+{\Sigma}_{k,j}\frac{\partial}{\partial x_{k}} \right)v\left( t_{m+1},\boldsymbol{x}+\boldsymbol{\mu}{\Delta} t+\boldsymbol{\Sigma}\boldsymbol{\zeta}\right)\exp\left( -\frac{{\zeta_{j}^{2}}}{2{\Delta} t}\right)d\zeta_{j}\\ &=&{\Delta} t\boldsymbol{\tilde{\Sigma}}_{j,(\cdot)}\mathbb{E}_{m}^{\boldsymbol{x}}\left[ D_{\boldsymbol{x}}v(t_{m+1},\boldsymbol{X}^{\Delta}_{m+1}) \right], \end{array} $$

where ζ_[j] is the vector ζ without the j th coordinate and $\boldsymbol {\tilde {\Sigma }}_{j,(\cdot )}$ is the j th row vector of $\boldsymbol {\tilde {\Sigma }}$. The third equation is obtained from integration by parts. □

Proof Theorem 2

We prove the following two properties by mathematical induction:

• (1) $y^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$, $\boldsymbol {z}^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$,

• (2) $D_{\boldsymbol {x}}y^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$,

for any m = M,M − 1,⋯ , 0. The results obviously hold for m = M in light of Assumption 2 and Eq. 3.1. Next, for any m = M − 1,⋯ , 1, we assume the above properties hold at m + 1. We denote $\mathcal {S}^{k}(K)$ as a k-dimensional ball with radius K and recall that f^m+ 1(x) is truncated to 0 out of this sphere (throughout the proof f^m+ 1 denotes the truncated version). From Assumption 1 (3), f^m+ 1(x) is continuous on the compact sphere $\mathcal {S}^{k}(K)$ and hence also uniformly continuous. Consequently f^m+ 1(x) and D_xf^m+ 1(x) are integrable. Together we have

$$ y^{m+1}(\boldsymbol{x}), {z}^{m+1}(\boldsymbol{x}), f^{m+1}(\boldsymbol{x}), D_{\boldsymbol{x}}y^{m+1}(\boldsymbol{x}), D_{\boldsymbol{x}}f^{m+1}(\boldsymbol{x})\in L^{1}(\mathbb{R}^{k}). $$

Theorem 1 shows $\mathcal {P}_{\Delta }$ preserves integrability. Thus, we have

$$ P_{\Delta}y^{m+1}(\boldsymbol{x}), P_{\Delta}\boldsymbol{z}^{m+1}(\boldsymbol{x}), P_{\Delta}f^{m+1}(\boldsymbol{x}), P_{\Delta} D_{\boldsymbol{x}}y^{m+1}(\boldsymbol{x}), P_{\Delta} D_{\boldsymbol{x}}f^{m+1}(\boldsymbol{x})\in L^{1}(\mathbb{R}^{k}). $$

Therefore, we have ${z}_{j}^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$ since all the terms on the right hand side of Eq. 3.2 are integrable. We next go to Eq. 3.3, where

$$ y^{m}(\boldsymbol{x})=P_{\Delta}y^{m+1}(\boldsymbol{x})+{\Delta} t \theta_{1}f(t_{m},\boldsymbol{x},y^{m}(\boldsymbol{x}),\boldsymbol{z}^{m}(\boldsymbol{x}))+ {\Delta} t(1-\theta_{1})P_{\Delta}f^{m+1}(\boldsymbol{x}). $$

With a initial guess by $y^{m,0}(\boldsymbol {x})=P_{\Delta }y^{m+1}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$, the Picard iteration Eq. 3.5 generates a sequence of functions {y^m,p(x)}, which are also in $L^{1}(\mathbb {R}^{k})$. Moreover,

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\mathbb{R}^{k}}|y^{m,p+1}(\boldsymbol{x})-y^{m,p}(\boldsymbol{x})|d\boldsymbol{x}\\ &=&{\Delta} t\theta_{1} {\int}_{\mathbb{R}^{k}}| f(t_{m},\boldsymbol{x},y^{m,p}(\boldsymbol{x}),\boldsymbol{z}^{m}(\boldsymbol{x}))-f(t_{m},\boldsymbol{x},y^{m,p-1}(\boldsymbol{x}),\boldsymbol{z}^{m}(\boldsymbol{x})) |d\boldsymbol{x}\\ &\leq&{\Delta} t\theta_{1} L{\int}_{\mathbb{R}^{k}}|y^{m,p}(\boldsymbol{x})-y^{m,p-1}(\boldsymbol{x})|dx \end{array} $$

where the last inequality is obtained by Assumption 1 and L is the Lipschitz constant for f. Since Δt𝜃₁L < 1, we have created a Cauchy sequence in $L^{1}(\mathbb {R}^{k})$ which converges to a $L^{1}(\mathbb {R}^{k})$ limit. The limit solves the equation and thus it is the y^m(x) we seek. This shows $y^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$.

Next we analyze D_xy^m(x). We have for i = 1,⋯ ,k,

$$ D_{\boldsymbol{x},i}y^{m}(\boldsymbol{x})=D_{\boldsymbol{x},i}\left( P_{\Delta}y^{m+1}(\boldsymbol{x})\right) +{\Delta} t \theta_{1}D_{\boldsymbol{x},i}f^{m}(\boldsymbol{x})+ {\Delta} t(1-\theta_{1})D_{\boldsymbol{x},i}\left( P_{\Delta}f^{m+1}(\boldsymbol{x})\right) , $$

As f^m(x) is truncated, D_x,if^m(x) is integrable. We show P_Δy^m+ 1(x) is integrable and P_Δf^m+ 1(x) can be treated in a similar way. Note that

$$ P_{\Delta}y^{m+1}(\boldsymbol{x})={\int}_{\mathbb{R}^{k}} y^{m+1}(\boldsymbol{\delta})p_{\Delta}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}, $$

where p_Δ(δ; x) is the conditional probability density function for $X^{\Delta }_{m+1}$ conditioned on $X^{\Delta }_{m}=\boldsymbol {x}$ and it is given by

$$ {p_{\Delta}}(\boldsymbol{\delta};\boldsymbol{x})=\frac{1}{\sqrt{(2\pi)^{k}\det(\boldsymbol{\Sigma}\tilde{\boldsymbol{\Sigma}}{\Delta} t)}}\exp\left[ -\frac{1}{2}(\boldsymbol{\delta}-(\boldsymbol{x}+\boldsymbol{\mu}{\Delta} t))^{\intercal}(\boldsymbol{\Sigma}\tilde{\boldsymbol{\Sigma}}{\Delta} t)^{-1}(\boldsymbol{\delta}-(\boldsymbol{x}+\boldsymbol{\mu}{\Delta} t))\right]. $$

It is straightforward to show that

$$ \frac{\partial {p_{\Delta}}}{\partial \delta_{i}}(\boldsymbol{\delta};\boldsymbol{x})=-\frac{\partial {p_{\Delta}}}{\partial x_{i}}(\boldsymbol{\delta};\boldsymbol{x}). $$

Then,

$$ \begin{array}{@{}rcl@{}} D_{\boldsymbol{x},i}\left( P_{\Delta}y^{m+1}(\boldsymbol{x})\right) &=&{\int}_{\mathbb{R}^{k}} y^{m+1}(\boldsymbol{\delta})\frac{\partial {p_{\Delta}}}{\partial x_{i}}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}\\ &=&-{\int}_{\mathbb{R}^{k}} y^{m+1}(\boldsymbol{\delta})\frac{\partial {p_{\Delta}}}{\partial \delta_{i}}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}\\ &=&{\int}_{\mathbb{R}^{k-1}}\!\left[\!- y^{m+1}(\boldsymbol{\delta}){p_{\Delta}}(\boldsymbol{\delta};\boldsymbol{x})\big|_{\delta_{i}=-\infty}^{\infty} + D_{\boldsymbol{\delta},i}y^{m+1}(\boldsymbol{\delta}){p_{\Delta}}(\boldsymbol{\delta};\boldsymbol{x})d\delta_{i}\!\right]\!d\boldsymbol{\delta}_{[i]}\\ &=&{\int}_{\mathbb{R}^{k}} D_{\boldsymbol{\delta},i}y^{m+1}(\boldsymbol{\delta}){p_{\Delta}}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta},\\ &=&P_{\Delta}\left( D_{\boldsymbol{x},i} y^{m+1}(\boldsymbol{x})\right), \end{array} $$

(A.1)

where δ_[i] with subscript [i] denotes the vector without the i th coordinate from δ. Since $D_{\boldsymbol {x},i} y^{m+1}\in L^{1}(\mathbb {R}^{k})$ and P_Δ preserves integrability, we know $D_{\boldsymbol {x},i}\left (P_{\Delta }y^{m+1}(\boldsymbol {x})\right )$ is also integrable from Eq. A.1. This shows $D_{\boldsymbol {x},i}y^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$, which completes the proof. □

Proof Lemma 1

The first part is obtained in Bertoin (1996). For the second part, we have

$$ \mathcal{F}_{k}\left( P_{\Delta}D_{\boldsymbol{x},j}v(\boldsymbol{x})\right)(\boldsymbol{\xi})=\phi_{\Delta}(-\boldsymbol{\xi})\mathcal{F}_{k}\left( D_{\boldsymbol{x},j}v(\boldsymbol{x})\right)(\boldsymbol{\xi}) $$

Using integration by parts, we obtain

$$ \begin{array}{@{}rcl@{}} &&\mathcal{F}_{k}\left( D_{\boldsymbol{x},j}v(\boldsymbol{x})\right)(\boldsymbol{\xi})\\ &=&{\int}_{\mathbb{R}^{k-1}}e^{i\boldsymbol{\xi}^{\intercal}_{[j]} \boldsymbol{x}_{[j]}}{\int}_{\mathbb{R}} e^{i{\xi}_{j}{x}_{j}}D_{\boldsymbol{x},j}v(\boldsymbol{x})dx_{j}d\boldsymbol{x}_{[j]}\\ &=&{\int}_{\mathbb{R}^{k-1}}e^{i\boldsymbol{\xi}^{\intercal}_{[j]} \boldsymbol{x}_{[j]}} \left( e^{i\xi_{j} x_{j}}v(\boldsymbol{x})\bigg|_{x_{j}=-\infty}^{\infty}-i\xi_{j}{\int}_{\mathbb{R}}e^{i\xi_{j} x_{j}}v(\boldsymbol{x})dx_{j} \right) d\boldsymbol{x}_{[j]}, \end{array} $$

where x_[j] is x without the j th coordinate and so is ξ_[j]. Since $v(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$, the first term in the parenthesis is 0 and we obtain the result in the statement. □

Proof Theorem 4

We prove by mathematical induction. At m = M, since $\boldsymbol {\tilde {\Sigma }}(\boldsymbol {x})$ is Lipschitz continuous in Assumption 1 and truncated outside a k-dimensional ball $\mathcal {S}^{k}(K)$, it is easy to see that y^M(x) and z^M(x) are integrable. Next, for any m = M − 1,⋯ , 1, we assume the $y^{m+1}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$ and ${\boldsymbol {z}}^{m+1}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$. In general, unlike Theorem 2, we no longer have $D_{\boldsymbol {x}}y^{m+1}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$ for the general case, and below we will use different arguments to prove P_ΔD_xy^m+ 1(x) is bounded.

Note that the weak derivative D_x,jy^m+ 1(x) is in $L^{1}_{loc}(\mathbb {R}^{k})$ and as y^m+ 1(x) is integrable, we can find a sufficient large region Ω, such that for any x ∈Ω^c, |D_x,jy^m+ 1(x)| < C. We have

$$ \begin{array}{@{}rcl@{}} |P_{\Delta}D_{\boldsymbol{x},j}y^{m+1}(\boldsymbol{x})|&=&\left|{\int}_{\mathbb{R}^{k}} D_{\boldsymbol{x},j}y^{m+1}(\boldsymbol{\delta})p_{\Delta}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}\right|\\ &\leq& \left|{\int}_{\Omega}D_{\boldsymbol{x},j}y^{m+1}(\boldsymbol{\delta})p_{\Delta}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}\right|+\left|{\int}_{{\Omega}^{\mathsf{c}}}D_{\boldsymbol{x},j}y^{m+1}(\boldsymbol{\delta})p_{\Delta}(\boldsymbol{\delta};\boldsymbol{x})d\boldsymbol{\delta}\right| \end{array} $$

where p_Δ(δ; x) is the conditional probability density function for $X^{\Delta }_{m+1}$ conditioned on $X^{\Delta }_{m}=\boldsymbol {x}$ with state dependent parameters μ(x) and Σ(x). Since D_x,jy^m+ 1(x) is in $L^{1}_{loc}(\mathbb {R}^{k})$ and p_Δ is a density function of a normal distribution, the first absolute value is bounded. Furthermore, |D_x,jy^m+ 1(x)| < C for any x ∈Ω^c and the integral for the density function is less than 1, thus the second absolute value is also bounded. Consequently, P_ΔD_x,jy^m+ 1(x) is bounded. Multiplying it with the truncated $\boldsymbol {\tilde {\Sigma }}(\boldsymbol {x})$, we obviously have $\boldsymbol {\tilde {\Sigma }}(\boldsymbol {x})P_{\Delta }D_{\boldsymbol {x}}y^{m+1}(\boldsymbol {x}) \in L^{1}(\mathbb {R}^{k})$. As f^m+ 1(x) is truncated, f^m+ 1(x) and D_x,if^m+ 1(x) are in $L^{1}(\mathbb {R}^{k})$. Using that $\mathcal {P}_{\Delta }$ preserves integrability, the induction assumption and Eq. 4.2, we obtain $\boldsymbol {z}^{m}(\boldsymbol {x})\in L^{1}(\mathbb {R}^{k})$. The integrability of y^m(x) is shown using the same arguments as in Theorem 2. □

Appendix B: Fractional Fast Fourier Transform (FrFFT)

1.1 B.1 One-Dimensional Fractional Fast Fourier Transform

Consider calculating the integral numerically:

$$ \begin{array}{@{}rcl@{}} \hat{f}(\xi_{\tilde{n}})&=&{{\int}_{a}^{b}}\exp{(i\xi_{\tilde{n}}x)} f(x)dx\\ &\approx& h_{x} \sum\limits_{n=0}^{N-1}\exp(i\xi_{\tilde{n}}x_{n})f(x_{n}) \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} x_{n}=a+ nh_{x}, \quad h_{x}&=\frac{b-a}{N}, \quad n=0,\cdots,N-1\\ \xi_{\tilde{n}}=\tilde{a}+ \tilde{n}h_{\xi}, \quad h_{\xi}&=\frac{\tilde{b}-\tilde{a}}{N}, \quad \tilde{n}=0,\cdots,N-1. \end{array} $$

Hence,

$$ \begin{array}{@{}rcl@{}} \hat{f}(\xi_{\tilde{n}})&\approx& h_{x} \sum\limits_{n=0}^{N-1}\exp(i(\tilde{a}+\tilde{n}h_{\xi})(a+nh_{x}))f(x_{n})\\ &=&h_{x} e^{ia\tilde{a}}e^{ia\tilde{n}h_{\xi}}\sum\limits_{n=0}^{N-1}e^{i\tilde{a}nh_{x}}e^{in\tilde{n}h_{\xi}h_{x}}f(x_{n}) \end{array} $$

Setting $f^{*}_{n}=e^{i\tilde {a}nh_{x}}f(x_{n})$ and $\theta =-\frac {h_{x}h_{\xi }}{2\pi }$, the equation above can be rewritten as follows:

$$ \hat{f}(\xi_{\tilde{n}})=h_{x} e^{ia\tilde{a}}e^{ia\tilde{n}h_{\xi}}\sum\limits_{n=0}^{N-1}e^{-2\pi in\tilde{n} \theta}f^{*}_{n}, \quad \tilde{n}=0,\cdots,N-1, $$

which is in the form of fractional discrete Fourier transform. The sum for all points on the ξ-grid can be computed efficiently using the fractional fast Fourier transform algorithm (see Bailey and Swarztrauber (1991)). We need to define a vector

$$ c=\left( e^{i\pi0^{2}\theta},e^{i\pi1^{2}\theta},\cdots,e^{i\pi(N-1)^{2}\theta},0,\cdots,0,e^{i\pi(N-1)^{2}\theta},\cdots e^{i\pi1^{2}\theta}\right)^{\intercal} $$

where c is a N^∗× 1 vector and N^∗ is the first power of 2 such that $N^{*}\geqslant 2N-1$. Denote

$$ \begin{array}{@{}rcl@{}} g_{n}&=&e^{-i\pi n^{2}\theta}f^{*}_{n},\quad n=0,\cdots,N-1\\ g^{*}&=&\left( g_{0},g_{1},\cdots,g_{N-1},0,0\cdots,0\right), \end{array} $$

where g^∗ is also a N^∗× 1 vector. Then we may numerically calculate the Fourier transform $\hat {f}$ by

$$ \hat{f}(\xi_{\tilde{n}})=h_{x}e^{ia\tilde{a}}e^{ia\tilde{n}h_{\xi}}e^{-i\pi \tilde{n}^{2}\theta}\left( \mathbb{F}^{-1}\left( \mathbb{F}(c)\cdot \mathbb{F}(g^{*}) \right) \right)_{\tilde{n}}, \quad \tilde{n}=0,1,\cdots,N-1, $$

which can reduce the computational complexity from $\mathcal {O}(N^{2})$ to $\mathcal {O}(N\log N)$.

1.2 B.2 Multidimensional Fractional Fast Fourier Transform

Consider the k-dimensional integral

$$ \begin{array}{@{}rcl@{}} \hat{f}(\boldsymbol{\xi}_{\boldsymbol{\tilde{n}}})&=&{\int}_{[\boldsymbol{a},\boldsymbol{b}]}e^{i\boldsymbol{\xi}_{\boldsymbol{\tilde{n}}}^{\intercal}\boldsymbol{x}}f(\boldsymbol{x})d\boldsymbol{x}\\ &\approx& h_{x_{1}}\cdots h_{x_{k}}\sum\limits_{n_{1}=0}^{N-1}\cdots\sum\limits_{n_{k}=0}^{N-1}e^{i(\xi^{1}_{\tilde{n}_{1}}x^{1}_{n_{1}}+\cdots+\xi^{k}_{\tilde{n}_{k}}x^{k}_{n_{k}})}f(x^{1}_{n_{1}},\cdots,x^{k}_{n_{k}}), \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} x^{i}_{n_{i}}=a_{i}+ n_{i}h_{x_{i}}, \quad h_{x_{i}}&=\frac{b_{i}-a_{i}}{N}, \quad n_{i}=0,\cdots,N-1, \\ \xi^{i}_{\tilde{n}_{i}}=\tilde{a}_{i}+ \tilde{n}_{i}h_{\xi_{i}}, \quad h_{\xi_{i}}&=\frac{\tilde{b}_{i}-\tilde{a}_{i}}{N}, \quad \tilde{n}_{i}=0,\cdots,N-1. \end{array} $$

To evaluate this k-dimensional sum, we can proceed as follows.

$$ \hat{f}(\xi^{1}_{\tilde{n}_{1}},\cdots,\xi^{k}_{\tilde{n}_{k}})\approx h_{x_{1}}\sum\limits_{n_{1}=0}^{N-1}e^{i\xi^{1}_{\tilde{n}_{1}}x^{1}_{n_{1}}}g(x^{1}_{n_{1}},\xi^{2}_{\tilde{n}_{2}},\cdots,\xi^{k}_{\tilde{n}_{k}}), $$

(B.1)

where

$$ g(x^{1}_{n_{1}},\xi^{2}_{\tilde{n}_{2}},\cdots,\xi^{k}_{\tilde{n}_{k}})=h_{x_{2}}{\cdots} h_{x_{k}}\sum\limits_{n_{2}=0}^{N-1}\cdots\sum\limits_{n_{k}=0}^{N-1}e^{i(\xi^{2}_{\tilde{n}_{2}}x^{2}_{n_{2}}+\cdots+\xi^{k}_{\tilde{n}_{k}}x^{k}_{n_{k}})}f(x^{1}_{n_{1}},x^{2}_{n_{2}},\cdots,x^{k}_{n_{k}}). $$

(B.2)

If we are given $g(x^{1}_{n_{1}},\xi ^{2}_{\tilde {n}_{2}},\cdots ,\xi ^{k}_{\tilde {n}_{k}})$, we can apply the one-dimensional FrFFT to compute Eqs. B.1, and B.2 is a (k − 1)-dimensional sum, which we can break down to a one-dimensional sum of functions defined by a (k − 2)-dimensional sum like Eq. B.2. Repeating this procedure would allow us to reduce the multidimensional problem into a sequence of one-dimensional problems, each of which can be computed by the one-dimensional FrFFT.

Let $\mathcal {O}(F_{k})$ be the computational complexity of the k-dimensional problem (k ≥ 2). Then, We have the following recursion

$$ \mathcal{O}(F_{k})=N\mathcal{O}(F_{k-1})+N^{k-1}\mathcal{O}(N\log(N)). $$

Since $\mathcal {O}(F_{1})=\mathcal {O}(N\log (N))$, we obtain $\mathcal {O}(F_{k})=\mathcal {O}(N^{k}\log (N))$.

Appendix C: Least Squares Monte Carlo (LSM) regression

We simulate X_t using the Euler scheme. Using the approximations in Bouchard and Touzi (2004) and Zhang et al. (2004), we have a backward scheme given by:

$$ \begin{array}{@{}rcl@{}} Y_{M}^{\Delta}&=&g\left( \boldsymbol{X}_{M}^{\Delta}\right), \end{array} $$

(C.1)

$$ \begin{array}{@{}rcl@{}} Z_{m,j}^{\Delta}&=&\frac{1}{\Delta t} \mathbb{E}_{m}\left[Y_{m+1}^{\Delta} {\Delta} {B}_{m+1,j}\right], \quad j=1,2,\cdots,d, \end{array} $$

(C.2)

$$ \begin{array}{@{}rcl@{}} Y_{m}^{\Delta}&=&\mathbb{E}_{m}\left[Y_{m+1}^{\Delta}+{\Delta} t f({t_{m}},\boldsymbol{X}_{m}^{\Delta},Y_{{m+1}}^{\Delta},\boldsymbol{Z}_{{m}}^{\Delta}) \right],\ \end{array} $$

(C.3)

where m = M − 1,…, 0. As $\boldsymbol {X}_{{m}}^{\Delta }$ is Markovian, there exist functions $y_{m}^{\Delta }(\boldsymbol {x})$ and $z_{m,j}^{\Delta }(\boldsymbol {x})$ such that

$$ Y_{m}^{\Delta}=y_{m}^{\Delta}(\boldsymbol{X}_{m}^{\Delta}), Z_{m,j}^{\Delta}=z_{m,j}^{\Delta}(\boldsymbol{X}_{m}^{\Delta}), \quad j=1,\cdots,d. $$

The conditional expectations can be computed by the least squares Monte Carlo method. Consider d + 1 sets of basis functions:

$$ \boldsymbol{\eta}_{0}(m, \boldsymbol{x})=\left( \eta_{0,1}(m, \boldsymbol{x}), \ldots, \eta_{0, K}(m, \boldsymbol{x})\right)^{\intercal} $$

(C.4)

for estimating $y_{m}^{\Delta }(\boldsymbol {x})$ and

$$ \boldsymbol{\eta}_{j}(m, \boldsymbol{x})=\left( \eta_{j,1}(m, \boldsymbol{x}), \ldots, \eta_{j, K}(m, \boldsymbol{x})\right)^{\intercal},\quad j=1,\cdots, d $$

(C.5)

for estimating $z_{m,j}^{\Delta }(\boldsymbol {x})$. After generating L independent samples of $({\Delta }{\boldsymbol {B}}_{m,\lambda }, {\boldsymbol {X}}^{\Delta }_{m,\lambda })_{m=1,}$ ⋯ ,M, λ = 1,⋯ ,L, we can follow Eqs. C.1, C.2 and C.3 to approximate $y_{m}^{\Delta }(\boldsymbol {x})$ and $z_{m,j}^{\Delta }(\boldsymbol {x})$ as

$$ \begin{array}{@{}rcl@{}} \tilde{y}_{M}^{\Delta, K, L}(\boldsymbol{x})&=&g(\boldsymbol{x}),\\ \tilde{z}_{m,j}^{\Delta, K, L}(\boldsymbol{x})&=&\boldsymbol{\eta}_{j}(m, \boldsymbol{x})^{\intercal} \boldsymbol{\alpha}_{m,j}^{\Delta, K, L}, \quad j=1, \ldots, d, \end{array} $$

(C.6)

$$ \begin{array}{@{}rcl@{}} \tilde{y}_{m}^{\Delta, K, L}(\boldsymbol{x})&=&\boldsymbol{\eta}_{0}(m, \boldsymbol{x})^{\intercal} \boldsymbol{\alpha}_{m, 0}^{\Delta, K, L}, \end{array} $$

(C.7)

for m = M − 1,…, 0,where

$$ \begin{array}{@{}rcl@{}} \boldsymbol{\alpha}_{m, j}^{\Delta, K, L}&=&\arg \min\limits_{\boldsymbol{\alpha} \in \mathbb{R}^{K}} \frac{1}{L} \sum\limits_{\lambda=1}^{L}\left[\boldsymbol{\eta}_{j}\left( m, {\boldsymbol{X}}^{\Delta}_{m,\lambda}\right)^{\intercal} \boldsymbol{\alpha}-\frac{\Delta{B}_{m+1,j,\lambda}}{\Delta t} \tilde{y}_{m+1}^{\Delta, K, L}\left( {\boldsymbol{X}}_{m+1,\lambda}^{\Delta}\right)\right]^{2} \end{array} $$

(C.8)

$$ \begin{array}{@{}rcl@{}} \boldsymbol{\alpha}_{m, 0}^{\Delta, K, L}&=&\arg \min\limits_{\boldsymbol{\alpha} \in \mathbb{R}^{K}} \frac{1}{L} \sum\limits_{\lambda=1}^{L}\left[\boldsymbol{\eta}_{0}\left( m, {\boldsymbol{X}}_{m,\lambda}^{\Delta}\right)^{\intercal} \boldsymbol{\alpha}-\tilde{y}_{m+1}^{\Delta, K, L}\left( {\boldsymbol{X}}_{m+1,\lambda}^{\Delta}\right)\right.\\ &&\left.-{\Delta} tf\left( t_{m}, {\boldsymbol{X}}_{m,\lambda}^{\Delta}, \tilde{y}_{m+1}^{\Delta, K, L}\left( {\boldsymbol{X}}_{m+1,\lambda}^{\Delta}\right), \tilde{\boldsymbol{z}}_{m}^{\Delta, K, L}\left( {\boldsymbol{X}}_{m,\lambda}^{\Delta}\right)\right) \right]^{2}. \end{array} $$

(C.9)

In general, there are various ways to choose the basis functions in Eqs. C.4 and C.5. Here, we consider the hypercube indicator basis functions used in Gobet et al. (2005). We choose a domain $H\subset \mathbb {R}^{k}$ centered on x₀ = (x_0,1,⋯ ,x_0,k) as

$$ H=[x_{0,1}-q_{1},x_{0,1}+q_{1}]\times\cdots\times[x_{0,k}-q_{k},x_{0,k}+q_{k}] $$

for some positive numbers q₁,⋯ ,q_k. Then we divide this domain H into K equal hypercubes, i.e., H = ∪_{k= 1,⋯ ,K}H_k. Thus we have the following indicator basis function:

where j = 0, 1,⋯ ,d and k = 1,⋯ ,K. We summarize the algorithm as follows:

In our implementation, we follow Lemor et al. (2006) to set the parameters:

$$ M=2\sqrt{2}^{J-1},\ L=2\sqrt{2}^{3(J-1)},\ K=(2q/\delta)^{k},\ \delta=C_{d}/\sqrt{2}^{J-1}, $$

(C.10)

where J is any positive integer, and C_d is a constant and we simply set q₁ = ⋯ = q_k = q. We also run Algorithm 3 for $\mathcal {R}$ times and use the average of their results to obtain more accurate approximations for the solutions. The computation time grows linearly in MLK.

Appendix D: The Backward COS Method

We summarize the BCOS method of Ruijter and Oosterlee (2015a) for the one-dimensional problem, which is the case considered in their paper. Although the results can be extended to multiple dimensions, we refrain from presenting them which involve more notations and they are not really needed for our numerical examples. Let [a,b] be a sufficiently large interval. Using the COS method, the conditional expectations are approximated as:

$$ \begin{array}{@{}rcl@{}} \mathbb{E}_{m}^{\boldsymbol{x}}\left[v\left( t_{m+1}, \boldsymbol{X}_{m+1}^{\Delta}\right)\right]&\approx&\frac{b-a}{2} {\sum\limits_{k=0}^{N-1}}{\prime} \mathcal{V}_{k}\left( t_{m+1}\right) {\Phi}_{k, m}(x),\\ \mathbb{E}_{m}^{\boldsymbol{x}}\left[v\left( t_{m+1}, \boldsymbol{X}_{m+1}^{\Delta}\right) {\Delta} \boldsymbol{B}_{m+1}\right] &=& \sigma\left( t_{m}, x\right) {\Delta} t \mathbb{E}_{m}^{x}\left[D_{x} v\left( t_{m+1}, X_{m+1}^{\Delta}\right)\right]\\ &\approx& \sigma\left( t_{m}, x\right) {\Delta} t \frac{b-a}{2} \sum\limits_{k=0}^{N-1}{\prime} \mathcal{V}_{k}\left( t_{m+1}\right) {\Phi}_{k,m}^{\prime}(x), \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} \mathcal{V}_{k}\left( t_{m+1}\right) &:=&\frac{2}{b-a} {{\int}_{a}^{b}} v\left( t_{m+1}, x \right) \cos \left( k \pi \frac{x-a}{b-a}\right) d x,\\ {\Phi}_{k,m}(x)&:=&\frac{2}{b-a} \Re\left( \phi_{\Delta}\left( \frac{k \pi}{b-a} ; x\right) e^{i k \pi \frac{x-a}{b-a}}\right), \end{array} $$

(D.1)

${\Phi }_{k,m}^{\prime }(x)$ is the derivative of Φ_k,m(x), ${\sum }{\prime }$ means the first term in the summation is 1/2, R(⋅) is the real part of a complex value, and ϕ_Δ(⋅; x) is the conditional characteristic function of $X^{\Delta }_{m+1}-X^{\Delta }_{m}$ given $X^{\Delta }_{m}=x$.

The COS series approximations can be applied to compute the conditional expectations in Eq. 2.3 to Eq. 2.5, and we must know $\mathcal {Y}_{k}\left (t_{m+1}\right )$, $\mathcal {Z}_{k}\left (t_{m+1}\right )$ and $\mathcal {F}_{k}\left (t_{m+1}\right )$ defined as

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!\!\mathcal{Y}_{k}\left( t_{m+1}\right)& = &\frac{2}{b-a} {{\int}_{a}^{b}} y\left( t_{m+1}, x\right) \cos \left( k \pi \frac{x-a}{b-a}\right) d x, \end{array} $$

(D.2)

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!\!\mathcal{Z}_{k}\left( t_{m+1}\right)& = &\frac{2}{b-a} {{\int}_{a}^{b}} z\left( t_{m+1}, x\right) \cos \left( k \pi \frac{x-a}{b-a}\right) d x, \end{array} $$

(D.3)

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!\!\mathcal{F}_{k}\left( t_{m+1}\right)& = &\frac{2}{b-a} {{\int}_{a}^{b}}\! f\left( t_{m+1}, x, y\left( t_{m+1}, x\right), z\left( t_{m+1}, x\right)\right) \cos \left( k \pi \frac{x - a}{b - a}\right) d x. \end{array} $$

(D.4)

Thus, from Eq. 2.4, z(t_m,x) can be approximated as

$$ \begin{array}{@{}rcl@{}} z\left( t_{m}, x\right) &\approx&-\frac{1-\theta_{2}}{\theta_{2}} \frac{b-a}{2} \sum\limits_{k=0}^{N-1}{\prime} \mathcal{Z}_{k}\left( t_{m+1}\right) {\Phi}_{k,m}(x) \\ &&+\frac{b-a}{2} \sum\limits_{k=0}^{N-1}{\prime}\left( \frac{1}{\Delta t \theta_{2}} \mathcal{Y}_{k}\left( t_{m+1}\right)+\frac{1-\theta_{2}}{\theta_{2}} \mathcal{F}_{k}\left( t_{m+1}\right)\right) \sigma\left( t_{m}, x\right) {\Delta} t {\Phi}_{k,m}^{\prime}(x). \end{array} $$

(D.5)

From Eq. 2.5, we have

$$ y\left( t_{m}, x\right)={\Delta} t \theta_{1} f\left( t_{m}, x, y\left( t_{m}, x\right), z\left( t_{m}, x\right)\right)+h\left( t_{m}, x\right), $$

(D.6)

where

$$ \begin{array}{@{}rcl@{}} h\left( t_{m}, x\right) &:=&\mathbb{E}_{m}^{x}\left[Y_{m+1}^{\Delta}\right]+{\Delta} t\left( 1-\theta_{1}\right) \mathbb{E}_{m}^{x}\left[f\left( t_{m+1}, \mathbf{X}_{m+1}^{\Delta}\right)\right] \\ & \approx& \frac{b-a}{2} \sum\limits_{k=0}^{N-1}{\prime}\left( \mathcal{Y}_{k}\left( t_{m+1}\right)+{\Delta} t\left( 1-\theta_{1}\right) \mathcal{F}_{k}\left( t_{m+1}\right)\right) {\Phi}_{k,m}(x). \end{array} $$

(D.7)

Integrals like Eq. D.1 are approximated using the mid-point rule as

$$ \begin{array}{@{}rcl@{}} \mathcal{V}_{k}\left( t_{m+1}\right) &\approx& \sum\limits_{n=0}^{N-1} \frac{2}{b-a} v\left( t_{m+1}, x_{n}\right) \cos \left( k \pi \frac{x_{n}-a}{b-a}\right) {\Delta} x\\ &=&\sum\limits_{n=0}^{N-1} v\left( t_{m+1}, x_{n}\right) \cos \left( k \pi \frac{2 n+1}{2 N}\right) \frac{2}{N} \end{array} $$

(D.8)

for k = 0, 1,⋯ ,N − 1, where

$$ x_{n}:=a+(n+\frac{1}{2}){\Delta} x, \quad {\Delta} x=\frac{b-a}{N}, $$

for n = 0, 1,⋯ ,N − 1. The sum Eq. D.8 is a discrete cosine transform (DCT), which can be computed by FFT with complexity $\mathcal {O}\left (N\log N\right )$.

1.1 D.1 Constant Drift μ and Volatility σ

In this case, ϕ_Δ(⋅; x) is independent of x. From Eq. D.5, we obtain

$$ \begin{array}{@{}rcl@{}} \mathcal{Z}_{k}\left( t_{m}\right) &=&\frac{2}{b-a} {{\int}_{a}^{b}} z\left( t_{m}, x\right) \cos \left( k \pi \frac{x-a}{b-a}\right) d x\\ & \approx& \Re\!\left( \sum\limits_{j=0}^{N-1}{\prime}\!\left[\!-\frac{1 - \theta_{2}}{\theta_{2}} \mathcal{Z}_{j}\!\left( t_{m+1}\right) + \frac{i j \pi}{b - a} \sigma {\Delta} t\!\left( \!\frac{1}{\Delta t \theta_{2}} \mathcal{Y}_{j}\left( t_{m+1}\right) + \frac{1 - \theta_{2}}{\theta_{2}} \mathcal{F}_{j}\left( t_{m+1}\right)\right)\!\right]\right. \\ && \left.\phi_{\Delta}\left( \frac{j \pi}{b-a}\right) \mathcal{M}_{k, j}\right), \end{array} $$

(D.9)

for k = 0, 1,⋯ ,N − 1, where

$$ \begin{array}{@{}rcl@{}} \phi_{\Delta}(u)&=&\exp\left( iu\mu{\Delta} t-\frac{1}{2}u^{2}\sigma^{2}{\Delta} t\right),\\ \mathcal{M}_{k, j} &:=&\frac{2}{b-a} {{\int}_{a}^{b}} e^{i j \pi \frac{x-a}{b-a}} \cos \left( k \pi \frac{x-a}{b-a}\right) d x. \end{array} $$

Similarly, from Eq. D.7 we have

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!\!\!\mathcal{H}_{k}\left( t_{m}\right) &=&\!\frac{2}{b-a} {{\int}_{a}^{b}} h\left( t_{m}, x\right) \cos \left( k \pi \frac{x-a}{b-a}\right) d x \\ & \approx&\! \Re\!\left( \sum\limits_{j=0}^{N-1}{\prime}\left[\mathcal{Y}_{j}\left( t_{m+1}\right) + {\Delta} t\left( 1 - \theta_{1}\right) \mathcal{F}_{j}\left( t_{m+1}\right)\right] \phi_{\Delta}\!\left( \!\frac{j \pi}{b - a}\right) \mathcal{M}_{k, j}\!\right). \end{array} $$

(D.10)

These Fourier cosine coefficients can be computed using FFT (see Fang and Oosterlee (2009)) and the complexity of obtaining all N coefficients is $\mathcal {O}\left (N\log N\right )$ instead of $\mathcal {O}\left (N^{2}\right )$.

To obtain y(t_m,x), we solve the equation given by Eq. D.6 by doing Picard iterations P times. We use the solution to calculate $\mathcal {F}_{k}\left (t_{m}\right )$ defined by Eq. D.4 via Eq. D.8 and then calculate $\mathcal {Y}_{k}\left (t_{m}\right )$ as

$$ \mathcal{Y}_{k}\left( t_{m}\right) = {\Delta} t \theta_{1} \mathcal{F}_{k}\left( t_{m}\right)+\mathcal{H}_{k}\left( t_{m}\right). $$

We summarize the algorithm as follows:

The complexity is as follows: $O(N\log (N))$ for the initialization, and for each step of recursion, $O(N\log (N))$ for (i), O(PN) for (ii) and $O(N\log (N))$ for (iii). The total complexity is $\mathcal {O}\left (M\left (N \log N+P N\right )\right )$.

1.2 D2. General Drift μ(x) and Volatility σ(x)

In the general case, ϕ_Δ(u; x) depends on x. With this change, in Step (i) of the recursion, $z\left (t_{m}, x\right )$ and $h\left (t_{m}, x\right )$ can no longer be computed with FFT and now it requires $\mathcal {O}\left (N^{2}\right )$ operations. Moreover, Eqs. D.9 and D.10 no longer hold, but DCT can still be applied in Step (iii) of the recursion to calculate $\mathcal {Z}_{k}(t_{m})$ and ${\mathscr{H}}_{k}(t_{m})$ with cost $\mathcal {O}\left (N\log (N)\right )$ by discretizing the integrals that define them (see Eq. D.8). The total computation complexity for this algorithm is $\mathcal {O}\left (M\left (N^{2}+N \log N+P N\right )\right )$.