Multidimensional Forward and Backward Equations

Two Directions of Evolution

A diffusion has two related descriptions. Starting from a state $x$, the backward equation describes how the expected value of a future payoff changes with the starting time and state. The forward equation describes how the distribution of the process changes over time. In several dimensions, the distinction matters: the backward operator differentiates the payoff in the starting state, while the forward operator differentiates the evolving density in the state variable where probability is observed.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying an $m$-dimensional standard Brownian motion $\mathbf{B}_t$, whose coordinates are independent. Consider a time-homogeneous diffusion in $\mathbb{R}^d$:

dX_t=b(X_t)\,dt+\sigma(X_t)\,d\mathbf{B}_t,\qquad X_0=x.

Here $b:\mathbb{R}^d\to\mathbb{R}^d$ is the drift vector, $\sigma:\mathbb{R}^d\to\mathbb{R}^{d\times m}$ is the diffusion coefficient, and the diffusion covariance matrix is $a(x)=\sigma(x)\sigma(x)^{\mathsf T}$. We use the usual Ito interpretation: $dB_t^k\,dB_t^\ell=\mathbf{1}_{\{k=\ell\}}dt$, $dB_t^k\,dt=0$, and $dt\,dt=0$. This is not the Stratonovich convention; converting a Stratonovich equation would change its drift.

For every finite $T$, Brownian integrals below are Ito integrals of progressively measurable matrix-valued integrands $H_s$ satisfying $\mathbb{E}\int_0^T\|H_s\|_{\mathrm F}^2\,ds<\infty$, where $\|\cdot\|_{\mathrm F}$ is the Frobenius norm. In particular, for the SDE we assume this integrability for $\sigma(X_s)$; standard global Lipschitz and linear growth hypotheses on $b,\sigma$ are sufficient when the initial state has finite second moment. Under those hypotheses the standard strong-existence and pathwise-uniqueness theorem applies; its proof is not needed for the operator calculations here. All expectations in this tutorial are under $\mathbb{P}$.

Definition: For a twice continuously differentiable function $f:\mathbb{R}^d\to\mathbb{R}$, the diffusion generator is $$ \mathcal{L}f(x)=\sum_{i=1}^d b_i(x)\,\partial_i f(x) +\frac12\sum_{i,j=1}^d a_{ij}(x)\,\partial_{ij}f(x). $$ The matrix $a$ is symmetric and nonnegative definite. Thus the second-order term records covariance, including correlations between coordinates.

The Backward Equation

Let $P_t f(x)=\mathbb{E}[f(X_t)\mid X_0=x]$ be the transition semigroup, and assume the coefficients and function class give the differentiability needed below. For terminal payoff $f$, set $u(t,x)=P_{T-t}f(x)$, for $0\le t\le T$. The backward equation is $\partial_tu+\mathcal{L}u=0$, with terminal condition $u(T,x)=f(x)$.

Theorem: Suppose the diffusion semigroup is differentiable on the functions being considered, its generator is $\mathcal{L}$, and $u(t,\cdot)=P_{T-t}f$ has the derivatives appearing in $\mathcal{L}$. Then $$ \partial_tu(t,x)+\mathcal{L}u(t,x)=0,\qquad u(T,x)=f(x). $$

Proof. Write $P_hg(x)=\mathbb{E}[g(X_h)\mid X_0=x]$. The generator identity means $\lim_{h\downarrow0}(P_hg-g)/h=\mathcal{L}g$ on the stated function class. By the semigroup property, for $h>0$, \[ u(t-h,x)=P_{T-t+h}f(x)=P_h(P_{T-t}f)(x)=P_hu(t,\cdot)(x). \] Consequently, the left time derivative is \[ \partial_tu(t,x)=\lim_{h\downarrow0}\frac{u(t,x)-u(t-h,x)}{h} =-\lim_{h\downarrow0}\frac{P_hu(t,\cdot)(x)-u(t,x)}{h} =-\mathcal{L}u(t,x). \] At $t=T$, $P_0f=f$, which gives the terminal condition. These limits use the assumed differentiability and generator identity. $\square$

The generator itself follows from Ito's formula. For a $C^2$ function $g$, the first derivatives multiply the drift and the second derivatives multiply the quadratic covariation $dX^i_t\,dX^j_t=a_{ij}(X_t)dt$. In particular, the coefficient is $\tfrac12 a_{ij}$, not $\tfrac12\sigma_{ij}$; it is the covariance matrix that enters the PDE.

The Forward Equation and the Adjoint

The forward statement can first be expressed without assuming a density. Let $\varphi$ be a smooth, compactly supported test function. Ito's formula and expectation give \[ \mathbb{E}[\varphi(X_t)]-\mathbb{E}[\varphi(X_0)] =\int_0^t\mathbb{E}[\mathcal{L}\varphi(X_s)]\,ds, \] provided the terms are integrable. For example, bounded coefficients and bounded first and second derivatives of $\varphi$ suffice. The stochastic integral has integrand $\nabla\varphi(X_s)^{\mathsf T}\sigma(X_s)$, which is progressively measurable and square-integrable on finite intervals under these conditions. By the Ito isometry it is square-integrable; its conditional mean-zero increment property, first for adapted step integrands and then by $L^2$ approximation, makes it a true martingale. Thus its expectation is zero. The forward equation in this test-function sense is often called the weak form.

Definition: The formal adjoint $\mathcal{L}^*$ is the operator satisfying $\int q\,\mathcal{L}f\,dx=\int f\,\mathcal{L}^*q\,dx$ for smooth compactly supported $f,q$, whenever integration by parts has no boundary terms. For the diffusion generator, $$ \mathcal{L}^*q =-\sum_{i=1}^d\partial_i(b_iq) +\frac12\sum_{i,j=1}^d\partial_{ij}(a_{ij}q). $$

Proof. For the drift term, integration by parts once gives \[ \int_{\mathbb{R}^d}q\,b_i\,\partial_i f\,dx =-\int_{\mathbb{R}^d}f\,\partial_i(b_iq)\,dx. \] For each second-order term, integration by parts twice gives \[ \int_{\mathbb{R}^d}q\,a_{ij}\,\partial_{ij}f\,dx =\int_{\mathbb{R}^d}f\,\partial_{ij}(a_{ij}q)\,dx. \] There are no boundary terms because the test functions are compactly supported. Multiplying the second identity by $\tfrac12$, summing over $i,j$, and combining the drift terms proves the displayed expression. The argument includes $d=1$, and if $a$ is constant it reduces to the familiar constant-coefficient second derivatives. $\square$

If $X_t$ has a sufficiently smooth density $p(t,y)$, its weak evolution becomes the forward Kolmogorov equation, also called the Fokker–Planck equation:

\partial_t p(t,y)=\mathcal{L}^*p(t,y) =-\sum_{i=1}^d\partial_{y_i}\!\big(b_i(y)p(t,y)\big) +\frac12\sum_{i,j=1}^d\partial_{y_i y_j}\!\big(a_{ij}(y)p(t,y)\big).

Its initial condition is the initial distribution; for a deterministic start $X_0=x$, this is a point mass at $x$, not generally an ordinary density at time zero. The density equation is a classical PDE only when the density and coefficients have enough regularity. Without that regularity, the test-function identity remains the appropriate formulation.

The adjoint relation explains why the two equations fit together. For compactly supported smooth $f$ and $p$, the integral of $f\mathcal{L}^*p$ equals the integral of $p\mathcal{L}f$. If boundary contributions are present, one must specify boundary conditions before using this identity globally; the detailed treatment of probability flux is a separate issue.

Worked Calculations

Worked Example: Independent Coordinates with Different Variances

Let $X^1_t=x_1+t+2B^1_t$ and $X^2_t=x_2-t+B^2_t$, where $B^1,B^2$ are independent Brownian coordinates. For terminal payoff $f(x)=x_1^2+x_2^2$, compute $u(s,x)=\mathbb{E}[f(X_T)\mid X_s=x]$, then verify the backward equation.

The drift is $b=(1,-1)$, the diffusion matrix is $\sigma=\operatorname{diag}(2,1)$, and therefore $a=\operatorname{diag}(4,1)$. Put $\tau=T-s$. From the independent increments, the future coordinates given $X_s=x$ have means $x_1+\tau$ and $x_2-\tau$, and variances $4\tau$ and $\tau$. Hence $$ u(s,x)=(x_1+\tau)^2+4\tau+(x_2-\tau)^2+\tau. $$ For verification, $\partial_su=-2x_1+2x_2-4\tau-5$, $\partial_1u=2(x_1+\tau)$, $\partial_2u=2(x_2-\tau)$, and $\partial_{11}u=\partial_{22}u=2$, with $\partial_{12}u=0$. Thus $$ \mathcal{L}u=2(x_1+\tau)-2(x_2-\tau)+\tfrac12(4\cdot2+1\cdot2) =2x_1-2x_2+4\tau+5, $$ so $\partial_su+\mathcal{L}u=0$. At $s=T$, $\tau=0$ and $u(T,x)=x_1^2+x_2^2$, as required.

Check: At $s=T$, the conditional variances vanish, and the formula gives exactly the terminal payoff. The coefficients $4$ and $1$ in the variance terms agree with the squared diffusion coefficients $2^2$ and $1^2$.

Worked Example: A Correlated Diffusion and a Cross-Term Payoff

Let $W^1,W^2$ be independent Brownian motions and define $X^1_t=x_1+W^1_t$, $X^2_t=x_2+\tfrac12W^1_t+\tfrac{\sqrt3}{2}W^2_t$. For terminal payoff $f(x)=x_1x_2$, find and verify the backward solution.

The two coordinates are driven by independent Brownian motions, but their increments are correlated: their variances over an interval of length $t$ are $t$ and $(\tfrac14+\tfrac34)t=t$, and their covariance is $\tfrac12t$. The covariance matrix and generator are therefore $$ a=\begin{pmatrix}1&\tfrac12\\[2pt]\tfrac12&1\end{pmatrix}, \qquad \mathcal{L}g=\tfrac12\partial_{11}g+\tfrac12\partial_{12}g+\tfrac12\partial_{22}g. $$ There is no drift. For $\tau=T-s$, the conditional expected product is the product of the means plus the covariance of the increments: \[ u(s,x)=x_1x_2+\tfrac12\tau. \] Here $\partial_su=-\tfrac12$, $\partial_{12}u=1$, and $\partial_{11}u=\partial_{22}u=0$. Thus $\mathcal{L}u=\tfrac12$, so $\partial_su+\mathcal{L}u=0$. Also $u(T,x)=x_1x_2$.

Check: The covariance contribution is $\operatorname{Cov}(W^1_\tau,\tfrac12W^1_\tau+\tfrac{\sqrt3}{2}W^2_\tau)=\tfrac12\tau$, because the independent $W^2$ term has zero covariance with $W^1$. This matches the correction in $u$.

Worked Example: The Adjoint with State-Dependent Covariance

In two dimensions, take $b(x_1,x_2)=(x_1,-x_2)$ and $a(x_1,x_2)=\operatorname{diag}(1+x_1^2,2)$. Compute both operators and expand the forward equation.

The generator is $$ \mathcal{L}f=x_1\partial_1f-x_2\partial_2f +\tfrac12(1+x_1^2)\partial_{11}f+\partial_{22}f. $$ The formal adjoint is $$ \mathcal{L}^*p=-\partial_1(x_1p)+\partial_2(x_2p) +\tfrac12\partial_{11}\big((1+x_1^2)p\big)+\partial_{22}p. $$ Expanding each term gives $-\partial_1(x_1p)=-p-x_1\partial_1p$, $\partial_2(x_2p)=p+x_2\partial_2p$, and \[ \tfrac12\partial_{11}\big((1+x_1^2)p\big) =\tfrac12(1+x_1^2)\partial_{11}p+2x_1\partial_1p+p. \] Therefore $$ \mathcal{L}^*p =\tfrac12(1+x_1^2)\partial_{11}p+x_1\partial_1p+p +x_2\partial_2p+\partial_{22}p. $$ The forward equation is $\partial_t p=\mathcal{L}^*p$.

Check: The coefficient of $\partial_1p$ is $-x_1+2x_1=x_1$, and the terms without derivatives combine to $-p+p+p=p$. In particular, differentiating $a_{11}p$, rather than multiplying $\partial_{11}p$ by $a_{11}$ alone, is essential when the covariance varies with state.

What the Equations Are For—and a Common Error

The backward equation is useful when the desired quantity is an expected payoff from a specified starting state. The forward equation is useful when the distribution or density at a later time is the target. They encode the same transition law from different sides: one evolves test functions, the other evolves distributions. For a transition density $p_t(x,y)$, when the needed regularity holds, the backward operator acts in $x$, while the adjoint forward operator acts in $y$.

A common error is to write the multidimensional forward diffusion term as $\tfrac12\sum_{i,j}a_{ij}\partial_{ij}p$ even when $a$ depends on position. That expression omits derivatives of the coefficients. The adjoint calculation instead gives $\tfrac12\sum_{i,j}\partial_{ij}(a_{ij}p)$. In one dimension this distinction is $\tfrac12\partial_{xx}(a p)$ versus $\tfrac12a\partial_{xx}p$; they agree only under additional restrictions, such as constant $a$. A second error is to omit off-diagonal covariance terms. In the correlated example, the two symmetric matrix entries combine with the factor $\tfrac12$ to produce the nonzero term $\tfrac12\partial_{12}g$.

Check Your Understanding

For each question, state any regularity or integrability assumptions you use.

For a diffusion with $b=(b_1,b_2)$ and $a=\begin{pmatrix}a_{11}&a_{12}\\a_{12}&a_{22}\end{pmatrix}$, write $\mathcal{L}f$ explicitly, collecting the mixed derivative term.
Explain why the forward diffusion term is $\tfrac12\sum_{i,j}\partial_{ij}(a_{ij}p)$, rather than $\tfrac12\sum_{i,j}a_{ij}\partial_{ij}p$, when $a$ varies with position.
For $a=\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}$ and zero drift in two dimensions, compute $\mathcal{L}(x_1x_2)$.
State the weak forward identity using a smooth compactly supported test function $\varphi$.
In what variables do the backward and forward operators act on a transition density $p_t(x,y)$?