Two PDEs, Two Different Unknowns
A diffusion gives rise to two Kolmogorov equations, and their differential operators are adjoints rather than interchangeable versions of one equation. The forward equation evolves a probability density in its destination variable and time. The backward equation evolves an expected future payoff in its starting point and time. A reliable diagnosis begins by asking what the unknown represents, which variable it depends on, and what data it must satisfy.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a one-dimensional diffusion \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), where \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. The Ito integral is understood for progressively measurable integrands \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); when coefficients do not meet this global condition, the usual stopped, local formulation is used. Write \(a(t,x)=\sigma(t,x)^2\). Its generator, acting on a twice differentiable function \(f\), is \(\mathcal{L}_t f(x)=\mu(t,x)f'(x)+\tfrac12a(t,x)f''(x)\).
The conditional notation in the definition means the expectation for the diffusion started from the specified state \(x\) at time \(s\); it does not require conditioning on a particular event \(\{X_s=x\}\) of positive probability. The forward equation instead starts from a point mass: \(p(s,x;s,\cdot)=\delta_x\) in the distributional sense. In particular, it is initial data at the starting time, while \(g\) is terminal data at the payoff time. These are the standard forward and backward Kolmogorov equations, derived in the earlier material on those equations.
Why the Operators and Data Differ
The forward operator acts on the density in \(y\); the backward operator acts on a function of the starting state \(x\). Their relationship can be checked by integration by parts. This also identifies the boundary terms that determine whether probability mass is conserved.
Proof. Write \(a=a(t,y)\) and \(\mu=\mu(t,y)\). Integrating the drift term by parts gives \(\int f(-\partial_y(\mu p))\,dy=\int f'\mu p\,dy\), because the boundary term \(-f\mu p\) vanishes. Integrating the diffusion term twice gives \(\int \tfrac12 f\partial_{yy}(ap)\,dy=\int \tfrac12 f''ap\,dy\): the boundary terms involving \(f\partial_y(ap)\) and \(f'ap\) vanish by the assumptions. Adding the two identities yields \(\int(\mu f'+\tfrac12af'')p\,dy=\int(\mathcal{L}_tf)p\,dy\), as required. \(\square\)
For a forward density, the probability current is \(J(t,y)=\mu(t,y)p(t,y)-\tfrac12\partial_y(a(t,y)p(t,y))\). Thus \(\partial_t p=-\partial_yJ\). On a finite interval \((\ell,r)\), integration gives \(\frac{d}{dt}\int_\ell^r p(t,y)\,dy=J(t,\ell)-J(t,r)\). In particular, zero flux at both endpoints preserves mass. On the real line, the corresponding conclusion requires the current to vanish at infinity and the density to be integrable. A backward solution has no requirement to integrate to one: its terminal values are the payoff \(g\), not a probability density.
Worked Example: Brownian Transition Density
For \(X_t=B_t\), started at \(x\) at time \(s\), the transition density for \(t>s\) is \(p(s,x;t,y)=(2\pi(t-s))^{-1/2}\exp(-(y-x)^2/(2(t-s)))\). Check the forward equation, normalization and initial data.
Set \(\tau=t-s>0\) and \(z=y-x\). Direct differentiation gives \(\partial_t p=p[-1/(2\tau)+z^2/(2\tau^2)]\), \(\partial_y p=-zp/\tau\), and \(\partial_{yy}p=p[z^2/\tau^2-1/\tau]\). Here \(\mu=0\), \(a=1\), so the forward equation is \(\partial_t p=\tfrac12\partial_{yy}p\). The displayed derivatives show that both sides equal \(p[-1/(2\tau)+z^2/(2\tau^2)]\).
To check normalization, substitute \(w=(y-x)/\sqrt{\tau}\), so \(dy=\sqrt{\tau}\,dw\). Then \(\int_{\mathbb{R}}p\,dy=(2\pi)^{-1/2}\int_{\mathbb{R}}e^{-w^2/2}\,dw=1\). As \(\tau\) decreases to zero, \(x+\sqrt{\tau}Z\) converges in probability to \(x\) for \(Z\sim N(0,1)\); consequently the densities converge to \(\delta_x\) in the distributional sense. These checks identify both the correct forward data and a normalized density.
Drift, Signs and Payoff Data
Worked Example: Brownian Motion with Constant Drift
Let \(X_t=x+\mu(t-s)+(B_t-B_s)\), where \(\mu\) is constant. Find its forward density and the backward solution for the payoff \(g(y)=y^2\), then check the equations and data.
For \(\tau=t-s>0\), the density is \(p(s,x;t,y)=(2\pi\tau)^{-1/2}\exp(-(y-x-\mu\tau)^2/(2\tau))\). Put \(z=y-x-\mu\tau\). Since \(\partial_t z=-\mu\), differentiation yields \(\partial_t p=p[-1/(2\tau)+\mu z/\tau+z^2/(2\tau^2)]\), \(\partial_y p=-zp/\tau\), and \(\partial_{yy}p=p[z^2/\tau^2-1/\tau]\). The forward operator is \(\mathcal{L}_t^*p=-\mu\partial_yp+\tfrac12\partial_{yy}p\). Substitution gives \(\mathcal{L}_t^*p=p[\mu z/\tau+z^2/(2\tau^2)-1/(2\tau)]=\partial_t p\).
For the backward problem, the increment \(X_T-x\) has mean \(\mu(T-s)\) and variance \(T-s\). Therefore \(u(s,x)=\mathbb{E}[X_T^2\mid X_s=x]=(x+\mu(T-s))^2+(T-s)\). Writing \(\tau=T-s\), its derivatives are \(u_s=-2\mu(x+\mu\tau)-1\), \(u_x=2(x+\mu\tau)\), and \(u_{xx}=2\). Since \(\mathcal{L}_su=\mu u_x+\tfrac12u_{xx}=2\mu(x+\mu\tau)+1\), we obtain \(u_s+\mathcal{L}_su=0\). At \(s=T\), \(u(T,x)=x^2=g(x)\). As an independent check, \(X_T=x+\mu\tau+Z\), with \(\mathbb{E}Z=0\) and \(\mathbb{E}Z^2=\tau\), gives \(\mathbb{E}X_T^2=(x+\mu\tau)^2+\tau\), exactly the stated solution.
The sign of the drift term in the two equations differs for a reason: the forward drift term is \(-\partial_y(\mu p)\), while the backward generator contains \(+\mu\partial_x\). The forward term redistributes mass; the backward term describes how the expected payoff changes with the starting state.
Ornstein–Uhlenbeck Checks and Boundary Flux
Worked Example: Ornstein–Uhlenbeck Density and Mean Payoff
Consider \(dX_t=-\kappa X_t\,dt+\sigma\,dB_t\), with \(\kappa>0\) and \(\sigma>0\), started from \(x\) at \(s\). Verify the transition density and the backward expectation for \(g(y)=y\).
For \(\tau=t-s>0\), set \(m=e^{-\kappa\tau}x\) and \(v=\sigma^2(1-e^{-2\kappa\tau})/(2\kappa)\). The candidate density is \(p=(2\pi v)^{-1/2}\exp(-(y-m)^2/(2v))\). It is normalized because it is a Gaussian density with variance \(v>0\). Let \(z=y-m\). The parameter derivatives are \(m_t=-\kappa m\) and \(v_t=\sigma^2-2\kappa v\), while the spatial derivatives are \(p_y=-zp/v\) and \(p_{yy}=p[z^2/v^2-1/v]\). Differentiating in time gives \(\partial_t p=p[-v_t/(2v)+zm_t/v+z^2v_t/(2v^2)]\).
The forward operator is \(\mathcal{L}^*p=\kappa\partial_y(yp)+\tfrac12\sigma^2p_{yy}\). Since \(\partial_y(yp)=p(1-yz/v)\), its value divided by \(p\) is \(\kappa(1-yz/v)+\tfrac12\sigma^2(z^2/v^2-1/v)\). Using \(y=m+z\), the constant term is \(\kappa-\sigma^2/(2v)=-v_t/(2v)\); the coefficient of \(z\) is \(-\kappa m/v=m_t/v\); and the coefficient of \(z^2\) is \(-\kappa/v+\sigma^2/(2v^2)=v_t/(2v^2)\). These are precisely the three coefficients in \(\partial_t p/p\). As \(\tau\) decreases to zero, \(m\) tends to \(x\) and \(v\) tends to zero, so the Gaussian density converges to \(\delta_x\).
For the backward payoff \(g(y)=y\), the expectation is \(u(s,x)=e^{-\kappa(T-s)}x\). Its derivatives are \(u_s=\kappa e^{-\kappa(T-s)}x\), \(u_x=e^{-\kappa(T-s)}\), and \(u_{xx}=0\). Thus \(\mathcal{L}_su=-\kappa x e^{-\kappa(T-s)}\), which cancels \(u_s\), and \(u(T,x)=x\). The independent probabilistic check is the mean of the Gaussian just computed: it is \(m=e^{-\kappa(T-s)}x\), agreeing with \(u\).
For this process, the current of the displayed density is \(J=-\kappa yp-\tfrac12\sigma^2p_y=-\kappa yp+\tfrac12\sigma^2(y-m)p/v\). The Gaussian tails make this current tend to zero at both ends of the real line. Its mass therefore remains one. On a bounded interval, however, simply integrating a candidate density to one at a single time would not establish conservation: the endpoint currents must also be checked, and nonzero boundary flux changes the mass inside the interval.
A Practical Diagnostic and a Common Sign Error
A transition density is a function of the destination variable and integrates to one when it describes the full state space. An expected payoff is a function of the starting state and is generally not normalized.
The forward density begins from a point mass at the starting time. The backward expectation ends at the payoff \(g\) at terminal time.
Use \(\mathcal{L}^*\) in the destination variable for the forward equation and \(\mathcal{L}\) in the starting variable for the backward equation. For a density, verify normalization and boundary flux.
A common mistake is to use the backward drift sign in the forward equation. For constant drift, writing \(\partial_t p=+\mu\partial_y p+\tfrac12\partial_{yy}p\) is wrong. The correct forward drift term is \(-\mu\partial_y p\). For the Gaussian in the constant-drift example, \(\partial_y p=-zp/\tau\), so the wrong expression contributes \(-\mu zp/\tau\), whereas the actual time derivative contains \(+\mu zp/\tau\). The corrected equation \(\partial_t p=-\mu\partial_y p+\tfrac12\partial_{yy}p\) gives exactly the required \(+\mu zp/\tau\). When \(\mu=0\), this particular sign error is invisible, so testing a nonzero drift is a useful diagnostic.
Check Your Understanding
For each question, distinguish carefully between a density and an expected payoff, and state which variable the equation differentiates.
- For Brownian motion with drift \(\mu\), which drift term belongs in the forward equation for \(p(s,x;t,y)\), and which belongs in the backward equation for \(u(s,x)\)?
- What initial data does a transition density satisfy as \(t\) decreases to \(s\), and what terminal data does a backward expected payoff satisfy at \(s=T\)?
- For the Ornstein–Uhlenbeck density above, verify directly that its variance parameter satisfies \(v_t=\sigma^2-2\kappa v\).
- On a finite interval, express the rate of change of total density mass in terms of the endpoint currents \(J(t,\ell)\) and \(J(t,r)\). What boundary condition makes this rate zero?
- Why does the condition \(\int p(t,y)\,dy=1\) apply to a transition density but not generally to \(u(s,x)=\mathbb{E}[g(X_T)\mid X_s=x]\)?