Deriving the Forward Kolmogorov Equation

From Transition Probabilities to a Density Equation

A diffusion evolves forward in time, so its transition probabilities describe how probability mass moves from an initial state to later states. The generator describes the short-time change in expectations of test functions. To obtain an equation for a transition density, first differentiate those expectations in time, then use integration by parts to move the generator from the test function onto the density. The integration by parts also exposes the boundary terms: the density equation alone does not determine what happens at a boundary unless those terms are controlled.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Consider the time-homogeneous one-dimensional diffusion $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$, where $\mu$ is the drift and $\sigma$ is the diffusion coefficient. The stochastic integral is the Ito integral; its integrand $\sigma(X_t)$ is assumed progressively measurable and in $\mathcal{L}^2$ on each finite time interval, so that $\mathbb{E}\int_0^T\sigma^2(X_s)\,ds<\infty$. Assume the SDE has a nonexplosive solution and that the transition law has a density $p(t,x,y)$ with respect to Lebesgue measure in $y$.

Write $a(y)=\sigma^2(y)$. For a twice continuously differentiable test function $f$, the diffusion generator is $\mathcal{L}f(y)=\mu(y)f'(y)+\tfrac12a(y)f''(y)$. The transition semigroup is $P_tf(x)=\mathbb{E}[f(X_t)\mid X_0=x]$, and the density represents it by $P_tf(x)=\int p(t,x,y)f(y)\,dy$.

Definition: The forward equation for the transition density is the equation obtained by differentiating $p(t,x,y)$ in time while holding the initial state $x$ fixed. In density form, it is $\partial_t p=\mathcal{L}^{*}p$, where $\mathcal{L}^{*}$ is the formal adjoint of the generator with respect to integration in the state variable $y$.

The Weak Equation: Differentiate First

The differentiation is most safely stated first in an integrated, or weak, form. Use test functions $f\in C_c^\infty$, the infinitely differentiable functions with compact support. The semigroup-generator identity says that, when $f$ is in the generator domain and the expectations are integrable, $\partial_tP_tf=P_t\mathcal{L}f$. This standard semigroup result applies here under the stated domain and regularity assumptions; it is the time-forward derivative of the expectation, not yet a pointwise statement about $p$.

Theorem: Suppose the semigroup-generator identity holds for $f\in C_c^\infty$, and the transition law has density $p(t,x,y)$. Then, for each fixed initial state $x$ and each such $f$, $$ \frac{d}{dt}\int p(t,x,y)f(y)\,dy =\int p(t,x,y)\mathcal{L}f(y)\,dy. $$ If the derivatives and integrations below are justified, the forward equation in density form is $\partial_t p=\mathcal{L}^{*}p$.

Proof. By the density representation, $P_tf(x)=\int p(t,x,y)f(y)\,dy$. Differentiate this identity in $t$, using the assumed regularity that permits differentiation under the integral, and then apply the semigroup-generator identity. Its right-hand side is $P_t\mathcal{L}f(x)$, which by the same density representation is $\int p(t,x,y)\mathcal{L}f(y)\,dy$. Thus the displayed weak identity holds. The density equation follows once the right-hand side is transferred from $f$ to $p$ by integration by parts; the next calculation gives that transfer and its boundary terms explicitly. $\square$

Integration by Parts and the Boundary Terms

Fix $t$ and $x$, and abbreviate $p(t,x,y)$ by $p(y)$. On a state interval $(\ell,r)$, the generator expectation is $\int_\ell^r [\mu p f'+\tfrac12ap f'']\,dy$. Integrate the drift term once and the second-order term twice. These are ordinary integrations by parts, not stochastic integral manipulations.

\begin{aligned} \int_\ell^r \mu p f'\,dy &=[\mu p f]_\ell^r-\int_\ell^r f(\mu p)'\,dy,\\ \frac12\int_\ell^r ap f''\,dy &=\frac12[ap f']_\ell^r-\frac12[(ap)'f]_\ell^r +\frac12\int_\ell^r f(ap)''\,dy. \end{aligned}

Combining the two identities gives the adjoint and the complete endpoint contribution.

Result: On an interval where the displayed derivatives exist, $$ \int_\ell^r p\,\mathcal{L}f\,dy =\int_\ell^r f\left[-(\mu p)'+\frac12(ap)''\right]dy +\left[\frac12ap f'+\left(\mu p-\frac12(ap)'\right)f\right]_\ell^r. $$ Thus the formal adjoint is $\mathcal{L}^{*}p=-(\mu p)'+\tfrac12(ap)''$. The bracket means the value at $r$ minus the value at $\ell$, interpreted as endpoint limits when necessary.

Proof. Apply the first integration-by-parts identity to $\int \mu p f'$, and the second to $\tfrac12\int ap f''$. Adding their integral terms gives $\int f[-(\mu p)'+\tfrac12(ap)'']\,dy$. Adding their boundary terms gives $[\mu p f+\tfrac12ap f'-\tfrac12(ap)'f]_\ell^r$, which is the bracket displayed in the result. This proves the identity and identifies the formal adjoint. $\square$

For compactly supported $f$ on the real line, both endpoint contributions vanish because $f$ and its derivatives are zero outside a bounded set. The weak equation therefore becomes $\frac{d}{dt}\int pf\,dy=\int f[-(\mu p)'+\tfrac12(ap)'']\,dy$. If $p$ has a continuous time derivative and the indicated spatial derivatives are continuous, equality against every compactly supported smooth $f$ implies equality of the integrands at each interior point. This gives the classical forward equation $\partial_t p=-(\mu p)'+\tfrac12(ap)''$.

On a finite interval, compactly supported tests still yield the interior equation, but do not by themselves specify boundary behavior. If one instead uses tests that need not vanish at the endpoints, the bracket must be accounted for: it must vanish under the chosen boundary conditions for the same unmodified weak identity to hold. Boundary mass, absorption, or reflection can require additional boundary information. One must not discard the bracket merely because the interior differential equation has been derived.

Worked Derivations

Worked Example: Brownian Motion and the Heat Kernel

For $X_t=x+B_t$, the drift is $\mu=0$, the diffusion coefficient is $\sigma=1$, and $a=1$. The transition density for $t>0$ is $p(t,x,y)=(2\pi t)^{-1/2}\exp(-(y-x)^2/(2t))$. Verify the forward equation.

Set $z=y-x$. Direct differentiation gives $p_y=-(z/t)p$, and differentiating once more gives $p_{yy}=(z^2/t^2-1/t)p$. Differentiation in time gives $p_t=(-1/(2t)+z^2/(2t^2))p$. The adjoint for these coefficients is $\mathcal{L}^{*}p=\tfrac12p_{yy}$, so

\mathcal{L}^{*}p =\frac12\left(\frac{z^2}{t^2}-\frac1t\right)p =\left(-\frac1{2t}+\frac{z^2}{2t^2}\right)p =p_t.

Check: integrating the density over $y$ and substituting $u=(y-x)/\sqrt{t}$ gives $\int_{\mathbb R}p(t,x,y)\,dy=(2\pi)^{-1/2}\int_{\mathbb R}e^{-u^2/2}\,du=1$. The density therefore preserves total probability, as expected for Brownian motion.

Worked Example: Constant Drift and Diffusion

Let $X_t=x+mt+sB_t$, where $m$ is constant and $s>0$. Its density has mean $x+mt$ and variance $s^2t$. Put $z=y-x$, $q=z-mt$, and write $p=(2\pi s^2t)^{-1/2}\exp(-q^2/(2s^2t))$. Here $\mathcal{L}^{*}p=-m p_y+\tfrac12s^2p_{yy}$. Verify this equation directly.

Since $q_y=1$, the spatial derivatives are $p_y=-q p/(s^2t)$ and $p_{yy}=(q^2/(s^4t^2)-1/(s^2t))p$. Since $q_t=-m$, the time derivative is $p_t=(-1/(2t)+mq/(s^2t)+q^2/(2s^2t^2))p$. On the adjoint side, substitution gives

-mp_y+\frac12s^2p_{yy} =\left(\frac{mq}{s^2t}+\frac{q^2}{2s^2t^2}-\frac1{2t}\right)p =p_t.

Check: at $y=x+mt$, $q=0$, so $p_y=0$, $p_{yy}=-p/(s^2t)$, and both sides equal $-p/(2t)$. The constant drift contributes the first-derivative term with a minus sign in the adjoint; the next pitfall shows why that sign matters.

Worked Example: Geometric Brownian Motion

For $dX_t=\alpha X_t\,dt+\beta X_t\,dB_t$, take $X_0=x>0$, $\beta>0$. The stochastic integrand $\beta X_t$ is progressively measurable and square-integrable on finite time intervals, since the explicit solution has finite second moments. The solution is $X_t=x\exp((\alpha-\beta^2/2)t+\beta B_t)$. Define $c=\alpha-\beta^2/2$, $q=\ln(y/x)-ct$, and, for $y>0$, its density is $p=(y\beta\sqrt{2\pi t})^{-1}\exp(-q^2/(2\beta^2t))$.

Here $\mu(y)=\alpha y$ and $a(y)=\beta^2y^2$, so $\mathcal{L}^{*}p=-(\alpha y p)_y+\tfrac12(\beta^2y^2p)_{yy}$. To compute it, put $k=q/(\beta^2t)$. The logarithmic derivative is $p_y/p=-(1+k)/y$. Also $k_y=1/(\beta^2ty)$, so differentiating once more gives $y^2p_{yy}/p=(1+k)(2+k)-1/(\beta^2t)$. Consequently

\begin{aligned} \frac{\mathcal{L}^{*}p}{p} &=-\alpha+\beta^2+(-\alpha+2\beta^2)\frac{y p_y}{p} +\frac12\beta^2\frac{y^2p_{yy}}p\\ &=-\alpha+\beta^2+(\alpha-2\beta^2)(1+k) +\frac12\beta^2\left((1+k)(2+k)-\frac1{\beta^2t}\right)\\ &=-\frac1{2t}+ck+\frac12\beta^2k^2. \end{aligned}

Directly differentiating the density in time gives $p_t/p=-1/(2t)+cq/(\beta^2t)+q^2/(2\beta^2t^2)$, which equals the last line because $k=q/(\beta^2t)$. Check: the change of variable $q=\ln(y/x)-ct$ has $dy/y=dq$, so $\int_0^\infty p\,dy=(\beta\sqrt{2\pi t})^{-1}\int_{\mathbb R}e^{-q^2/(2\beta^2t)}\,dq=1$. Thus the density is normalized as well as satisfying the equation.

A Common Sign Error

When moving the drift term $\mu f'$ onto the density, it is tempting to leave the drift derivative with a positive sign. The integration-by-parts identity shows the error: $\int \mu p f'=[\mu pf]-\int f(\mu p)'$, so the adjoint drift contribution is $-(\mu p)'$, not $+(\mu p)'$.

For a concrete check, take drift $m=1$, diffusion coefficient $s=1$, $t=1$, and $y=x+2$ in the constant-coefficient example. Then $q=1$, $p_y=-p$, and $p_{yy}=0$. The correct right-hand side is $-p_y+\tfrac12p_{yy}=p$, matching $p_t=p$ from the time derivative. The incorrect expression $+p_y+\tfrac12p_{yy}$ gives $-p$, the opposite value. The mistake reverses the direction of drift in the density equation.

The forward equation matters because it evolves distributions rather than individual sample paths. Its derivation also separates three issues that are easy to conflate: differentiating transition expectations, transferring the generator by integration by parts, and specifying boundary behavior. The interior equation follows from the first two under regularity assumptions; the boundary bracket records what must additionally be checked at an endpoint.

Check Your Understanding

Use the semigroup identity and the displayed integration-by-parts formula to organize your answers.

For drift $\mu(y)=m$ and diffusion coefficient $\sigma(y)=s$, write the formal adjoint and state the sign of its first-derivative term.
Why do compactly supported test functions remove endpoint contributions on the real line?
On an interval $(\ell,r)$, write the complete boundary bracket that arises when transferring the generator to the density.
For Brownian motion, calculate $p_t$ and $\tfrac12p_{yy}$ using $z=y-x$, and verify their equality.
Explain why deriving the interior forward equation does not, by itself, specify behavior at a finite boundary.