Deriving the Backward Kolmogorov Equation

From Transition Expectations to a PDE

A diffusion started at $x$ has a distribution at each later time. Its transition expectation records how that distribution acts on a test function $f$. The backward Kolmogorov equation describes the change in this expectation as elapsed time varies, with the starting point $x$ held fixed. Its spatial operator is the infinitesimal generator of the diffusion.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider a time-homogeneous diffusion satisfying $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$, with $X_0=x$. Here $\mu$ is the drift and $\sigma$ is the diffusion coefficient. One common set of sufficient conditions for a unique nonexplosive strong solution is that $\mu$ and $\sigma$ are globally Lipschitz and satisfy linear growth bounds: for some constant $C$, $|\mu(x)|+|\sigma(x)|\le C(1+|x|)$. The standard existence-and-uniqueness theorem for Lipschitz SDEs gives that conclusion; its proof is not needed for the semigroup derivation below. We also assume the solution has the time-homogeneous Markov property.

Definition: For a measurable test function $f$ for which the expectation is finite, the transition semigroup is $P_t f(x)=\mathbb{E}[f(X_t)\mid X_0=x]$. Its infinitesimal generator $A$ has domain $D(A)$ consisting of functions for which the limit $Af=\lim_{h\downarrow0}(P_hf-f)/h$ exists in the chosen function-space norm. For a diffusion, the generator acts on twice continuously differentiable functions in its domain by $\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12\sigma^2(x)f''(x)$.

The semigroup property $P_{t+h}=P_tP_h=P_hP_t$ expresses evolution in two consecutive time intervals. It follows from the time-homogeneous Markov property. The backward equation is an infinitesimal consequence of this composition rule: advancing the elapsed time by a small $h$ applies $P_h$ to the current transition expectation.

The Semigroup-Generator Identity

Theorem: Let $(P_t)_{t\ge0}$ be a strongly continuous semigroup of bounded linear operators on a Banach space $E$, with generator $A$ defined by the norm limit above. If $f\in D(A)$, then $P_t f\in D(A)$ for every $t\ge0$, $A P_t f=P_t Af$, and $t\mapsto P_t f$ is differentiable for $t>0$, with right derivative at $t=0$, satisfying $\frac{d}{dt}P_t f=A P_t f=P_t Af$. In particular, $P_0f=f$.

Proof. Fix $t\ge0$. By the semigroup property and boundedness of $P_t$,

\lim_{h\downarrow0}\frac{P_hP_tf-P_tf}{h} =P_t\lim_{h\downarrow0}\frac{P_hf-f}{h} =P_tAf.

The limit exists in $E$, so the definition of the generator gives $P_tf\in D(A)$ and $AP_tf=P_tAf$. For the right derivative of $P_tf$, the quotient $(P_{t+h}f-P_tf)/h$ equals $P_t(P_hf-f)/h$, and therefore converges to $P_tAf$. For $t>0$, the left difference quotient is $(P_tf-P_{t-h}f)/h=P_{t-h}(P_hf-f)/h$. Strong continuity and the generator limit show that it also converges to $P_tAf$. Thus the derivative exists and equals both $AP_tf$ and $P_tAf$. Finally $P_0$ is the identity operator by the definition of a semigroup, so $P_0f=f$. $\square$

To apply the theorem to a diffusion, take $E$ to be a function space on which its transition semigroup is strongly continuous. For example, one may use $C_0$, the continuous functions vanishing at infinity, with the supremum norm, when the semigroup has that property. To interpret the result as a classical PDE, also assume that $u(t,\cdot)=P_tf$ is twice continuously differentiable in $x$ and that the generator acts on it by the diffusion expression $\mathcal{L}$. These are substantive domain and regularity assumptions: the semigroup identity alone does not assert that every transition expectation is classically differentiable in space.

Theorem: Suppose $f\in D(A)$, the semigroup assumptions of the preceding theorem hold, and $u(t,x)=P_tf(x)$ has the spatial regularity needed for $A u(t,\cdot)=\mathcal{L}u(t,\cdot)$. Then $u$ solves the backward Kolmogorov equation in elapsed time: $$ \partial_t u(t,x)=\mathcal{L}u(t,x) =\mu(x)\,\partial_xu(t,x)+\tfrac12\sigma^2(x)\,\partial_{xx}u(t,x), \qquad u(0,x)=f(x). $$

Proof. Apply the semigroup-generator identity to $f$. It gives $\partial_t u(t,\cdot)=A P_tf$. Since $P_tf=u(t,\cdot)$ and the stated domain assumption identifies $A$ with $\mathcal{L}$ on this function, $\partial_t u(t,x)=\mathcal{L}u(t,x)$. The semigroup identity at zero gives $u(0,x)=P_0f(x)=f(x)$. Each equality is pointwise wherever the assumed classical regularity holds. $\square$

The initial condition is the test function itself, because at elapsed time zero the process has not moved from its starting state. This is an initial-value equation. A PDE written using calendar time and a fixed future maturity may have a different time orientation; the equation here uses $t$ for elapsed time after starting at $x$.

Worked Transition Expectations

Worked Example: Brownian Motion with a Quadratic Test Function

Let $X_t=x+B_t$, and set $f(y)=y^2$. Compute $u(t,x)=P_tf(x)$ and verify the backward equation.

Since $B_t$ has mean zero and second moment $t$, expand the square and take expectations term by term:

u(t,x)=\mathbb{E}[(x+B_t)^2] =x^2+2x\mathbb{E}[B_t]+\mathbb{E}[B_t^2] =x^2+t.

Here $\mu=0$ and $\sigma=1$, so $\mathcal{L}g=\tfrac12g''$. For $u(t,x)=x^2+t$, the derivatives are $\partial_tu=1$, $\partial_xu=2x$, and $\partial_{xx}u=2$. Therefore $\mathcal{L}u=\tfrac12(2)=1=\partial_tu$. At $t=0$, $u(0,x)=x^2=f(x)$. Check: setting $t=0$ in the expectation gives $\mathbb{E}[(x+B_0)^2]=x^2$, since $B_0=0$, in agreement with the initial condition.

Worked Example: Ornstein–Uhlenbeck Transition Mean

Consider $dX_t=-2X_t\,dt+dB_t$ with $X_0=x$, and use $f(y)=y$. Find the transition expectation and check its PDE.

The explicit solution is $X_t=e^{-2t}x+\int_0^t e^{-2(t-s)}\,dB_s$. The stochastic integral is an Itô integral with deterministic, hence progressively measurable, integrand $H_s=e^{-2(t-s)}$. On $[0,t]$, it belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^t H_s^2\,ds=\int_0^t e^{-4(t-s)}\,ds=(1-e^{-4t})/4<\infty$. This integral has mean zero, so

u(t,x)=\mathbb{E}[X_t\mid X_0=x]=e^{-2t}x.

The generator is $\mathcal{L}g(x)=-2xg'(x)+\tfrac12g''(x)$. The derivatives are $\partial_tu=-2e^{-2t}x$, $\partial_xu=e^{-2t}$, and $\partial_{xx}u=0$. Thus $\mathcal{L}u=-2x e^{-2t}+\tfrac12(0)=-2e^{-2t}x=\partial_tu$. At zero, $u(0,x)=x$. Check: the explicit solution has expectation $e^{-2t}x+0$, exactly the same expression as the semigroup calculation.

Worked Example: A Second Moment for Geometric Brownian Motion

Let $dS_t=\tfrac14S_t\,dt+\tfrac12S_t\,dB_t$, $S_0=x>0$, and take $f(y)=y^2$. Compute $P_tf(x)$ and verify the equation.

The explicit solution is $S_t=x\exp((\tfrac14-\tfrac18)t+\tfrac12B_t)$. Squaring gives $S_t^2=x^2\exp(\tfrac14t+B_t)$. Since $B_t$ is normal with mean zero and variance $t$, its exponential moment is $\mathbb{E}[e^{B_t}]=e^{t/2}$. Consequently,

u(t,x)=\mathbb{E}[S_t^2\mid S_0=x] =x^2e^{t/4}e^{t/2}=x^2e^{3t/4}.

For this diffusion, $\mathcal{L}g(x)=\tfrac14xg'(x)+\tfrac12(\tfrac12)^2x^2g''(x)$. The derivatives of $u$ are $\partial_tu=\tfrac34x^2e^{3t/4}$, $\partial_xu=2xe^{3t/4}$, and $\partial_{xx}u=2e^{3t/4}$. Substitution gives $\mathcal{L}u=\tfrac14x(2xe^{3t/4})+\tfrac18x^2(2e^{3t/4})=(\tfrac12+\tfrac14)x^2e^{3t/4}=\partial_tu$. At $t=0$, $u(0,x)=x^2$. Check: the normal exponential-moment calculation gives the same factor $e^{t/4}e^{t/2}=e^{3t/4}$; in particular, the drift contribution and the Brownian variance contribution have both been included.

Why the Direction and Assumptions Matter

The generator acts on the starting-state variable $x$, while the derivative on the left acts on elapsed time $t$. The equation is called backward because it describes how a transition expectation changes as one varies the state from which the process is started. For a transition density, the corresponding operator acts on the starting-state coordinate; the equation for evolution of the density in its destination coordinate is a different equation.

Common mistake: reversing the sign. For the Ornstein–Uhlenbeck expectation $u(t,x)=e^{-2t}x$, the correct time derivative is $-2e^{-2t}x$. The generator calculation gives $\mathcal{L}u=-2e^{-2t}x$, so the equation is $\partial_tu=\mathcal{L}u$. A mistaken sign choice $\partial_tu=-\mathcal{L}u$ would give $+2e^{-2t}x$, which disagrees with the derivative whenever $x\ne0$. The case $x=0$ alone would not reveal this sign error, so checking a nonzero starting state matters.

There is also a distinction between deriving the equation and proving existence or uniqueness for a classical PDE solution. The semigroup argument proves that a sufficiently regular transition expectation satisfies the equation under the stated domain assumptions. It does not, without additional hypotheses, establish that every proposed initial function produces such a classical solution, nor that every classical solution equals the transition expectation. Those questions require further existence, growth, and uniqueness conditions.

Takeaway: For a time-homogeneous diffusion with a suitable strongly continuous transition semigroup, a test function in the generator domain yields $u(t,x)=P_tf(x)$ with $\partial_tu=\mathcal{L}u$ and $u(0,x)=f(x)$, provided the generator and spatial regularity assumptions permit the differential interpretation.

Check Your Understanding

Use the semigroup definition and the generator expression to organize each answer.

Why does the transition semigroup satisfy $P_{t+h}=P_tP_h$ for a time-homogeneous Markov process?
State the generator-domain condition needed to define $Af$ through a norm limit, and explain why spatial $C^2$ regularity is separately relevant to a classical PDE.
For Brownian motion with $f(x)=x$, compute $P_tf(x)$, its time derivative, and its generator expression.
For $dX_t=-\kappa X_t\,dt+dB_t$ and $f(x)=x$, use the transition mean to check the backward equation for $\kappa>0$.
Explain why the initial condition for $u(t,x)=P_tf(x)$ is $f(x)$, rather than a terminal payoff specified at a later time.

Deriving the Backward Kolmogorov Equation

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