Semigroups of Diffusion Processes

From Transition Laws to Operators

Transition kernels describe where a diffusion may be after a given amount of time. A semigroup packages the same information as an operation on functions: instead of asking for the distribution of $X_t$, we ask for the expected value of a test function evaluated at $X_t$. The semigroup law says that evolving for two successive time intervals has the same effect as evolving for their combined length. Its proof is a direct use of the Markov property.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. A useful class of examples is a time-homogeneous diffusion solving $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$. Here $\mu$ is the drift and $\sigma$ is the diffusion coefficient. For instance, if $\mu$ and $\sigma$ are globally Lipschitz and satisfy a linear growth bound, the standard existence and uniqueness theorem gives a strong solution for each deterministic initial state $x$. The integrand $\sigma(X_s)$ is progressively measurable and, under these conditions, satisfies $\mathbb{E}_x\int_0^T\sigma^2(X_s)\,ds<\infty$ for finite $T$; thus it belongs to the class $\mathcal{L}^2$ for the Ito integral. The semigroup definition and proof below require the time-homogeneous Markov property, not this particular SDE representation.

Let $E$ be the state space, and suppose that for each $x\in E$ there is a law $\mathbb{P}_x$ under which the process starts at $X_0=x$. Write $\mathbb{E}_x$ for expectation under $\mathbb{P}_x$. The notation $\mathbb{E}[f(X_t)\mid X_0=x]$ is understood as expectation under this starting-state law (or a regular conditional distribution), not as conditioning on an event of positive probability. For general state spaces, the event $\{X_0=x\}$ may have probability zero.

Definition: For a Borel measurable function $f:E\to\mathbb{R}$ for which the expectation is defined, the transition operator at time $t\ge0$ is $$ P_t f(x)=\mathbb{E}_x[f(X_t)]. $$ For bounded Borel $f$, this is finite for every $x$. The family $(P_t)_{t\ge0}$ is the transition semigroup if $P_0f=f$ and $P_{t+s}f=P_t(P_sf)$ for all $s,t\ge0$ and bounded Borel $f$.

The order in $P_t(P_sf)$ matters for reading the definition: first $P_s$ turns $f$ into a function of the current state; then $P_t$ averages that function over the state reached after time $t$. The Markov property states that, for bounded Borel $f$, $\mathbb{E}_x[f(X_{t+s})\mid\mathcal{F}_t]=P_sf(X_t)$. This property is what makes the operator identity hold.

The Semigroup Law

Theorem: Let $(X_t)_{t\ge0}$ be a time-homogeneous Markov process with starting-state laws $(\mathbb{P}_x)_{x\in E}$. For every bounded Borel function $f:E\to\mathbb{R}$ and $s,t\ge0$, $$ P_{t+s}f(x)=P_t(P_sf)(x),\qquad x\in E. $$ Also, $P_0f=f$. The same identity holds for nonnegative Borel $f$, allowing extended expectations, and for signed $f$ whenever the relevant absolute expectations are finite.

Proof. Fix $x\in E$, $s,t\ge0$, and bounded Borel $f$. By definition, $P_{t+s}f(x)=\mathbb{E}_x[f(X_{t+s})]$. Apply the tower property of conditional expectation with respect to $\mathcal{F}_t$, then use the time-homogeneous Markov property:

\begin{aligned} P_{t+s}f(x) &=\mathbb{E}_x\!\left[\mathbb{E}_x[f(X_{t+s})\mid\mathcal{F}_t]\right]\\ &=\mathbb{E}_x[P_sf(X_t)]\\ &=P_t(P_sf)(x). \end{aligned}

All conditional expectations are integrable because $f$ is bounded, and $P_sf$ is bounded by the same bound. At time zero, $X_0=x$ under $\mathbb{P}_x$, so $P_0f(x)=\mathbb{E}_x[f(X_0)]=f(x)$. For nonnegative $f$, the same conditioning identity follows by applying conditional expectation to nonnegative random variables, with extended values permitted. For signed $f$, apply the identity to its positive and negative parts when the stated absolute integrability makes both expectations finite. $\square$

The operators also preserve order and constants. If $f\le g$, then $P_tf\le P_tg$, because expectation preserves inequalities. For a constant $c$, $P_tc=c$, since the expectation of a constant is that constant. In particular, for bounded $f$, $\lVert P_tf\rVert_\infty\le\lVert f\rVert_\infty$. These facts follow directly from the definition; they do not require an additional property of the diffusion.

Brownian Motion: The Gaussian Semigroup

For Brownian motion started at $x$, set $X_t=x+B_t$. Its increment $B_t$ is independent of the initial state and has distribution $N(0,t)$. Thus, for $t>0$, its transition density from $x$ to $y$ is $p_t(x,y)=(2\pi t)^{-1/2}\exp(-(y-x)^2/(2t))$. Consequently, for bounded Borel $f:\mathbb{R}\to\mathbb{R}$,

P_tf(x)=\int_{\mathbb{R}} f(y)\,p_t(x,y)\,dy =\int_{\mathbb{R}} f(x+z)\frac{1}{\sqrt{2\pi t}}e^{-z^2/(2t)}\,dz,\qquad t>0.

At $t=0$, define $P_0f(x)=f(x)$; the density formula is only for positive $t$. The same integral formula applies to unbounded $f$ whenever the absolute integral is finite. For Brownian motion, the semigroup property can also be seen from its independent increments: $B_{t+s}=B_t+(B_{t+s}-B_t)$, where the second term is independent of $\mathcal{F}_t$ and has distribution $N(0,s)$. The theorem above makes this reasoning precise for functions of the state.

Worked Example: An Affine Function

For $f(x)=3x-2$, calculate $P_tf(x)$ and verify the semigroup identity. The function is unbounded, but its absolute expectation is finite under every Gaussian starting-state law.

Since $X_t=x+B_t$, linearity of expectation and $\mathbb{E}[B_t]=0$ give

P_tf(x)=\mathbb{E}[3(x+B_t)-2]=3x+3\mathbb{E}[B_t]-2=3x-2.

Applying the operator again, $P_s f$ is the same function $3x-2$, so $P_t(P_sf)(x)=3x-2=P_{t+s}f(x)$. Check: if $x=2$, the initial value of the function is $3\cdot2-2=4$. At every time $t$, the expected state is still $2$, so the expected function value remains $3\cdot2-2=4$, matching the formula.

Worked Example: A Quadratic Function

For $f(x)=x^2$, compute $P_tf(x)$ and compose two operators.

Expand the square and use $\mathbb{E}[B_t]=0$ and $\mathbb{E}[B_t^2]=t$:

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

For $s\ge0$, $P_sf(x)=x^2+s$. Applying $P_t$ to this entire function gives

P_t(P_sf)(x)=\mathbb{E}[(x+B_t)^2+s]=x^2+t+s=P_{t+s}f(x).

Check: starting at $x=2$ and taking $t=1$, the formula gives $P_1f(2)=2^2+1=5$. Directly, $X_1$ has mean $2$ and variance $1$, so $\mathbb{E}[X_1^2]=\operatorname{Var}(X_1)+(\mathbb{E}[X_1])^2=1+4=5$. The two-time formula at $s=t=1$ gives $P_2f(2)=2^2+2=6$, and composition gives $P_1(x^2+1)(2)=5+1=6$.

Worked Example: An Exponential Function

For $f(x)=e^{ax}$, where $a\in\mathbb{R}$, calculate the Brownian semigroup and check composition. Gaussian exponential moments are finite for every real $a$.

Using $X_t=x+B_t$ and completing the square in the normal density, $\mathbb{E}[e^{aB_t}]=e^{a^2t/2}$. Therefore

P_tf(x)=e^{ax}\mathbb{E}[e^{aB_t}] =e^{ax+a^2t/2}.

Now $P_sf(x)=e^{ax+a^2s/2}$, so

P_t(P_sf)(x) =e^{a^2s/2}\mathbb{E}[e^{a(x+B_t)}] =e^{ax+a^2s/2+a^2t/2} =e^{ax+a^2(t+s)/2} =P_{t+s}f(x).

Check: for $a=2$, $P_tf(x)=e^{2x+2t}$, since $a^2t/2=4t/2=2t$. Composing at $s=1$ and $t=2$ yields $e^{2x+2}e^{4}=e^{2x+6}$, while the single operator at time $3$ gives $e^{2x+2\cdot3}=e^{2x+6}$.

What the Semigroup Viewpoint Adds

A transition kernel gives probabilities for sets of future states. The operator $P_t$ instead answers a broad class of expectation questions at once: choosing $f=\mathbf{1}_A$ gives $P_t\mathbf{1}_A(x)=\mathbb{P}_x(X_t\in A)$, while other functions encode moments or payoffs. Composition then becomes a compact statement about successive evolution. This operator viewpoint is also the starting point for studying how $P_tf$ changes with $t$, although identifying that time derivative and its domain is a separate question.

The semigroup uses the laws of the process, rather than a particular coupling of sample paths. A strong solution is constructed on a specified probability space using its Brownian motion; a weak solution specifies a probability space and Brownian motion on which the equation holds. Once the starting-state transition laws and the Markov property are given, the semigroup identity concerns those laws. No identification of solutions path by path is needed in its proof.

Common pitfall: applying an operator only to part of a function. For Brownian motion and $f(x)=x^2$, one has $P_1f(x)=x^2+1$. A mistaken composition might apply $P_1$ only to $x^2$, report $x^2+1$, and forget that the constant $1$ is also part of the function. The correct calculation is $P_1(P_1f)(x)=P_1(x^2+1)(x)=x^2+1+1=x^2+2$, which agrees with $P_2f(x)=x^2+2$. Each operator acts on the whole function supplied to it.

Check Your Understanding

Use the starting-state interpretation and the semigroup law in each answer.

For Brownian motion and $f(x)=x^3$, calculate $P_tf(x)$, stating which Gaussian moments you use.
For $f(x)=x^2$, calculate $P_2(P_3f)(x)$ and compare it with $P_5f(x)$.
Explain why $P_t\mathbf{1}_A(x)$ is a transition probability for a Borel set $A$.
What integrability condition allows the semigroup identity to extend from bounded $f$ to a signed unbounded function?
In the quadratic example, identify the precise term that is omitted when an operator is incorrectly applied only to the nonconstant part of $P_sf$.

Semigroups of Diffusion Processes

What You'll Learn