Brownian Transition Densities and the Semigroup

From Brownian Increments to Transition Densities

Brownian motion moves randomly, but its distribution after a fixed time interval has a precise Gaussian form. A transition density records that distribution as a function of the starting and ending points. Its usefulness is not limited to finding probabilities: integrating a bounded function against the density gives its expected value after the transition. The densities also compose consistently when a time interval is split into two parts.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. In particular, $B_0=0$, and for $0\le s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. For a starting point $x\in\mathbb{R}$, write $X^x_t=x+B_t$. All integrals in this tutorial are ordinary Lebesgue integrals, not stochastic integrals.

For $t>0$, the increment $B_t$ has density $z\mapsto (2\pi t)^{-1/2}\exp(-z^2/(2t))$. Translating by $x$ gives the density at $y$ of $X^x_t$.

Definition: For $t>0$ and $x,y\in\mathbb{R}$, the Brownian transition density is $$ p_t(x,y)=\frac{1}{\sqrt{2\pi t}}\exp\!\left(-\frac{(y-x)^2}{2t}\right). $$ Thus, for every Borel set $A\subseteq\mathbb{R}$, $\mathbb{P}(X^x_t\in A)=\int_A p_t(x,y)\,dy$. At time $t=0$, the transition law is the point mass at $x$, rather than a density with respect to Lebesgue measure.

The kernel is nonnegative and integrates to one: the substitution $z=(y-x)/\sqrt{t}$ reduces its integral over $y\in\mathbb{R}$ to the standard normal density integral. For $t>0$, $p_t(x,y)=p_t(y,x)$, since the formula depends on $x,y$ through $(y-x)^2$. This symmetry does not mean that the transition distribution is independent of its starting point; its mean is $x$.

Worked Example: Evaluating a Transition Density

Find the density at $y=3$ after time $t=4$ for Brownian motion started at $x=1$.

Substitute $t=4$, $x=1$, and $y=3$ in the definition. The time factor in the denominator is $\sqrt{2\pi\cdot4}=\sqrt{8\pi}$, and the squared displacement is $(3-1)^2=4$. Therefore

p_4(1,3)=\frac{1}{\sqrt{8\pi}}\exp\!\left(-\frac{4}{2\cdot4}\right) =\frac{e^{-1/2}}{\sqrt{8\pi}}.

Check: the density is positive and has the same value at $y=-1$, because $(-1-1)^2=4=(3-1)^2$. These points are equally far from the starting point $1$, as the symmetric Gaussian formula requires.

The Convolution Semigroup

Suppose a transition lasts for time $s+t$, with $s,t>0$. The density at the endpoint can be obtained by integrating over all possible intermediate positions $z$: first move from $x$ to $z$ in time $s$, then from $z$ to $y$ in time $t$. This leads to the convolution identity below. The identity is a statement about Gaussian densities; it does not, by itself, assert a conditional Markov property.

Theorem: For $s,t>0$ and $x,y\in\mathbb{R}$, $$ \int_{\mathbb{R}}p_s(x,z)p_t(z,y)\,dz=p_{s+t}(x,y). $$ If $P_0$ is taken to be the identity operator, the corresponding transition kernels also compose when $s=0$ or $t=0$.

Proof. Fix $s,t>0$ and $x,y\in\mathbb{R}$. To establish the identity, complete the square in the exponent. Define $m=(tx+sy)/(s+t)$. Expanding both sides gives $$ \frac{(z-x)^2}{2s}+\frac{(y-z)^2}{2t} =\frac{(y-x)^2}{2(s+t)}+\frac{s+t}{2st}(z-m)^2. $$ The product of the two density prefactors is $1/(2\pi\sqrt{st})$. Hence the integral on the left is $$ \frac{\exp\!\left(-\frac{(y-x)^2}{2(s+t)}\right)}{2\pi\sqrt{st}} \int_{\mathbb{R}}\exp\!\left(-\frac{s+t}{2st}(z-m)^2\right)\,dz. $$ The Gaussian integral in this expression is $\sqrt{2\pi st/(s+t)}$, by translating by $m$ and rescaling to the standard normal integral. Multiplying prefactors gives $1/\sqrt{2\pi(s+t)}$, so the result is $p_{s+t}(x,y)$. When one time is zero, its transition leaves the state unchanged; composing with that identity kernel leaves the other kernel unchanged. $\square$

For bounded Borel functions $f:\mathbb{R}\to\mathbb{R}$, define the transition operator $$ P_t f(x)=\int_{\mathbb{R}}p_t(x,y)f(y)\,dy\quad(t>0),\qquad P_0f(x)=f(x). $$ The integral exists absolutely: if $|f(y)|\le M$, then $\int p_t(x,y)|f(y)|\,dy\le M$. The operator also gives the expectation $\mathbb{E}[f(X^x_t)]$, since $X^x_t$ has density $p_t(x,\cdot)$.

Corollary: For bounded Borel $f$, $P_s(P_t f)(x)=P_{s+t}f(x)$ for all $s,t\ge0$ and $x\in\mathbb{R}$.

Proof. First take $s,t>0$. The iterated integral is absolutely integrable, since $$ \int_{\mathbb{R}}\int_{\mathbb{R}}p_s(x,z)p_t(z,y)|f(y)|\,dy\,dz \le M\int_{\mathbb{R}}p_s(x,z)\,dz=M. $$ Fubini’s theorem therefore permits exchanging the order of integration. Applying the theorem to the inner integral over $z$ yields $$ P_s(P_t f)(x)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}p_s(x,z)p_t(z,y)\,dz\right)f(y)\,dy =\int_{\mathbb{R}}p_{s+t}(x,y)f(y)\,dy. $$ This is $P_{s+t}f(x)$. If $s=0$, $P_0$ is the identity, so the equality reads $P_t f=P_t f$; if $t=0$, it reads $P_s f=P_s f$. These cases include $s=t=0$. $\square$

Worked Example: Composing Two One-Unit Transitions

Check the convolution formula for $s=t=1$, $x=0$, and $y=0$ by evaluating its integral.

Here $p_1(0,z)=(2\pi)^{-1/2}e^{-z^2/2}$ and $p_1(z,0)=(2\pi)^{-1/2}e^{-z^2/2}$. Their product is $(2\pi)^{-1}e^{-z^2}$. Consequently, using $\int_{\mathbb{R}}e^{-z^2}\,dz=\sqrt{\pi}$,

\int_{\mathbb{R}}p_1(0,z)p_1(z,0)\,dz =\frac{1}{2\pi}\sqrt{\pi} =\frac{1}{2\sqrt{\pi}}.

The combined transition has time $1+1=2$, so $p_2(0,0)=1/\sqrt{2\pi\cdot2}=1/\sqrt{4\pi}=1/(2\sqrt{\pi})$, agreeing with the integral. Check: both calculations give the density at the mean of a centered Gaussian with variance $2$.

Transition Probabilities and Bounded Expectations

Write $\Phi(a)=\int_{-\infty}^{a}(2\pi)^{-1/2}e^{-z^2/2}\,dz$ for the standard normal CDF. For $t>0$, a change of variable $z=(y-x)/\sqrt t$ shows that $$ \mathbb{P}(X^x_t\in(a,b])= \Phi\!\left(\frac{b-x}{\sqrt t}\right) -\Phi\!\left(\frac{a-x}{\sqrt t}\right),\qquad a<b. $$ The formula also gives the probability for $[a,b]$, $(a,b)$, or $[a,b)$: a Gaussian density assigns probability zero to either individual endpoint. The displayed interval convention is explicit, but endpoint choices do not alter this probability.

Worked Example: A Symmetric Interval Probability

For Brownian motion started at $x=2$, calculate the probability that its position at time $1$ belongs to $(1,3]$.

At time $1$, the standard deviation is $\sqrt1=1$. The standardized endpoints are $(1-2)/1=-1$ and $(3-2)/1=1$. Thus $$ \mathbb{P}(X^2_1\in(1,3])=\Phi(1)-\Phi(-1)=2\Phi(1)-1. $$ The last equality uses the symmetry $\Phi(-1)=1-\Phi(1)$.

Check: the interval is centered at the starting point $2$, so its probability is the probability that a standard normal variable lies between $-1$ and $1$. Including $1$ or excluding either endpoint does not change the result, because each endpoint has probability zero under a density.

More generally, if $f$ is bounded and Borel measurable, its expectation after time $t$ is an integral against the transition density. This lets us handle functions that are not continuous, including indicators of sets, without requiring a separate formula for each function.

Worked Example: Expectation of a Bounded Step Function

Let $X^0_t=B_t$, take $t=4$, and define $f(y)=1$ on $[-1,0)$, $f(y)=2$ on $[0,1]$, and $f(y)=0$ elsewhere. Find $\mathbb{E}[f(B_4)]$.

The standard deviation at time $4$ is $\sqrt4=2$. The probability of $[-1,0)$ is $\Phi(0)-\Phi(-1/2)$, and the probability of $[0,1]$ is $\Phi(1/2)-\Phi(0)$. Since $\Phi(0)=1/2$ and $\Phi(-1/2)=1-\Phi(1/2)$, both probabilities equal $\Phi(1/2)-1/2$. Weighting these probabilities by the two values of $f$ gives $$ \mathbb{E}[f(B_4)] =\big(\Phi(1/2)-1/2\big)+2\big(\Phi(1/2)-1/2\big) =3\big(\Phi(1/2)-1/2\big). $$

Check: the function takes values between $0$ and $2$, and the calculated expectation is positive and less than $2$, since $1/2<\Phi(1/2)<1$. The two interval probabilities are equal by symmetry, while the second interval contributes twice as much because $f$ has value $2$ there.

Why the Semigroup Matters—and a Common Mistake

The semigroup identity says that averaging over a time interval can be performed in stages without changing the result. For example, $P_s(P_t f)(x)=P_{s+t}f(x)$ means that first applying the time-$t$ averaging operation and then the time-$s$ operation gives the same answer as applying the single time-$(s+t)$ operation. This consistency is a basic tool for studying how expectations evolve over time.

It is important to separate this identity from the Markov property. The calculation above proves the composition rule for the Gaussian transition kernels and their operators. A conditional statement about the process given its filtration is a different claim; the distribution of Brownian increments supplies the ingredients for that claim, but the composition identity alone is not its proof.

Common mistake: using time as the standard deviation. A tempting but incorrect formula is $$ \frac{1}{\sqrt{2\pi t}}\exp\!\left(-\frac{(y-x)^2}{2\sqrt t}\right). $$ It has the correct-looking prefactor but puts $\sqrt t$, rather than $t$, in the exponent’s denominator. For $t=4,x=1,y=3$, this wrong expression gives $(8\pi)^{-1/2}e^{-1}$, because $(3-1)^2/(2\sqrt4)=4/4=1$. The correct variance is $4$, so the exponent denominator is $2t=8$, giving $(8\pi)^{-1/2}e^{-1/2}$. The standard deviation $\sqrt t$ appears when standardizing an endpoint; the variance $t$ appears in the density’s exponent denominator.

A second error is to multiply two density values and call the result the density for the combined time. The intermediate position is not fixed. It must be integrated out, which is precisely why the convolution has an integral over $z$. The Gaussian convolution theorem shows that this integration, not pointwise multiplication, produces the correct combined transition density.

Check Your Understanding

Use the density and semigroup identities to answer these questions.

Write $p_3(x,y)$ explicitly and identify the mean and variance of the corresponding transition law.
For $t>0$, explain why replacing $(a,b]$ by $[a,b]$ does not change its transition probability.
State the convolution identity for $s,t>0$ and describe what the identity kernel does when one time is zero.
For a bounded Borel function $f$, why is the integral defining $P_t f(x)$ absolutely convergent?
In the step-function example, why are the probabilities of $[-1,0)$ and $[0,1]$ equal?