Conditional Gaussian Increments and the Markov Property

From Independent Increments to Conditional Laws

Brownian motion is built from Gaussian increments that are independent of the past. This gives a direct way to answer a basic prediction question: once the current value is known, what is the distribution of the value after an additional time $h$? The answer depends on the present value and the time gap, but not on the earlier path. Expressing that answer as a transition kernel makes the Markov property precise and provides a reusable tool for conditional probabilities and expectations.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(B_t)_{t\ge0}$ be standard Brownian motion with state space $S=\mathbb{R}$, time index $t\ge0$, and $B_0=0$. Use the natural filtration $\mathcal{F}_t=\sigma(B_s:0\le s\le t)$, completed by null sets. Brownian motion has independent increments relative to this filtration: for $h\ge0$, $B_{t+h}-B_t$ is independent of $\mathcal{F}_t$, and for $h>0$ it has the normal distribution with mean $0$ and variance $h$. The case $h=0$ is the constant increment zero.

Definition: For $h\ge0$, the Brownian transition kernel $K_h(x,A)$ gives the probability of being in the Borel set $A\subseteq\mathbb{R}$ after time $h$, when the present state is $x$. For $h>0$, it is the normal probability measure with mean $x$ and variance $h$: $$ K_h(x,A)=\int_A \frac{1}{\sqrt{2\pi h}}\exp\!\left(-\frac{(y-x)^2}{2h}\right)\,dy. $$ At $h=0$, define $K_0(x,A)=\mathbf{1}_A(x)$, the point mass at $x$.

A kernel is a probability measure in its set argument for each fixed starting point, and is measurable in the starting point for each fixed Borel set. The Gaussian density integrates to one, so $K_h(x,\cdot)$ is a probability measure when $h>0$; the point mass has the same property when $h=0$. Measurability in $x$ follows from the integral formula: the integrand is a nonnegative Borel function of $(x,y)$, and integrating a nonnegative measurable function in $y$ gives a measurable function of $x$.

Theorem (Conditional Gaussian law and deterministic-time Markov property): For every deterministic $t,h\ge0$ and every Borel set $A\subseteq\mathbb{R}$, $$ \mathbb{P}(B_{t+h}\in A\mid\mathcal{F}_t)=K_h(B_t,A)\quad\text{almost surely}. $$ Equivalently, for every bounded Borel function $f:\mathbb{R}\to\mathbb{R}$, $$ \mathbb{E}[f(B_{t+h})\mid\mathcal{F}_t]=\int_{\mathbb{R}} f(y)\,K_h(B_t,dy)\quad\text{almost surely}. $$

Proof. Put $Z=B_{t+h}-B_t$. If $h>0$, the independent-increment property says that $Z$ is independent of $\mathcal{F}_t$ and has density $\frac{1}{\sqrt{2\pi h}}\exp(-z^2/(2h))$. Since $B_{t+h}=B_t+Z$, for a bounded Borel function $f$, conditioning on $\mathcal{F}_t$ leaves $B_t$ fixed and averages over the independent variable $Z$: $$ \mathbb{E}[f(B_{t+h})\mid\mathcal{F}_t] =\int_{\mathbb{R}}f(B_t+z)\frac{1}{\sqrt{2\pi h}}\exp\!\left(-\frac{z^2}{2h}\right)\,dz =\int_{\mathbb{R}}f(y)\,K_h(B_t,dy). $$ The last equality is the change of variable $y=B_t+z$. Taking $f=\mathbf{1}_A$ proves the assertion for Borel sets. If $h=0$, then $B_{t+h}=B_t$, and both sides of the set formula equal $\mathbf{1}_A(B_t)$. Thus the statement holds for every $h\ge0$. $\square$

The theorem conditions on all the information in the Brownian path up to deterministic time $t$, not merely on a particular observed value $B_t=x$. It says that this larger information set gives the same conditional law as the present state alone. When one writes “given $B_t=x$,” the corresponding formula is understood as a version of a regular conditional distribution; the event $B_t=x$ itself generally has probability zero, so ordinary conditioning on that singleton is not the justification.

Calculating with the Kernel

Worked Example: A Future Value Given the Present

Find the conditional distribution of $B_5$ given $\mathcal{F}_2$, and use it to compute the conditional mean, variance, and a symmetric interval probability.

The time gap is $5-2=3$. The increment $B_5-B_2$ is independent of $\mathcal{F}_2$ and is normal with mean $0$ and variance $3$. Therefore $$ B_5\mid\mathcal{F}_2\ \sim\ N(B_2,3), $$ where the notation means that the conditional distribution has the displayed random mean and fixed variance. In particular, $$ \mathbb{E}[B_5\mid\mathcal{F}_2]=B_2,\qquad \operatorname{Var}(B_5\mid\mathcal{F}_2)=3. $$

For the interval centered at the current value with radius $\sqrt{3}$, standardize the increment by dividing by its standard deviation $\sqrt{3}$: $$ \mathbb{P}(B_2-\sqrt{3}\le B_5\le B_2+\sqrt{3}\mid\mathcal{F}_2) =\mathbb{P}(-1\le Z\le1)=\Phi(1)-\Phi(-1), $$ where $Z\sim N(0,1)$ and $\Phi$ is its cumulative distribution function. By symmetry, this is $2\Phi(1)-1$. The endpoints are included; because a normal distribution has no point masses, using strict rather than non-strict endpoints gives the same probability.

Check: The interval is one conditional standard deviation on either side of $B_2$, since the conditional variance is $3$. Its standardized endpoints are therefore exactly $-1$ and $1$, as used in the calculation.

Worked Example: A Conditional Tail Probability

Calculate $\mathbb{P}(B_3>4\mid\mathcal{F}_1)$, and evaluate the resulting conditional probability at present value $B_1=2$.

The time gap is $3-1=2$, so $B_3-B_1$, given $\mathcal{F}_1$, is normal with mean $0$ and variance $2$. Subtract $B_1$ from both sides of the event and divide by the standard deviation $\sqrt{2}$: $$ \mathbb{P}(B_3>4\mid\mathcal{F}_1) =\mathbb{P}\!\left(\frac{B_3-B_1}{\sqrt{2}}>\frac{4-B_1}{\sqrt{2}}\,\middle|\,\mathcal{F}_1\right) =1-\Phi\!\left(\frac{4-B_1}{\sqrt{2}}\right). $$ The inequality remains strict after division because $\sqrt{2}>0$. At a present value $B_1=2$, the standardized threshold is $(4-2)/\sqrt{2}=\sqrt{2}$, giving the conditional probability $1-\Phi(\sqrt{2})$. This value at $B_1=2$ describes the kernel evaluated at that state; it does not require the event $\{B_1=2\}$ to have positive probability.

Check: If the present value increases, the threshold $4-B_1$ decreases, so the displayed upper-tail probability increases. This agrees with the fact that a larger current value makes $B_3>4$ more likely.

Kernel Composition and the Markov Property

The transition kernels are consistent across successive time intervals. This composition rule is also called the Chapman–Kolmogorov identity. It says that transitioning for time $h$, then for time $k$, gives the same distribution as transitioning once for time $h+k$.

Proposition (Semigroup identity): For all $h,k\ge0$, $x\in\mathbb{R}$, and Borel sets $A\subseteq\mathbb{R}$, $$ \int_{\mathbb{R}}K_k(y,A)\,K_h(x,dy)=K_{h+k}(x,A). $$

Proof. If $h=0$, the left side is $K_k(x,A)$, because $K_0(x,\cdot)$ puts all its mass at $x$; the right side is also $K_k(x,A)$. If $k=0$, then $K_0(y,A)=\mathbf{1}_A(y)$, so the left side is $K_h(x,A)$, again equal to the right side. Now suppose $h,k>0$. A transition from $x$ for time $h$ can be represented as $x+U$, where $U\sim N(0,h)$. The next transition can be represented as adding an independent $V\sim N(0,k)$. Their sum has density obtained by convolution. Writing $z$ for the final value, that density is $$ \int_{\mathbb{R}}\frac{1}{\sqrt{2\pi h}}e^{-(y-x)^2/(2h)} \frac{1}{\sqrt{2\pi k}}e^{-(z-y)^2/(2k)}\,dy =\frac{1}{\sqrt{2\pi(h+k)}}e^{-(z-x)^2/(2(h+k))}. $$ To verify the exponent, complete the square: $$ \frac{(y-x)^2}{h}+\frac{(z-y)^2}{k} =\frac{h+k}{hk}\left(y-\frac{kx+hz}{h+k}\right)^2 +\frac{(z-x)^2}{h+k}. $$ Integrating the centered Gaussian term in $y$ gives the normalizing factor $1/\sqrt{2\pi(h+k)}$. Thus the two-step density is the density defining $K_{h+k}(x,\cdot)$; integrating it over $A$ proves the identity. $\square$

Worked Example: Two Transitions Give One Longer Transition

Start at $x$, transition for time $2$, and then transition for time $3$. Find the distribution after both steps and the probability that the final value lies between $x$ and $x+\sqrt{5}$, including endpoints.

The first increment has distribution $N(0,2)$, and the second independent increment has distribution $N(0,3)$. Their sum is normal with mean $0+0=0$ and variance $2+3=5$. Consequently the final value has distribution $N(x,5)$, which is exactly the kernel $K_5(x,\cdot)$. Standardizing the requested interval gives $$ \mathbb{P}(x\le B_{\text{final}}\le x+\sqrt{5}) =\mathbb{P}\!\left(0\le\frac{B_{\text{final}}-x}{\sqrt{5}}\le1\right) =\Phi(1)-\Phi(0)=\Phi(1)-\frac12. $$ The inclusion of the endpoints does not alter the result, since this normal distribution assigns probability zero to each endpoint.

Check: The time gaps add to $2+3=5$, and the variance of the sum of independent increments is $2+3=5$. Both calculations give the same one-step kernel.

The theorem and semigroup identity together give the deterministic-time Markov description: the conditional law at a future time is obtained by applying a transition kernel to the current state, and those kernels compose consistently. The kernel is time-homogeneous because it depends on the time difference $h$, not on the absolute time $t$. Brownian motion itself is not a stationary process: for example, $B_t$ has variance $t$, which changes with $t$.

For a measurable function $f$ for which the integral exists, the same kernel gives the conditional expectation $$ \mathbb{E}[f(B_{t+h})\mid\mathcal{F}_t] =\int_{\mathbb{R}}f(y)\,K_h(B_t,dy). $$ For bounded $f$, existence is automatic. For unbounded $f$, integrability must be checked. For example, taking $f(y)=y$ and $f(y)=y^2$ is valid because Brownian values have finite second moments; the resulting conditional mean is $B_t$, and the conditional second moment is $B_t^2+h$.

Common Pitfall: Using the Wrong Variance

A frequent error is to use the unconditional variance of the future value when conditioning on the present. For instance, at $t=2$ and $h=3$, one might incorrectly write $B_5\mid\mathcal{F}_2\sim N(B_2,5)$, using $\operatorname{Var}(B_5)=5$. The correction is $B_5=B_2+(B_5-B_2)$: conditional on $\mathcal{F}_2$, the first term is already known, while the second is an independent $N(0,3)$ increment. Hence the conditional distribution is $N(B_2,3)$, not $N(B_2,5)$.

A useful check comes from total variance. Since $B_2$ has variance $2$, and the conditional variance of $B_5$ given $\mathcal{F}_2$ is $3$, $$ \operatorname{Var}(B_5) =\mathbb{E}[\operatorname{Var}(B_5\mid\mathcal{F}_2)] +\operatorname{Var}(\mathbb{E}[B_5\mid\mathcal{F}_2]) =3+\operatorname{Var}(B_2)=3+2=5. $$ Thus $5$ is the unconditional variance of $B_5$, while $3$ is the conditional variance of the future increment after time $2$.

Finally, this result has been stated for deterministic times $t$. A random time, such as a hitting time, is not interchangeable with a deterministic time in the conditioning argument: independent increments after a deterministic time alone do not establish the corresponding assertion after an arbitrary random time. That requires a separate result about the strong Markov property.

Check Your Understanding

Use the kernel and the independent-increment property to answer each question.

For $h>0$, state the conditional mean and conditional variance of $B_{t+h}$ given $\mathcal{F}_t$.
Write $K_0(x,A)$. Why is it not appropriate to use the positive-time Gaussian density formula at $h=0$?
Given $\mathcal{F}_4$, what is the conditional distribution of $B_7$? Identify both its mean and variance.
Explain why the conditional probability at a specified value $B_t=x$ is interpreted through a regular conditional distribution rather than ordinary conditioning on $\{B_t=x\}$.
For $h,k>0$, what property of independent Gaussian increments makes two successive transitions agree with one transition of length $h+k$?

Conditional Gaussian Increments and the Markov Property

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