Proving the Brownian Markov Property

From Independent Increments to a Markov Property

Independent increments describe how Brownian motion changes over disjoint time intervals. The Markov property is a different-looking statement: once the present value is known, information from the past gives no further information about the future. To prove this carefully, one must first pass from independence of finitely many increments to independence from the entire past, and then from finitely many future values to events involving the whole future path.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(B_t)_{t\ge0}$ be standard Brownian motion with state space $\mathbb{R}$: $B_0=0$, its paths are continuous almost surely, and its increments over disjoint intervals are independent, with $B_t-B_r\sim N(0,t-r)$ for $0\le r<t$. Use the completed natural filtration $\mathcal{F}_s=\sigma(B_u:0\le u\le s)$, completed by the $\mathbb{P}$-null events. All Markov statements below concern deterministic times $s\ge0$.

Definition: Brownian motion has the Markov property at deterministic times if, for every $s,t\ge0$ and bounded Borel function $g:\mathbb{R}\to\mathbb{R}$, the conditional expectation $\mathbb{E}[g(B_{s+t})\mid\mathcal{F}_s]$ is a function of $B_s$ alone. Its transition kernel is $P_t(x,A)=\mathbb{P}(x+Z_t\in A)$, where $Z_t\sim N(0,t)$; for $t=0$, $Z_0=0$.

In this definition, “a function of $B_s$ alone” means that the conditional expectation can be written as $P_tg(B_s)$, where $P_tg(x)=\mathbb{E}[g(x+Z_t)]$. The claim concerns conditioning on all information in $\mathcal{F}_s$, not just conditioning on the single value $B_s$. We will prove a stronger statement: the conditional law of the entire path after time $s$ is determined by $B_s$.

Independence from the Entire Past

Fix $s\ge0$ and define the future-increment process $W^{(s)}_u=B_{s+u}-B_s$, $u\ge0$. It has continuous paths and starts at zero. Its finite collections of increments have the same distributions and independence properties as those of Brownian motion. The additional fact needed is that this whole process is independent of $\mathcal{F}_s$.

Proposition: For each deterministic $s\ge0$, the process $(W^{(s)}_u)_{u\ge0}$ is independent of $\mathcal{F}_s$ and has the same law on continuous path space as standard Brownian motion.

Proof. First take finitely many past times $0\le u_1<\cdots<u_m\le s$ and finitely many future times $0\le v_1<\cdots<v_k$. The past vector $(B_{u_1},\ldots,B_{u_m})$ is a function of increments over intervals ending no later than $s$. The future vector $(W^{(s)}_{v_1},\ldots,W^{(s)}_{v_k})$ is a function of increments over intervals starting at or after $s$. Brownian independent increments therefore make these two vectors independent. The increments defining the future vector have the standard Brownian distributions, so its finite-dimensional distributions agree with those of Brownian motion. By continuity, the sigma-field of the past is generated, up to completion, by coordinates at rational times in $[0,s]$; likewise, the Borel sigma-field on continuous path space is generated by rational-time coordinates. The finite-dimensional independence just shown extends to these sigma-fields by the $\pi$-$\lambda$ theorem. Completion does not change independence, since adjoining null events does not change probabilities. Thus the future-increment path is independent of $\mathcal{F}_s$, and its continuous-path law is Brownian law. $\square$

This proposition gives the reason the present value matters: for every $u\ge0$, $B_{s+u}=B_s+W^{(s)}_u$. The part of the future not already specified by $B_s$ is a fresh Brownian path independent of the past.

Finite Collections of Future Times

Theorem: Let $s\ge0$, let $0\le t_1<\cdots<t_k$, and let $g:\mathbb{R}^k\to\mathbb{R}$ be bounded and Borel. Then $$ \mathbb{E}\!\left[g(B_{s+t_1},\ldots,B_{s+t_k})\mid\mathcal{F}_s\right] = \mathbb{E}\!\left[g(B_s+Z_{t_1},\ldots,B_s+Z_{t_k})\right], $$ where $(Z_{t_1},\ldots,Z_{t_k})$ has the joint law of Brownian motion at those times and is independent of $\mathcal{F}_s$. The right-hand side is evaluated with $B_s$ held fixed, so it is a Borel function of $B_s$.

Proof. Write $B_{s+t_i}=B_s+W^{(s)}_{t_i}$ for each $i$. By the proposition, the vector of future increments is independent of $\mathcal{F}_s$ and has the stated Brownian joint law. For a fixed value $x$, define $h(x)=\mathbb{E}[g(x+Z_{t_1},\ldots,x+Z_{t_k})]$. Independence gives the conditional expectation of $g(B_s+W^{(s)}_{t_1},\ldots,B_s+W^{(s)}_{t_k})$ as $h(B_s)$. One can verify this conditional-expectation identity first for products of bounded functions of $B_s$ and of the future-increment vector, using independence; bounded Borel functions follow by the monotone-class theorem. This proves the formula. $\square$

For a single future time, $B_{s+t}-B_s$ is $N(0,t)$. When $t>0$, the transition kernel has density $$ p_t(x,y)=\frac{1}{\sqrt{2\pi t}}\exp\!\left(-\frac{(y-x)^2}{2t}\right), $$ so $P_tg(x)=\int_{\mathbb{R}}g(y)p_t(x,y)\,dy$. At $t=0$, the kernel is concentrated at $x$, rather than having a density.

Worked Example: Two Future Values Given the Present

Take $s=2$, and consider $B_3$ and $B_5$. Set $U=B_3-B_2$ and $V=B_5-B_3$. The intervals $[2,3]$ and $[3,5]$ have lengths $1$ and $2$, so $U\sim N(0,1)$, $V\sim N(0,2)$, and $U,V$ are independent of each other and of $\mathcal{F}_2$. Thus, given the past, $B_3=B_2+U$ and $B_5=B_2+U+V$.

Writing $b$ for the present value in the conditional transition kernel, the conditional mean vector is $(b,b)$. The variances are $\operatorname{Var}(U)=1$ and $\operatorname{Var}(U+V)=1+2=3$, while the covariance is $\operatorname{Cov}(U,U+V)=1+0=1$. Therefore the conditional covariance matrix is $$ \begin{bmatrix}1&1\\1&3\end{bmatrix}. $$ Its determinant is $1\cdot3-1\cdot1=2$, and its inverse is $\frac12\begin{bmatrix}3&-1\\-1&1\end{bmatrix}$. Equivalently, the joint conditional density at $(z_1,z_2)$ is $$ \frac{1}{2\pi\sqrt{2}}\exp\!\left(-\frac{(z_1-b)^2}{2}-\frac{(z_2-z_1)^2}{4}\right). $$ Indeed, the two factors are the $N(b,1)$ density of $B_3$ and, given that value, the $N(z_1,2)$ density of $B_5$.

Check: The density integrates to one by the change of variables $u=z_1-b$, $v=z_2-z_1$, whose Jacobian has absolute value $1$. It becomes the product of the $N(0,1)$ and $N(0,2)$ densities. The covariance matrix is positive definite because its leading entry is $1>0$ and its determinant is $2>0$.

Worked Example: A Conditional Second Moment

Take $s=1$ and $t=2$. The increment $D=B_3-B_1$ is independent of $\mathcal{F}_1$, has mean $0$, and has variance $2$. Expanding the square gives $$ B_3^2=(B_1+D)^2=B_1^2+2B_1D+D^2. $$

Conditioning on $\mathcal{F}_1$, the first term stays $B_1^2$; the second has conditional expectation $2B_1\mathbb{E}[D\mid\mathcal{F}_1]=0$; and the last has conditional expectation $\mathbb{E}[D^2\mid\mathcal{F}_1]=2$. Hence $$ \mathbb{E}[B_3^2\mid\mathcal{F}_1]=B_1^2+2. $$

Check: The conditional mean of $B_3$ is $B_1$, and its conditional variance is $2$. The second moment is therefore $B_1^2+2$, agreeing with the direct expansion. The calculation is valid because $B_3^2$ is integrable.

Extending the Result to the Whole Future Path

Finite collections of future values are not the only events of interest. A path event might ask whether Brownian motion crosses a barrier at any time in an interval. To handle such events, regard a continuous future trajectory as one random element in $C([0,\infty),\mathbb{R})$, equipped with the Borel sigma-field for uniform convergence on compact intervals. Let $\nu$ denote the law of standard Brownian motion on this space.

Theorem (Conditional law of the future path): For fixed deterministic $s\ge0$ and any bounded Borel function $H:C([0,\infty),\mathbb{R})\to\mathbb{R}$, $$ \mathbb{E}\!\left[H\big((B_{s+u})_{u\ge0}\big)\mid\mathcal{F}_s\right] = \int H\big((B_s+w(u))_{u\ge0}\big)\,\nu(dw). $$ In particular, for any Borel set $A$ of continuous paths, the conditional probability that the future path belongs to $A$ is the probability that a fresh Brownian path, started at $B_s$, belongs to $A$.

Proof. The future path is the path-valued sum of the constant path at $B_s$ and $W^{(s)}$. By the proposition, $W^{(s)}$ is independent of $\mathcal{F}_s$ with law $\nu$. For functions of the form $f(x,w)=a(x)c(w)$, with $a,c$ bounded and Borel, independence gives $$ \mathbb{E}[f(B_s,W^{(s)})\mid\mathcal{F}_s] =a(B_s)\int c(w)\,\nu(dw). $$ Finite linear combinations of such products satisfy the corresponding integral identity. The monotone-class theorem extends the identity to every bounded Borel function $f$ of $(B_s,W^{(s)})$. Apply it to $f(x,w)=H((x+w(u))_{u\ge0})$. The addition map from $\mathbb{R}\times C([0,\infty),\mathbb{R})$ to continuous path space is Borel, so this is an allowable function. The resulting identity is the theorem. Taking $H$ to be the indicator of a Borel path set gives the conditional-probability statement. $\square$

Worked Example: Crossing a Fixed Barrier After Time Two

Consider the event that the path reaches level $1$ at some time between $2$ and $3$, including the initial time $2$. Its indicator is a Borel function of the future path on the interval $[0,1]$. If $B_2=b\ge1$, the event already holds at time $2$, so its conditional probability is $1$. If $b<1$, reaching level $1$ is equivalent to $\sup_{0\le u\le1}W^{(2)}_u\ge1-b$.

For $a>0$, the reflection principle gives $\mathbb{P}(\sup_{0\le u\le1}B_u\ge a)=2\mathbb{P}(B_1\ge a)=2(1-\Phi(a))$, where $\Phi$ is the standard normal distribution function. Substituting $a=1-b>0$, the conditional probability is therefore $$ \mathbb{P}\!\left(\sup_{2\le r\le3}B_r\ge1\mid\mathcal{F}_2\right) = \begin{cases} 1,&B_2\ge1,\\ 2\bigl(1-\Phi(1-B_2)\bigr),&B_2<1. \end{cases} $$ The cases use the non-strict barrier condition $\ge1$; at $B_2=1$, the event holds at the starting time.

Check: In the second case, $1-B_2>0$, so the reflection-principle formula applies, and $0\le 2(1-\Phi(1-B_2))\le1$. At the boundary $B_2=1$, the first case gives $1$. This is a conditional-probability formula as a function of the random variable $B_2$; it does not require conditioning on the zero-probability event $B_2=b$.

Scope and a Common Pitfall

The proved property is a deterministic-time Markov property. The argument fixes $s$ first and uses independent increments after that fixed time. It does not establish that the increments after an arbitrary random time are independent of the information available at that time; that distinct question requires the strong Markov property.

A common error is to treat the future value $B_{s+t}$ itself as an increment independent of the past. For example, taking $s=2$ and $t=1$, a mistaken calculation might assert $\operatorname{Cov}(B_2,B_3)=0$ and hence $\mathbb{E}[B_3\mid\mathcal{F}_2]=0$. But $B_3=B_2+(B_3-B_2)$, and the second term is independent of $\mathcal{F}_2$ with mean zero. Consequently, $$ \operatorname{Cov}(B_2,B_3) =\operatorname{Cov}(B_2,B_2)+\operatorname{Cov}(B_2,B_3-B_2) =2+0=2, \qquad \mathbb{E}[B_3\mid\mathcal{F}_2]=B_2. $$

The fresh, independent object is the increment path $(B_{s+u}-B_s)_{u\ge0}$, not the absolute future path $(B_{s+u})_{u\ge0}$. Adding the present value back is exactly what makes the conditional future law depend on the past only through $B_s$.

Check Your Understanding

For each question, distinguish the future increments from the future values and state the conditioning information precisely.

For fixed $s$ and $t>0$, what is the conditional distribution of $B_{s+t}-B_s$ given $\mathcal{F}_s$?
For fixed $s$ and $t>0$, express $\mathbb{E}[B_{s+t}^2\mid\mathcal{F}_s]$ in terms of $B_s$ and $t$.
Let $U=B_{s+1}-B_s$ and $V=B_{s+3}-B_{s+1}$. Find their variances and covariance, and say which information they are independent of.
Why does proving the conditional formula for finitely many future coordinates not, by itself, establish the formula for every Borel event of the future path? What sigma-field argument completes the proof?
In the barrier example, why must the case $B_2\ge1$ be treated separately from the reflection-principle case?