# Strong Markov Property for BM

The Strong Markov Property is the most important result to demonstrate for any Markov process, such as Brownian Motion. It is also probably the most widely requested item of bookwork on the Part III Advanced Probability exam. I feel it is therefore worth practising writing as quickly as possible.

Theorem (SMP): Take $(B_t)$ a standard $(\mathcal{F}_t)$-BM, and T an a.s. finite stopping time. Then $(B_{T+t}-B_T,t\geq 0)$ is a standard BM independent of $\mathcal{F}_T$.

Proof: We write $B_t^{(T)}=B_{T+t}-B_T$ for ease of notation. We will show that for any $A\in\mathcal{F}_T$ and F bounded, measurable:

$\mathbb{E}[1_AF(B_{T+t_1}-B_T,\ldots,B_{T+t_n}-B_T)]=\mathbb{P}(A)\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

This will suffice to establish independence, and taking $A=\Omega\in\mathcal{F}_t$ shows that $B_t^T$ is a standard BM since (Levy), BM is uniquely characterised by its finite joint distributions.

To prove the result, we approximate discretely, and apply the Markov property.

$\mathbb{E}[1_AF(B_{t_1}^{(T)},\ldots)]=\lim_{m\rightarrow\infty}\sum_{k=1}^\infty \mathbb{E}[1_{A\cap\{T\in((k-1)2^{-m},k2^{-m}]\}}F(B_{t_1}^{(k2^{-m})},\ldots)]$

by bounded convergence, using continuity of F, right-continuity of B, and that $T<\infty$ a.s. (so that $1_A=\sum 1_{A\cap \{T\in(-,-]\}}$)

$\stackrel{\text{WMP}}{=}\lim_{m\rightarrow\infty}\sum_{k=1}^\infty \mathbb{P}[A\cap\{T\in((k-1)2^{-m},k2^{-m}]\}]\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

$\stackrel{\text{DOM}}{=}\mathbb{P}(A)\mathbb{E}F(B_{t_1},\ldots,B_{t_n})$

which is exactly what we required.

Remarks: 1) We only used right-continuity of the process, and characterisation by joint marginals, so the proof works equally well for Levy processes.

2) We can in fact show that it is independent of $\mathcal{F}_T^+$, by considering $T+\frac{1}{n}$ which is still a stopping time, then taking a limit in this as well in the above proof. For details of a similar result, see my post on Blumenthal’s 0-1 Law.

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