Dubins-Schwarz Theorem

In developing the stochastic integral, much of our motivation has come from considering integrals with respect to Brownian Motion. In this section, we develop some results which justify that Brownian Motion is the canonical stochastic process with non-zero quadratic variation (which is related, but not directly equivalent to the property of infinite total variation). In particular, we shall observe the Dubins-Schwarz theorem, which shows that martingales with unbounded (as time \rightarrow\infty) quadratic variation ARE Brownian Motion, up to a (stochastic) time change.

Recall Levy’s characterisation of a d-dimensional BM, which allows us to avoid considering independent normal increments. Given X^1,\ldots,X^d\in\mathcal{M}_{c,loc}:

X=(X^1,\ldots,X^d) a BM iff [X^i,X^j]_t=\delta_{ij}t

Obviously, one direction has been shown as part of the construction and properties of quadratic variation. For the other direction,, because laws are precisely defined by characteristic functions, it suffices to show that

\mathbb{E}\left[\exp(i\langle \theta,X_t-X_s\rangle)|\mathcal{F}_s\right]=\exp(-\frac12||\theta||^2(t-s))

We set Y_t:=\langle \theta,X_t\rangle, and deduce [Y]=t||\theta||^2 and Z=\mathcal{E}(iY)=\exp(iY_t+\frac12[Y]_t)\in\mathcal{M}_{c,loc}, and furthermore is bounded on compact [0,t], hence is a true martingale. So \mathbb{E}\left(\frac{Z_t}{Z_s}|\mathcal{F}_s\right)=1 which is pretty much what was required.

Now, Dubins-Schwarz states

Theorem: Given M\in\mathcal{M}_{c,loc}, M_0=0, [M]_\infty=\infty almost surely, if we set \tau_s:=\inf\{t:[M]_t>s\}, then B_s:=M_{\tau_s} is a (\mathcal{F}_{\tau_s})-BM, with M_t=B_{[M]_t}.

This final result is clear if [M]_t is almost surely strictly increasing in t: just take s=[M]_t in the definition.

We know B is cadlag: we first show B as defined is almost surely continuous. It remains to show B_{s-}=B_s\,\forall s>0\iff M_{\tau_{s-}}=M_{\tau_s}, noting that \tau_{s-}=\inf\{t\geq 0:[M]_t=s\} (by continuity) is a stopping time also.

The only interesting case is if \tau_{s-}<\tau_s, for which need to show [M] is constant. This is intuitively obvious, but formally, we must appeal to (M^2-[M])^{\tau_s} which is UI, since \mathbb{E}[M^{\tau_s}]_\infty<\infty. Now may apply OST to obtain \mathbb{E}[M_{\tau_s}^2-M_{\tau_{s-}}^2|\mathcal{F}_{\tau_{s-}}]=\mathbb{E}[(M_{\tau_s}-M_{\tau_{s-}})^2|\mathcal{F}_{\tau_{s-}}]=0 which implies M is almost surely constant on [\tau_{s-},\tau_s]. We need to lift this to the case where it holds for all s simultaneously almost surely. Note that cadlag almost surely plus almost surely continuous at each point does not implies almost surely continuous everywhere (eg consider H(U(0,1)) with H the Heaviside function and U a uniform distribution). Instead, we record intervals of constancy of both M_t,[M]_t. That is, we set

T_r=\inf\{t>r:M_t\neq M_r\},\quad S_r=\inf\{t>r:[M]_t\neq [M]_r\}

Then these are cadlag, and by above T_r=S_r\,\forall r\in\mathbb{Q}^+ a.s. therefore T_r=S_r\,\forall r almost surely. Thus M, [M] are constant on the same intervals.

We also check B is adapted to \mathcal{G}_t=\mathcal{F}_{\tau_t}. STP X_T1_{\{T<\infty\}} is \mathcal{F}_T-measurable for X cadlag adapted. Approximating T discretely from above gives the result, exploiting that the result is clear if T has countable support. Now, obtain M^{\tau_s}\in\mathcal{M}_c^2, so M_{t\wedge \tau_s} UI by Doob, so by OST, get \mathbb{E}[M_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}, to get B a martingale. The finally:

\mathbb{E}[B_s^2-s|\mathcal{G}_r]=\mathbb{E}[(M^2-[M])_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}^2-[M]_{\tau_r}=B_r^2-r

And so we can apply Levy’s characterisation to finish the result.

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3 thoughts on “Dubins-Schwarz Theorem

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