26 | May | 2012 | Eventually Almost Everywhere

We have defined the stochastic integral $(H\cdot A)_t$ for a finite variation, continuous adapted process A, and H a previsible process. We want to extend this to integrals with respect to continuous local martingales. Analogous to defining the Lebesgue integral by extending linearly from indicator functions, we define integrals of simple processes with respect to martingales. A simple process is a disjointly (time-)supported linear combination of the canonical previsible process $Z_s1_{(s,t]}$ . It will be convenient to demand that the variables of integration are in $\mathcal{M}^2$ . Then the space $\mathcal{M}^2$ is preserved under this integral operation, that is $H\cdot M\in\mathcal{M}^2$ , and in particular

$\mathbb{E}(H\cdot M)_\infty^2\leq ||H||_\infty^2\mathbb{E}(M_\infty-M_0)^2$

By Doob’s $L^2$ inequality, $X\in\mathcal{M}^2\Rightarrow ||\sup_t|X_t|||_2<\infty$ , so it makes sense to define $\mathcal{C}^2$ as the set of cadlag adapted processes s.t. $|||X|||:=||\sup_t|X_t|||<\infty$ . Of particular interest is that $\mathcal{C}^2$ is complete, by finding limits for subsequences and lifting to the whole sequence by Fatou.

We aim to construct integrals over $\mathcal{M}_{c,loc}$ by taking limits of integrals over $\mathcal{M}_c^2$ . For this space, we have that $M^2-[M]$ is not just $\in\mathcal{M}_{c,loc}$ but in fact a UI true martingale. Then, as a natural extension of the orthogonality trick (remembering that $M^2-[M]\in\mathcal{M}_{c,loc}$ ), we obtain

$\mathbb{E}[M]_\infty=\mathbb{E}(M_\infty-M_0)^2$

We then define for $M\in\mathcal{M}_c^2$ the norm $||\cdot||_M$ by $||H||_M^2:=\mathbb{E}\left(\int_0^tH_s^2d[M]_s\right)$ with $L^2(M)$ the space of H s.t. that this is finite. Then for $H\in \mathcal{S}$ , we have the Ito isometry:

$I: (L^2(M),||\cdot||_M)\rightarrow(M_c^2,||\cdot||)$ with $I(H)=H\cdot M$

Now, we also have S dense in $L^2(M)$ . Why? We know linear combinations of $1_A, A\in\mathcal{P}$ are dense, so it STP $A\in\mathcal{P}\Rightarrow A\in\bar{S}$ . Use a typical Dynkin’s lemma argument on $\mathcal{A}=\{A\in\mathcal{P}:1_A\in \bar{S}\}$ , which contains $\pi$ -system generating $\mathcal{P}$ . So extend to $L^2(M)$ generally. We recover the original motivating result that $[H\cdot M]=H^2\cdot[M]$ . Take a sequence of stopping times reducing both H and M to force boundedness of integrand, and $M\in\mathcal{M}_c^2$ . Stopping the integral is equivalent to stopping the integrand, and checking limits in stopping times allows us to lift the Ito Isometry result to this one about quadratic variations.

There is a natural definition of ‘pathwise’ stochastic integrals of a certain type of ‘simple’ process with respect to cadlag non-decreasing processes. It can be a shown that a function is of finite variation iff it can be expressed as the difference of two such functions. Hence, these finite variation processes can be used as variable of integration via an obvious linear extension. One direction of this result is obvious; the other is fiddly. To proceed, we show that the total valuation process is cadlag (and, obviously, increasing), and then check that $a'=\frac12(v+a),a''=\frac12(v-a)$ are processes satisfying the conditions of the result.

Our overall aim is to define integrals with respect to Brownian Motion since that is (in a sense to be made precise through the Dubins-Schwarz theorem later) the canonical non-trivial stochastic process with non-zero quadratic variation. The result we demonstrate shows that it is not possible to define the integral with respect to BM through pathwise finite variation integrals.

Theorem: $M\in\mathcal{M}_{c,loc},M_0=0$ a.s. is of finite variation. Then M is indistinguishable from 0.

We will show this for M a bounded martingale with bounded variation. Why does this suffice? In general, set $S_n:=\inf\{t,V_t\leq n\}$ , noting that V is continuous adapted non-decreasing. If $M^{S_n}\equiv 0\,\forall n$ , then we are done, as the $S_n$ s are increasing. But this is a bounded martingale with bounded variation.

To prove this, we make use of the orthogonality relation which is a key trick for this sort of result: If M is a martingale, with $M_s,M_t\in L^2$ , for s<t, then just by multiplying out:

$\mathbb{E}[(M_t-M_s)^2|\mathcal{F}_s]=\mathbb{E}[M_t^2-M_s^2|\mathcal{F}_s]$ a.s.

Now, for this particular result, we decompose

$\mathbb{E}[M_t^2]=\mathbb{E}\left[\sum_{k=0}^{2^n-1}(M_{(k+1)2^{-n}t}^2-M_{k2^{-n}t}^2)\right]=\mathbb{E}[\sum (M_{(k+1)2^{-n}t}-M_{k2^{-n}t})^2]$

and then we bound this last term as

$\leq \mathbb{E}\left[\sup_k [M_+-M_-]\sum_k |M_+-M_-|\right]$

Now, as $n\uparrow\infty$ , we have $\sum_k |M_+-M_-|\uparrow V_t\leq N$ by the boundedness assumption. Furthermore, M is almost surely continuous on [0,t] and so it is in fact uniformly continuous, which allows us to conclude that

$\sup_k |M_+-M_-|\downarrow 0$

By bounded convergence, this limit applies equally under the expectation, and so conclude that $\mathbb{E}M_t^2=0$ for each time t, and so for each time t the martingale is almost surely equal to 0. In the usual, can lift this to rational points by countability, then to all points by continuity.

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Daily Archives: May 26, 2012

Working towards the Ito Isometry

Brownian Motion is not finite variation