Brownian Motion is not finite variation

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

Brownian Motion is not finite variation

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