Almost Sure

25 March 10

Preservation of the Local Martingale Property

Now that it has been shown that stochastic integration can be performed with respect to any local martingale, we can move on to the following important result. Stochastic integration preserves the local martingale property. At least, this is true under very mild hypotheses. That the martingale property is preserved under integration of bounded elementary processes is straightforward. The generalization to predictable integrands can be achieved using a limiting argument. It is necessary, however, to restrict to locally bounded integrands.

Theorem 1 Let X be a local martingale and {\xi} be a locally bounded predictable process. Then, {\int\xi\,dX} is a local martingale.

Proof: By localization, we may suppose that {\xi} is uniformly bounded and that X is a proper martingale. So, {\vert\xi\vert\le K} for some constant K. Then, as previously shown there exists a sequence of elementary predictable processes {\vert\xi^n\vert\le K} such that {Y^n\equiv\int\xi^n\,dX} converges to {Y\equiv\int\xi\,dX} in the semimartingale topology and, hence, converges ucp. Being elementary integrals of a martingale, {Y^n} will be martingales. Also, {\vert\Delta Y^n\vert=\vert\xi^n\Delta X\vert\le K\vert\Delta X\vert}. Recall that a cadlag adapted process X is locally integrable if and only its jump process {\Delta X} is locally integrable, and all local martingales are locally integrable. So,

\displaystyle  \sup_n\vert\Delta Y^n_t\vert\le K\vert\Delta X_t\vert

is locally integrable. Then, by ucp convergence for local martingales, Y will satisfy the local martingale property. ⬜

This result can immediately be extended to the class of local {L^p}-integrable martingales, denoted by {\mathcal{M}^p_{\rm loc}}.

Corollary 2 Let {X\in\mathcal{M}^p_{\rm loc}} for some {0< p\le\infty} and {\xi} be a locally bounded predictable process. Then, {\int\xi\,dX\in\mathcal{M}^p_{\rm loc}}.

Proof: The previous theorem shows that {Y\equiv\int\xi\,dX} is a local martingale. By localization, we may suppose that {\xi} is uniformly bounded, so {\vert\xi\vert\le K} for some constant K. As {\Delta X} is locally {L^p}-integrable,

\displaystyle  \vert\Delta Y\vert\le K\vert\Delta X\vert

will also be locally {L^p}-integrable, as required. ⬜

Moving on, it seems natural to ask if Theorem 1 also applies to proper martingales. That is, if X is a cadlag martingale and {\xi} is a uniformly bounded predictable process, then is the integral {\int\xi\,dX} necessarily a martingale? Unfortunately, this is not true. Theorem 1 shows that the integral will be a local martingale, but there are examples where it is not a proper martingale. However, by placing some restriction on X, it is possible to get a positive result. In particular, the following result says that {\int\xi\,dX} will be a martingale if X is also square integrable. In fact, the result generalizes to {L^p}-integrable martingales for all {p>1}, although the proof of that more general statement will have to wait until after we have introduced the Burkholder-Davis-Gundy inequalities.

Lemma 3 Let X be a cadlag square integrable martingale and {\xi} be a bounded predictable process. Then, {\int\xi\,dX} is a square integrable martingale.

Proof: Suppose that {\vert\xi\vert\le K} for some constant K and set {Y=\int\xi\,dX}. Then, choose elementary predictable {\vert\xi^n\vert\le K} such that {Y^n\equiv\int\xi^n\,dX\rightarrow Y} in the semimartingale topology and, hence, {Y^n_t\rightarrow Y_t} in probability for each time t. Then, {Y^n} are martingales and, by the inequality for elementary integrals of square integrable martingales given in the previous post,

\displaystyle  {\mathbb E}\left[(Y^n_t)^2\right]\le K^2{\mathbb E}\left[X_t^2\right].

So, {\{Y^n\colon n=1,2,\ldots\}} is {L^2}-bounded and, hence, is uniformly integrable. As convergence in probability of a uniformly integrable sequence implies {L^1}-convergence, {{\mathbb E}\vert Y^n_t-Y_t\vert\rightarrow 0} as n goes to infinity. Therefore, Y is a martingale. Finally, passing to a subsequence so that {Y^n_t\rightarrow Y_t} almost surely, Fatou’s lemma can be applied

\displaystyle  {\mathbb E}\left[Y^2_t\right] = {\mathbb E}\left[\lim_{n\rightarrow\infty} (Y^n_t)^2\right]\le\liminf_{n\rightarrow\infty}{\mathbb E}\left[(Y^n_t)^2\right]\le K^2{\mathbb E}\left[X_t^2\right].

So Y is square integrable as required. ⬜

Finally, let us consider how Theorem 1 can be extended to arbitrary X-integrable integrands. In order for {Y\equiv\int\xi\,dX} to be a local martingale it must, at the very least, be locally integrable. In fact, it only needs to be shown that either the positive part {Y^+} or negative part {Y^-} of Y is locally integrable, as the following result shows.

Recall that we say that a process Y is locally integrable if there is a sequence of stopping times {\tau_n\uparrow\infty} such that the stopped processes {1_{\{\tau_n>0\}}(Y^*)^{\tau_n}} are all integrable, where {Y^*_t\equiv\sup_{s\le t}\vert Y_s\vert} is the maximum process of Y. The standard definition is often taken to apply only to nonnegative increasing processes, for which {Y^*=Y}. The more general definition is used here, as it seems to give slightly cleaner statements and proofs.

Theorem 4 Let {Y=\int\xi\,dX} for a local martingale X and X-integrable process {\xi}. Then, the following are equivalent.

  1. Y is a local martingale.
  2. Y is locally integrable.
  3. {Y^+} is locally integrable.
  4. {Y^-} is locally integrable.

Proof: Property 1 implies 2, because all local martingales are locally integrable. Then, 3 and 4 follow directly from 2. It only remains to show that 3 implies 1 because, applying the same result to -Y, 4 would then also imply 1, showing that all the conditions are equivalent.

So, suppose that {Y^+} is locally integrable. Consider the bounded predictable processes {\xi^n\equiv (\xi\wedge n)\vee(-n)}. These satisfy {\vert \xi^n\vert\le\vert\xi\vert} and have the same sign as {\xi} at all times. Therefore, setting {Y^n=\int\xi^n\,dX} gives {\Delta Y^n=\xi^n\Delta X} and {\Delta Y=\xi\Delta X}. So, {\vert\Delta Y^n\vert\le\vert\Delta Y\vert} and {\Delta Y^n}, {\Delta Y} have the same sign. In particular, {(\Delta Y^n)^+\le(\Delta Y)^+}. Furthermore, by Theorem 1, {Y^n} are local martingales and, by dominated convergence, tend ucp to Y. Passing to a subsequence, we may suppose that {Y^n} converge uniformly to Y on bounded intervals, with probability one.

Set

\displaystyle  M_t\equiv\sup_n(Y^n_t)^+

which is a cadlag adapted process. Being left-continuous, {Y_{t-}} is locally integrable. So {(\Delta Y)^+\le Y^+ - Y_{\cdot-}} is also locally integrable. Then,

\displaystyle  \Delta M_t \le \sup_n (\Delta Y^n)^+ \le (\Delta Y )^+

is locally integrable, and so is M. By localizing, we may suppose that {M^*\equiv\sup_t M_t} has finite expectation.

As convex functions of martingales are submartingales, {a\vee Y^n} are local submartingales, for any constant a. Furthermore, as {\vert a\vee Y^n\vert} is bounded by the integrable random variable {\vert a\vert\vee M^*}, they are proper submartingales. Then, by the dominated convergence theorem for convergence of random variables, {{\mathbb E}\vert a\vee Y^n_t- a\vee Y_t\vert\rightarrow 0} as n goes to infinity. It follows that {a\vee Y} is a submartingale. For any {s\le t}, monotone convergence gives

\displaystyle  Y_s = \lim_{a\rightarrow-\infty}a\vee Y_s\le\lim_{a\rightarrow-\infty}{\mathbb E}[a\vee Y_t\mid\mathcal{F}_s] ={\mathbb E}[Y_t\mid\mathcal{F}_s].

This shows that, after localizing, Y becomes a submartingale. Therefore, in general, Y is a local submartingale.

As local submartingales are locally integrable, it follows that Y and, in particular, {Y^-} are locally integrable. Then, the same argument as above can be applied to –Y to show that it is also a local submartingale. Finally, we have shown that locally, Y and –Y are both submartingales, so Y is a local martingale. ⬜

Example: Loss of the Local Martingale Property

Theorem 4 is as far as we can go in establishing the preservation of the local martingale property. There do exist integrals, with respect to local martingales, which are not locally integrable and, hence, cannot themselves be local martingales. To demonstrate this, consider the following simple example.

Let {U, \tau} be independent random variables such that {\tau} is uniformly distributed over the interval [0,1] and U has the discrete distribution {{\mathbb P}(U=1)={\mathbb P}(U=-1)=1/2}. Then define the process

\displaystyle  X_t = 1_{\{t\ge\tau\}}U.

With respect to its (completed) natural filtration {\{\mathcal{F}_t\}_{t\ge 0}}, X is a martingale. The only stopping times {\sigma\le\tau} defined on this filtration are of the form {\sigma=\tau\wedge t} for a fixed time {t\in\bar{\mathbb R}_+}. To see this, let t be the supremum of all times s satisfying {{\mathbb P}(\sigma>s)>0}. Then, for any {s<t}, the random variables {\{X_u\colon u\le s\}} are all zero when restricted to the set {\{\tau>s\}} and, therefore, any set {A\in\mathcal{F}_s} satisfies {{\mathbb P}(A\mid\tau>s)=0} or 1. In particular, {{\mathbb P}(\sigma>s\mid\tau>s)=1} and, therefore, {\{\tau>s\ge\sigma\}} has zero probability. This holds for all {s<t}, giving {\sigma\ge\tau\wedge t}. By construction, {\sigma\le t} almost surely, giving {\sigma=\tau\wedge t}.

Now consider the integral,

\displaystyle  Y_t = \int_0^t s^{-1}\,dX_s = 1_{\{t\ge\tau\}}\tau^{-1}U.

If {\sigma} is any stopping time with a positive probability of being non-zero, then {\sigma\wedge\tau=t\wedge\tau} for some fixed positive time t. So,

\displaystyle  {\mathbb E}\vert Y_\sigma\vert = {\mathbb E}\left[1_{\{\tau\le t\}}\tau^{-1}\vert U\vert\right] =\int_0^{t\wedge 1} s^{-1}\,ds = \infty.

Therefore Y is not locally integrable, and cannot be a local martingale.

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2 Comments »

  1. locally bounded predictable process

    Does this just mean that the predictable process is finite?

    Or in other words, what is an example for a predictable process that is not locally bounded?

    Thanks a lot. Not just for a prospective answer but for the blog in general. It’s the best textbook on stochastic calculus that I’ve ever seen.

    Comment by nutshell — 7 April 11 @ 7:06 PM | Reply

    • Hi. For a process \xi to be locally bounded means that there is a sequence of stopping times \tau_n increasing to infinity such that the stopped processes 1_{\{\tau_n > 0\}}\xi^{\tau_n} are uniformly bounded (as defined here). Equivalently, 1_{(0,\tau_n]}\xi are uniformly bounded.

      This certainly implies that \xi is finite. In fact, it implies that \xi^*_t\equiv\sup_{s\le t}\vert\xi_s\vert is almost surely finite for each time t. The converse does not hold in general, but for a right-continuous filtration and predictable process \xi, the converse does hold (its not easy to prove this without using some advanced results though).

      An example of a predictable but not locally bounded processes is \xi_t = 1_{\{t > 0\}}t^{-1}. An example of a predictable process for which \xi^* is finite but \xi is not locally bounded can be constructed as follows. Let U be an unbounded random variable (e.g., standard normal). Set \xi_t=1_{\{t > 0\}}\min(\vert U\vert,1/t). With respect to its natural filtration, this is not locally bounded. As \mathcal{F}_0 is trivial, any stopping time \tau with positive probability of being positive is almost-surely positive. Then, the supremum of 1_{\{\tau > 0\}}\xi^\tau is U, which is not locally bounded. Note that if we passed to the right-continuous filtration \mathcal{F}_{t+} then \xi will be locally bounded. As |U| is \mathcal{F}_{0+}-measurable, you can choose \tau_n to be equal to zero whenever |U| is greater than n and infinity otherwise.
      Examples of non-predictable processes (but adapted to a right-continuous filtration) which are not locally bounded but for which \xi^* is finite are given by Lévy processes with unbounded jump size.

      Comment by George Lowther — 8 April 11 @ 12:58 AM | Reply


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