For some time $u$ and positive continuous process $a_t$ adapted to $\mathcal{F}_t$ I have a (continuous-time) martingale defined as:

$$M_t(u) = \mathbb{E}[a_u | \mathcal{F}_t]$$

for $t\leq u$. I have a few questions about properties of $M_t$:

  1. Is it a uniformly integrable martingale? (or under what conditions it is)
  2. According to the martingale representation theorem it can be written as:

$$M_t(u) = M_0(u) + \int_0^t v_s(u) M_s(u) dW$$

where $W$ is the Brownian motion generating the filtration. What are the properties of $v_s(u)$? Does it need to be square-integrable? Bounded?

EDIT: I assume standard Brownian filtration. $u$ is regarded as a parameter, so I am only interested in behaviour in $t$.

  • $\begingroup$ If I understand 1. correctly, the answer is yes (assuming of course that $a_u$ is integrable), because $a_u$ is just a fixed random variable an you can use Jensen for conditional probabilities. $\endgroup$ – Wolfgang Loehr Sep 10 '12 at 12:46
  • $\begingroup$ Please clarify what is your filtration (are you assuming indirectly the Brownian?) and for which variable you want uniformity (just t up to u -then Wolfgang Loehr is right) or in both variables (on infinite time horizon)? $\endgroup$ – Stephan Sturm Sep 10 '12 at 13:14
  • $\begingroup$ @Stephan, see edits. $\endgroup$ – Grzenio Sep 10 '12 at 14:04
  • $\begingroup$ If $u$ is fixed, why do you specify that $a_t$ is a continuous process? $\endgroup$ – Wolfgang Loehr Sep 10 '12 at 15:19

I think question 1) is reasonably answered by Wolfgang Loehr in his comment. To get a counterexample for your claims in 2), just set $a_u=W_u^2-u$ for your Brownian motion. Ito's formula gives you the martingale representation $$ W_t^2-t = 0+\int_0^t 2W_s dW_s = 0 + \int_0^t \frac{2W_s}{W_s^2-s}(W_s^2-s) dW_s$$ Thus for in your terms you have $v_s(u) = \frac{2W_s}{W_s^2-s}$. Trying to integrate this you get as antiderivative an exponential integral with pole at $\sqrt{s}$, thus $v_s(u)$ is not integrable.

On the positive side you have always (even when you have just a local martingale) that $v_s(u) M_s(u)$ is predictable locally in $L^2$ (cf. Revuz/Yor, Continuous Martingales and Brownian Motion, Theorem V.3.4) and it is square integrable if your $a_u$ is square integrable (Proposition V.3.2). Under some regularity conditions in terms of Malliavin calculus you may calculate $v_s(u) M_s(u)$ even explicitly by means of the Clark-Ocone formula (see e.g. the Lecture notes of Eulalia Nualart, Section 1.5.3.)


In the book of Meyer, Probability and Potential ther is the following result (Theorem 19 in Chapter 5) Let $\mathscr{F}$ be a $\sigma$-algebra and $X\in L^1(P)$. Then the family $$(\{E[X|\mathscr{G}],\quad\mathscr{G}\subseteq\mathscr{F}\quad\ sub-\sigma-algebra\})$$ is uniformly integrable.

The predictable representation property of the Wiener process states that every martingale $M$ adapted to the filtration generated by $W$ can be represented as $$ M_t=M_0+\int_0^t\phi_sdW_s $$ where $\phi$ is a predictable process. Every martingale of the Wiener filtration is then conctinuous and therefore locally square integrable. Because of the definition of stochastic integral with respect to $W$ this means that $\phi$ satisfies the following condition $$ E[\int_0^t\phi_s^2 ds]<+\infty\quad\forall t\geq0. $$

  • $\begingroup$ I do not think that you can get better properties on $\phi$. $\endgroup$ – Paolo Sep 11 '12 at 16:06

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