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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 $t$.

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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. –  Wolfgang Loehr Sep 10 '12 at 12:46
    
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)? –  Stephan Sturm Sep 10 '12 at 13:14
    
@Stephan, see edits. –  Grzenio Sep 10 '12 at 14:04
    
If $u$ is fixed, why do you specify that $a_t$ is a continuous process? –  Wolfgang Loehr Sep 10 '12 at 15:19

2 Answers 2

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.)

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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 $$ E[\int_0^t\phi_s^2 ds]<+\infty\quad\forall t\geq0. $$

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

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