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(Stochastic calculus and Brownian motion, LeGall, page 80).

Suppose $M = (M_t)$ is a martingale. Also, let $Z$ be a bounded random variable which is $\mathcal{F}_r$ adapted. Then we like to show that for any $0 \leq r < s$, $$N_t = Z(M_{t\wedge s} - M_{t \wedge r})$$ is a martingale.

My attempt: $Z(M_{t\wedge s} - M_{t \wedge r}) \in L_1$, should be fine since both $Z$ and $M$ are bounded. I am not sure how to show it is adapted. Finally we need to show the martingale identity. Suppose $v \geq r$, then we have $$\mathbb{E} \{Z(M_{t\wedge s} - M_{t \wedge r}) \mid \mathcal{F}_v \} = Z \mathbb{E} \{(M_{t\wedge s} - M_{t \wedge r}) \mid \mathcal{F}_v \} = Z (M_{v\wedge s} - M_{v \wedge r})$$ where in the first equality we use the fact that $Z \in \mathcal{F}_r$, and $v \geq r$. This proves the result for this case. However, I don't know how to get the result for general $v$.

Thanks for you helps in advance.

saz
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Nick
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  • "and hence the bound is in $L_1$" could you elaborate why you believe that this is true? Note that $X<\infty$ does not imply $X \in L_1$. – saz Jun 08 '18 at 04:20
  • My mistake. thanks for reminding me. I edited the response. In fact, I couldn't show even the $L_1$ yet. – Nick Jun 08 '18 at 06:57
  • That's not surprising because it is in general wrong. If your assertion was correct, it would imply that $(M_r)_{r \leq t}$ is a martingale for any local martingale $M$... which is clearly wrong. – saz Jun 08 '18 at 12:53
  • Do you impose $r \leq t$ ? Or could $Z$ be rather $F_t$-adapted ? I cannot imagine how $N_t$ is adapted (with respect to $F_t$, right?) as you state it. – Leon Jun 08 '18 at 13:10
  • It seems too much for the general local martingale. I have modified the statement of the problem according to LeGall (page 80). The integrability should be fine in this case, but still addaptedness and martingale identity is not. Thanks again. – Nick Jun 09 '18 at 01:56

1 Answers1

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First of all, we note that

$$N_t = \begin{cases} 0, & t \leq r, \\ Z (M_t-M_r) & t \in (r,s), \\ Z (M_s-M_r), & t \geq s. \end{cases} \tag{1}$$

Since $Z$ is, by assumption, $\mathcal{F}_r$-measurable and $(M_t)_{t \geq 0}$ is $(\mathcal{F}_t)_{t \geq 0}$-adapted, this implies, in particular, that $(N_t)_{t \geq 0}$ is $(\mathcal{F}_t)_{t \geq 0}$-adapted. Moreover, $M_t \in L^1$ together with the boundedness of $Z$ entails $N_t \in L^1$ for all $t \geq 0$. It remains to check that

$$\mathbb{E}(N_t \mid \mathcal{F}_u) = N_u \quad \text{for all $u \leq t$}. \tag{2}$$

We consider two cases separately. If $t \geq u \geq r$ then it follows from the $\mathcal{F}_r$-measurability of $Z$ that

$$\mathbb{E}(N_t \mid \mathcal{F}_u) = Z \mathbb{E}(M_{t \wedge s}-M_{t \wedge r} \mid \mathcal{F}_u).$$

Because of the martingal property of $(M_t)_{t \geq 0}$ we have

$$ \mathbb{E}(M_{t \wedge s}-M_{t \wedge r} \mid \mathcal{F}_u) = M_{t \wedge s \wedge u} - M_{t \wedge r \wedge u} \stackrel{t \geq u}{=} M_{u \wedge s} - M_{u \wedge r},$$

and so

$$\mathbb{E}(N_t \mid \mathcal{F}_u) = N_u, \qquad r \leq u \leq t. $$

In particular,

$$\mathbb{E}(N_t \mid \mathcal{F}_r) = N_r \stackrel{(1)}{=} 0. \tag{3}$$

It remains to consider the case $u < r$. If $t \leq r$ then, by $(1)$, $N_t = 0$ and so

$$\mathbb{E}(N_t \mid \mathcal{F}_u)=0=N_u.$$

Suppose now that $t \geq r$. Since the tower property gives

$$\mathbb{E}(N_t \mid \mathcal{F}_u) = \mathbb{E} \bigg[ \mathbb{E}(N_t \mid \mathcal{F}_r) \mid \mathcal{F}_u \bigg], $$

we conclude from $(3)$ that

$$\mathbb{E}(N_t \mid \mathcal{F}_u) = 0 = N_u.$$

This finishes the proof of $(2)$.

saz
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