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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathbb F=(\mathcal F)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
  • $B=(B_t)_{t\ge 0}$ be an $\mathbb F$-adapted Brownian motion with respect to $\mathbb F$
  • $h_i$ be a bounded, real-valued, $\mathbb F$-adapted random variable on $(\Omega,\mathcal A)$

Let $H=(H_t)_{t\ge 0}$ be of the form $$H_t(\omega)=\sum_{i=1}^nh_{i-1}(\omega)1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)$$ and $$I_t^B(H):=\sum_{i=1}^nh_{i-1}\left(B_{t_i\wedge t}-B_{t_{i-1}\wedge t}\right)\;\;\;\text{for }t\ge 0$$


I want to show, that $\left(I_t^B(H)\right)_{t\ge 0}$ is a $\mathbb F$-martingale. Please note, that $B$ is a $\mathbb F$-martingale. Hence, $$\operatorname E\left[I_\tau^B(H)\right]=0$$ by the optional stopping theorem, for any $\mathbb F$-stopping time $\tau$:

This might help to prove the desired statement. We need to show, that $$\operatorname E\left[I_t^B(H)\mid\mathcal F_s\right]=I_s^B(H)\;\;\;\text{for all }s<t\;.$$ How can we do that?

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You should also mention that $h_i$ is $\mathbb{F}_{t_i}$-measurable. Since $i$ is not a time index, it is not clear what it means for $(h_i)$ to be adapted. That being said, note that linear combinations of martingales are again martingales. So then we are done if we just show that whenever $s < t$

$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} (B_{t_i\wedge s} - B_{t_{i-1}\wedge s})$$

First consider the case $s < t_{i-1}$. Clearly, the RHS is $0$. For the LHS use the tower property as follows

$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = E[E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_{t_{i-1}}]\mid \mathbb{F}_s] = E[h_{i-1} E[ (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_{t_{i-1}}]\mid \mathbb{F}_s] = 0$$

To see why the last step is true note that $E[B_{t_i\wedge t} \mid \mathbb{F}_{t_{i-1}}] = B_{t_{i-1}\wedge t}$.

Now consider the case $s \geq t_{i-1}$. Doob helps a lot here. Remember that a stopped martingale is again a martingale. That is what I use below.

$$E[ h_{i-1} (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} E[ (B_{t_i\wedge t} - B_{t_{i-1}\wedge t})\mid \mathbb{F}_s] = h_{i-1} E[ (B^t_{t_i} - B^t_{t_{i-1}})\mid \mathbb{F}_s] = h_{i-1} (B^s_{t_i} - B^s_{t_{i-1}}) = h_{i-1}(B_{t_i\wedge s} - B_{t_{i-1}\wedge s})$$

This finishes the proof I think.

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  • $\begingroup$ Yeah, one can even show (using the OST), that an $\mathbb F$-adapted, integrable and right-continuous process $X$ is a $\mathbb F$-martingale if and only if $$\operatorname E[X_\tau]=\operatorname E[X_0]\;\;\;\text{for all bounded }\mathbb F\text{-stopping times }\tau\;.$$ I'd forgotten that for a moment. $\endgroup$ – 0xbadf00d Jul 7 '15 at 10:09
  • $\begingroup$ @0xbadf00d I don't immediately see why that result is relevant here. Perhaps you could elaborate a bit. $\endgroup$ – Calculon Jul 7 '15 at 10:58
  • $\begingroup$ We've got $$\operatorname E\left[I_\tau^B(H)\right]=0=\operatorname E\left[I_0^B(H)\right]\;\;\;\text{for all }\mathbb F\text{-stopping times }\tau\;.$$ $\left(I_t^B\right)_{t\ge 0}$ is $\mathbb F$-adapted and by boundedness of the $h_i$ integrable. Since $B$ is continuous, we can conclude that $\left(I_t^B\right)_{t\ge 0}$ is a $\mathbb F$-martingale. $\endgroup$ – 0xbadf00d Jul 7 '15 at 11:45
  • $\begingroup$ @0xbadf00d How do you know that $E[I^B_{\tau}(H)] = 0$ for an arbitrary stopping time $\tau$? $\endgroup$ – Calculon Jul 7 '15 at 11:51
  • $\begingroup$ By the OST, the stopped process $B^\tau$ is a $\mathbb F$-martingale. Especially, $$\operatorname E\left[B^\tau_t-B^\tau_s\mid\mathcal F_s\right]=0\;\;\;\text{for all }s<t\;.$$ Since $H$ is $\mathbb F$-adapted, we've got $$\operatorname E\left[H_{t_{i-1}}\left(B^\tau_{t_i}-B^\tau_{t_{i-1}}\right)\mid\mathcal F_{t_{i-1}}\right]=H_{t_{i-1}}\operatorname E\left[B^\tau_{t_i}-B^\tau_{t_{i-1}}\mid\mathcal F_{t_{i-1}}\right]=0\;$$ Thus, $$\operatorname E\left[I_\tau^B(H)\right]\stackrel{\text{def}}{=}\operatorname E\left[\sum_{i=1}^nH_{t_{i-1}}\left(B^\tau_{t_i}-B^\tau_{t_{i-1}}\right)\right]=0$$ $\endgroup$ – 0xbadf00d Jul 7 '15 at 12:14

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