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This is an exercise from Karatzas and Shreve.

Find a $(Y_s)_{s \in [0,1]}$ progressively measurable such that

  1. $ 0 < \int_0^1 Y_s ^2 ds < \infty$ almost surely, and

  2. $\int _0^1 Y_s dW_s = 0$ almost surely

where $W$ is a Brownian Motion.

Not having much luck so far. I know $Y$ cannot be deterministic and that the integral cannot be a true martingale but that is about it.

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1 Answer 1

up vote 2 down vote accepted

Here is a solution sketch. By making a time change, it is enough to find an $X$ where $\int_0^\infty X_r\,dB_r=0$ a.s. The geometric Brownian motion $dX = X\,dB$ goes to zero, but the problem is that when we write it as an integral, $X_t=X_0+\int_0^tX_r\,dB_r$, and set $X_0=0$, then $X_t=0$ for all $t$. So we might try a modification of this SDE: \[ dX = \left({\frac1{1+t} + X}\right)\,dB,\quad X_0=0. \] If $Z=1/(1+t)+X$, then we should have $\int_0^\infty Z_r\,dB_r=0$ a.s.

We then want to make a time change: \[ U_t = X_{t/(1-t)} = \int_0^{t/(1-t)} Z_r\,dB_r = \int_0^t V_s\,dM_s, \] where \[ V_t = Z_{t/(1-t)} = 1 - t + X_{t/(1-t)}, \] and $M_t=B_{t/(1-t)}$. Lastly, we observe that we can write \[ M_t = \int_0^t \frac1{1-s}\,dW_s, \] where $W$ is another Brownian motion. Hence, if we define \[ Y_t = \frac1{1-t}V_t = 1 + \frac1{1-t}X_{t/(1-t)}, \] then we have $U_t=\int_0^t Y_s\,dW_s$, and $U_1=0$ a.s.

Edit:

I cannot think of an easy proof that $X_t\to0$. For purposes of the exercise, there are other nonzero integrands satisfying $\int_0^\infty Z_r\,dB_r=0$ a.s. A simpler, although less interesting example, is $\int_0^\infty 1_{[0,T]}(r)\,dB_r=0$ a.s., where $T=\inf\{t\ge1:B_t=0\}$.

As for proving $X_t\to0$, the intuition was based on the fact that the solution to $dX = (\varepsilon + X)\,dB$ is $X=(X_0+\varepsilon)G-\varepsilon$, where $G_t = \exp(B_t - t/2)$ is the geometric Brownian motion. In this case, the solution tends to $-\varepsilon$. However, the solution to $dX = (1/(1+t) + X)\,dB$ with $X_0=0$, is \[ X_t = G_t - \frac1{1+t} - G_t\int_0^t \frac1{(1+s)^2}G_s\,ds. \] From here the proof would be easy if the process $H_t=G_t\int_0^t G_s^{-1}\,ds$ were almost surely bounded above. Unfortunately, this does not seem to be the case. Writing $dH=dt+H\,dB$ and doing some crude mental calculations with Feller's test for explosions, I believe it can be shown that $\liminf H=0$ a.s. and $\limsup H=\infty$ a.s. So it seems a more delicate analysis is required to prove that $X_t\to0$ (assuming it is even true).

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Thanks for your answer. Do you have a proof that $X_t \to 0$ as $t \to \infty$ ? –  mathman Mar 2 '12 at 11:46
    
I am sorry, but no. In the edit above, I offer an alternative solution to the problem which avoids this, and also some commentary on the question about proving $X_t\to0$. –  Jason Swanson Mar 6 '12 at 7:32
    
Thanks again for your effort. One point, though, if we go down your route of trying to change time by mapping $[0,1]$ to $[0, \infty)$, will we not end up with a process which is not progressively measurable? For example, if I understand you correctly, you think that $Y_s := 1_{[0,T]} (\frac{s}{1-s}) \frac{1}{1-s}$ is a solution, but I don't think it is progressively measurable. –  mathman Mar 8 '12 at 10:32
    
If $\tau=\inf\{s\ge1/2:M_s=0\}$, then $Y_s=1_{[0,\tau]}(s)\frac1{1-s}$. This process is adapted to $\{\mathcal{F}^M_t\}$, and therefore also adapted to $\{\mathcal{F}^W_t\}$. It also has left-continuous sample paths. It is therefore progressively measurable with respect to $\{\mathcal{F}^W_t\}$. –  Jason Swanson Mar 8 '12 at 19:43
    
I agree that $Y_t$ is $\{\mathcal{F}^M_t\}$ adapted but if $M_t = W_{t/(1-t)}$ and $t/(1-t) >t$ then why is $Y_t$ adapted to $\{\mathcal{F}^W_t\}$ ? –  mathman Mar 9 '12 at 14:40
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