Stochastic Integral which is almost surely zero at fixed time This is an exercise from Karatzas and Shreve.
Find a $(Y_s)_{s \in [0,1]}$ progressively measurable such that


*

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

*$\int _0^1 Y_s dW_s  = 0$ 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.
 A: 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).
