Hi kenneth

Have a look at the following document http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf (In particular to Propositions 16.24,16.25, 16.26, and 16.30)

First $W_^1$ will be of class DL as soon as it is a martingale by proposition 16.25. So showing that $W^1$ is a martingale is sufficient to prove your claim (which follows from Proposition 16.30 for example if you know that a Brownian Motion is a martingale)

Second here is why Solution 2 doesn't work By proposition 16.26 if $M_\tau$ is in L^1 for every bounded stopping times (which is the case here).

We have $X_t$ is a martingale if $E[M_\tau]=E[M_0]$ for every bounded stopping time $\tau$.

This is the criteria you are trying to apply to get your contradiction.

The problem with this, is that $T_1$ is not bounded almost surely so you cannot apply the preceding criteria to show that $W^1$ is not of class DL.

I hope I didn't make any mistake

Regards

Hi kenneth

Have a look at the following document http://www.ma.utexas.edu/users/gordanz/teaching/10_Spring_M385D/lecture16.pdf (In particular to Propositions 16.24,16.25, 16.26, and 16.30)

First $W^1$ will be of class DL as soon as it is a martingale by proposition 16.25. So showing that $W^1$ is a martingale is sufficient to prove your claim (which follows from Proposition 16.30 for example if you know that a Brownian Motion is a martingale)

Second here is why Solution 2 doesn't work By proposition 16.26 if $M_\tau$ is in L^1 for every bounded stopping times (which is the case here).

We have $X_t$ is a martingale if $E[M_\tau]=E[M_0]$ for every bounded stopping time $\tau$.

This is the criteria you are trying to apply to get your contradiction.

The problem with this, is that $T_1$ is not bounded almost surely so you cannot apply the preceding criteria to show that $W^1$ is not of class DL.

I hope I didn't make any mistake

Regards