A Stopped Martingale is still a Martingale?

[quotes]

Textbooks proved this.

Wt∧Ta, where Ta = inf{t: Wt=a}, is also a martingale when Wt is a martingale.

Does it prove anything? To price barriers?
Any explanation? I am totally confused on its usage to price barrier options.
Thanks in advance.

 

[stochan]

Consider a knock-and out barrier option written on a stock, let (B>S_0) be the barrier. You could study the discounted price of the stock or its stopped version, i.e. the (e^{-rt}S_{t\wedge T_B}). They are both martingale thanks to the theorem, but the stopped one is much better since you could think of the payoff of the option as something that depends only on the value of the process at time T.

 

[quotes]

Consider a knock-and out barrier option written on a stock, let (B>S_0) be the barrier. You could study the discounted price of the stock or its stopped version, i.e. the (e^{-rt}S_{t\wedge T_B}). They are both martingale thanks to the theorem, but the stopped one is much better since you could think of the payoff of the option as something that depends only on the value of the process at time T.

So you want to say that a Freezed martingale is still a martingale?
If (e^{-rt}S_{t\wedge T_B}) is a martingale, (e^{-rT}S_{t\wedge T_B}) is not.

 

[stochan]

sorry I meant (e^{-r(t\wedge T_B)}S_{t\wedge T_B}) does it sound correct now ?

 

[quotes]

The stock price S does not stop at the barrier, but does the replication portfolio of a barrier option.
Shreve illustrate that the barrier option stopped is still a martingale, but I still have not understood that.

 

[stochan]

The stock price S does not stop at the barrier, but does the replication portfolio of a barrier option.

You can assume that the stock price stops at the barrier. This doesn't change anything, as long as there is no point in trading the stock after it has reached the barrier (for the hedging purpose I mean..).
So now you have a new martingale (M_t=e^{-rt\wedge T_B}S_{t\wedge T_B}). Consider the payoff of the option as a random variable measurable al time (T). By the martingale representation theorem, you can represent it as an Ito integral, and you can do the same about the gain of a strategy on the stopped stock (since the latter is a mgale). Now equate the 2 integral and you get the hedging strategy.

Is it a bit better now?

 

[quotes]

For example, an up-and-out call option with Barrier J and Strike K<J. The stock price stops at barrier J at time t <=T.
In a risk-neutral measure, J(s=inf{t St=J})*exp(-rs) & S(T)*exp(-rT) is a martingale, correct?
Then we have a stop boundary for the martingale S(t)*exp(-rt) a near flat curve above and straight vertex on the right.

The option pay off is dependent on the martingale of discounted stock price on the right boundary. In fact, it only depends on part of the boundary. Any payoff on a martingale can be priced by risk-neutral measure by martingale representation theorem.

In this way the call is priced.

However, consider this one touch digital: money 1 is paid at terminal if price touches J during 0<J<T
payoff is at ending time T but the martingale is stopped halfway. I doubt if similar pricing can work.