Chapter 16 - Q16.1(ii)

[Darragh Kelly]

Hi,

Regarding Q16.1(i) from the notes I have a question on proving Xt is a martingale. I follow the proof right until the last line/statement.

For proving Xt is a martingale, how can we say v^T*ER[C|Fs]=Xs? Can someone explain this? C is a random variable at T>t ?

Thanks,

Darragh

 

[Steve Hales]

Hi
The definition of Xt is given in the question as:

Xt = v^T*ER[C|Ft]

Therefore (by changing t to s) it must also be true that:

Xs = v^T*ER[C|Fs]

This is required in the last line of the proof.
Hope that helps.

 

[Darragh Kelly]

Hi Steve,

Thanks for your help.

Ok so we start with EQ(Xt|Fs) = Xs (1) -> this is what we want to prove, equation for a martingale.

We sub Xt for Xt = v^T*ER[C|Ft], and by using the tower law we can rewrite L.H.S. of equation (1) EQ(Xt|Fs) = v^T*ER[C|Fs] which is equal to Xs.

We know v^T*ER[C|Fs] = Xs is a martingale, as it is written in the martingale equation form, and C is the random variable occurring at T>t>s, the future value give filtration Fs. This fits definition of a martingale? Finally V^T is just a constant, and multiplying a martingale by a constant doesnt affect the martingale property?

Therefore using the above argument, we can say v^T*ER[C|Fs] = Xs is a martingale and as v^T*ER[C|Fs] = EQ(Xt|Fs), then Xt must also be a martingale?

Thanks,

Darragh