Application of Bayes rule / Girsanov density

Hello,
I'm struggling with an application of bayes rule in a proof and I hope you can help me.

At first some preliminaries:
**P(t,T) is the price of a Zero-Coupon-Bond and B(t) is a discount factor<<--I think this isnt necessary to know in order to understand the problem below. But just for the completeness.


We have a prop measure Q and define an equivalent prop. measure $\displaystyle Q^T$by
$\displaystyle \frac{dQ^T}{dQ}=(P(0,T)B(T))^{-1}$
on a Filtration F_T, generated by Brownian Motions.

In my context we have in particulary for $\displaystyle t\leq T$
$\displaystyle \frac{dQ^T}{dQ}|F_t=\frac{P(t,T)}{P(0,T)B(t)}$


Problem:
Now one can show, that $\displaystyle \frac{P(t,S)}{P(t,T)}$ is a martingale.
But I do not understand the key step in the argumentation where bayes rule is used. The argumentation is for $\displaystyle u\leq t\leq T\wedge S$

$\displaystyle \mathbb{E}_{Q^T}\bigg[\frac{P(t,S)}{P(t,T)}\bigg| \mathcal{F}_u\bigg]=\frac{\mathbb{E}_\mathbb{Q}[\frac{P(t,T)}{P(0,T)B(t)}\frac{P(t,S)}{P(t,T)}| \mathcal{F}_u]}{\frac{P(u,T)}{P(0,T)B(u)}} = \frac{\frac{P(u,S)}{B(u)}}{\frac{P(u,T)}{B(u)}}= \frac{P(u,S)}{P(u,T)}$

The first equation is justified by bayes rule. But how?

I think it should be something like:

Represent the left Expectation through the Radon-Nikodym-Derivative and then use bayes, but I just can't complete it.


P.s. If some more "context" is needed, just say. But, I think the problem above is just a pure stochastic question.

Hello,

I'm trying and I can't really complete it... I think we have to go with multiplications and divisions of dQ's (and their conditionals, but that's the point where I'm in trouble).
What do you call the Bayes rule ? I mean which version is that here ? And what's the rv in the first conditional expectation (the one you're calling "left"), it's t, isn't it ? I'm just making sure of it, as I've studied sto calc for not a so long time.

Hey Moo,
thanks for your answer.

Yeah, P(.,T)_t is the stochastic Process.
I have the same understanding problem of which Bayes is meant. In the few proofs which all use the justification "Bayes rule", it is nowhere specified.

I have the same troubles with the R-N-derivative, in particulary with the conditionals.

Without conditions it is easier for the comprehension for me, when I write the exp.value as an integral. But I'm not sure, how to do this in the conditional case. That's my problem.

P.s. Treffende Signatur für ein englischsprachiges Mathe-Forum.;)

Hi again !

Sorry, actually ich sproche nicht Deutsch ! "Treffende" ??

I happened to have a document about sto calc since yesterday (a course by Tomas Björk, for your information), for a different purpose than solving your problem and I told myself this morning : "Hey, what if I searched for Bayes in there, because Zeroy's problem seems easy yet unreachable ??" and it gave something great, called "the abstract bayes' formula". In our case, it exactly gives :
(while noticing that $\displaystyle \mathcal F_u\subseteq \mathcal F_t$)

$\displaystyle E_{Q^T}\left[X\mid\mathcal F_u\right]=\frac{E_Q\left[\left(\frac{dQ^T}{dQ}|\mathcal F_u\right) X |\mathcal F_t \right]}{E_Q\left[\left(\frac{dQ^T}{dQ}|\mathcal F_t\right) |\mathcal F_u \right]}$

(I can't view the LaTeX images at work, I hope they're correct)
And X is P(t,S)/P(t,T).

And I think that to finish, you just have to deal with the denominator, which gives the answer since dQ^T/dQ | F_t shall be a martingale under the measure Q (thanks to Girsanov ?).

If by tonight you still can't find the answer, just tell me, because with this Bayes' rule I'm a.s. that we can find the answer a.s. (Rofl)

Auf wiedersen !

Hey Moo,

yes, that's GREAT. It just looks like exactly the formula we need.

Unfortunately I haven't much time these days(till sunday) for math. But I think maybe at least tonight I will take a little bit time and try to complete the proof, if it's not so hard anymore. And I think the rest shouldn't be.

Again, big thanks for this Bayes Version. I have "Arbitrage theory" of thomas björk and will take a look on it, if I can find this Bayes in it. I'm pretty sure it should be in it.

best regards!

P.s. "Treffend" -> fitting, accurate, to the point.
-> you have a well fitting signature for a math-forum ^^

Quote:

---

Originally Posted by Zeroy

Hey Moo,

yes, that's GREAT. It just looks like exactly the formula we need.

Unfortunately I haven't much time these days(till sunday) for math. But I think maybe at least tonight I will take a little bit time and try to complete the proof, if it's not so hard anymore. And I think the rest shouldn't be.

Again, big thanks for this Bayes Version. I have "Arbitrage theory" of thomas björk and will take a look on it, if I can find this Bayes in it. I'm pretty sure it should be in it.

best regards!

P.s. "Treffend" -> fitting, accurate, to the point.
-> you have a well fitting signature for a math-forum ^^

---

No, now I'm sure, it's pretty straightforward with this formula (which is, with the adapted notations, the very formula of this Bayes' rule).
I found it on the 4th part of a pdf entitled "continuous time finance", which corresponds to some "chapter 10-12", so maybe this refers to the book you have ?

The only point remaining is to show that dQ^T/dQ | F_t is a martingale under Q, which should be the case because Q is the risk neutral probability, or something like that ? That would be left to you :D

I'm not sure if it's fitting, people aren't dumb, they may just ask dumb questions (which is not the case here, don't worry (Rofl))

See ya & good luck !

haha, I'm stupid, I hadn't invest the 5 seconds, to look again at the Proof in the initial Post, because this bayes is exactly what stands there in the first equation. The denominaters are equal, because as you had seen properly, Q is the risk neutral measure, hence per definition the discounted Price Process P/B is a martingal.

My comment regarding your signature was just ironcal. (Wink)

thanx again,
gz