Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**lindah****Member**- Registered: 2010-01-25
- Posts: 121

Hi guys,

Sorry for the raft of questions, my exam is in a couple of days and I think my brain has imploded, so apologies if this is a very silly question

Currently I am assessing whether the Radon-Nikodym derivative is a martingale. A sufficient condition (Novikov's) for this is:

Does this condition simply requires the function to be finite. I've always taken this definition (integrable functions) for granted, and have never needed analysis until now.

For example if the expectation I obtain is

then this is not ?Thank you in advance for any feedback

*Last edited by lindah (2012-06-25 00:46:05)*

Offline

**lindah****Member**- Registered: 2010-01-25
- Posts: 121

Thanks to bobbym, bob bundy and gAr for your help with my questions recently!!

My exams have finished and I'll know results in a couple of weeks

Linda

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi lindah;

I remember reading that Girsanov's theorem says that sometimes that is a Martingale.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**lindah****Member**- Registered: 2010-01-25
- Posts: 121

Hi bobbym,

The Radon-Nikodym always has to be a martingale for Girsanov's change of measure to be valid w.r.t Brownian motion.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

How would you get e^(ax)? Where does the a come from?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**lindah****Member**- Registered: 2010-01-25
- Posts: 121

Hi bobbym;

I made up the e^ax, the a is supposed to be a constant

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Then I would assume that it is less then infinity. The integral is said to exist and is thus a Martingale. The H(s) would also have to reduce to a constant? So shouldn't that be an indefinite integral?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Pages: **1**