MHB Brownian Motion: Martingale Property

[mathmari]

Hi!
I need some help at the following exercise...
Let $$B$$ be a typical brownian motion with $$μ>0$$ and $$x$$ ε $$R$$. $$ X_{t}:=x+B_{t}+μt$$, for each $$t>=0$$, a brownian motion with velocity $$μ$$ that starts at $$x$$. For $$r$$ ε $$R$$, $$T_{r}$$:=inf{$$s>=0:X_{s}=r$$} and $$φ(r):=exp(-2μr)$$. Show that $$M_{t}:=φ(X_{t})$$ for t>=0 is martingale.

Could you tell me the purpose of $$T_{r}$$??

 

[tarantella]

Hi!

The definition of $T_r$ will be probably used latter, but it seems it's not needed to solve the first question. Did you manage to do it?

 

[mathmari]

To show that M_t is martingale, I have to show that:
1. M_t is adapted to the filtration {F_t}_t>=0
2. For every t>=0, E(|M_t|)<oo
3. E(M_t|F_s)=M_s, for every 0<=s<=t
Right?

For the property 2. what I've done until now is:

E(|e^{-2μ(x+B_t+μt}|)=E(|e^{-2μ(x+B_t+B_s-B_s+μt)}|)=E(|e^-2μ(x+B_s)e^{-2μ(B_t-B_s)}e^{-2$$μ^2$$t}|)=e^-2μ(x+B_s)E(|e^{-2μ(B_t-B_s)}|)E(|e^{-2$$μ^2$$t}|).

Since B is a typical brownian motion, is the mean value E(|e^{-2μ(B_t-B_s)}|) equal to e⁰=1?

 

[tarantella]

I don't understand why you put $e^{-2\mu(x+B_s)}$ outside the expectation.

To see that $E(|M_t|)$ is finite for each $t$, it's enough to show that $E(e^{-2\mu B_t})$ is finite. Note that $B_t$ is normally distributed, hence the term $e^{-2\mu x}$ in the computation of the expectation should not be problematic.

 

[mathmari]

I put $$e^{-2μ(x+B_s)}$$ outside the expectation, because I thought that since it has no t, it's like a constant...
And how can I show the property 3?

 

[tarantella]

I put $$e^{-2μ(x+B_s)}$$ outside the expectation, because I thought that since it has no t, it's like a constant...

The problem is that what you get is a random variable, whereas an expectation should be a (deterministic number).

And how can I show the property 3?

You can use the idea you used for point 2 (it was not needed there, but it is now). Indeed, one can write $M_t=\exp(-2\mu(x+B_t+\mu t))=e^{-2\mu(x+\mu t) }\exp(-2\mu(B_t-B_s))\exp(-2\mu B_s)$. Computing the conditional expectation of this term goes like that: $\exp(-2\mu(B_t-B_s))$ is independent of $\mathcal F_s$ and $\exp(-2\mu B_s)$ is $\mathcal F_s$-measurable.

 

[mathmari]

Ok... Thank you very much! ;)