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Brownian Motion: Martingale Property

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hi!!!
I need some help at the following exercise...
Let \(\displaystyle B\) be a typical brownian motion with \(\displaystyle μ>0\) and \(\displaystyle x\) ε \(\displaystyle R\). \(\displaystyle X_{t}:=x+B_{t}+μt\), for each \(\displaystyle t>=0\), a brownian motion with velocity \(\displaystyle μ\) that starts at \(\displaystyle x\). For \(\displaystyle r\) ε \(\displaystyle R\), \(\displaystyle T_{r}\):=inf{\(\displaystyle s>=0:X_{s}=r\)} and \(\displaystyle φ(r):=exp(-2μr)\). Show that \(\displaystyle M_{t}:=φ(X_{t})\) for t>=0 is martingale.

Could you tell me the purpose of \(\displaystyle T_{r}\)??
 
Last edited:

girdav

Member
Feb 1, 2012
96
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
To show that Mt is martingale, I have to show that:
1. Mt is adapted to the filtration {Ft}t>=0
2. For every t>=0, E(|Mt|)<oo
3. E(Mt|Fs)=Ms, for every 0<=s<=t
Right???

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

E(|e-2μ(x+Bt+μt|)=E(|e-2μ(x+Bt+Bs-Bs+μt)|)=E(|e-2μ(x+Bs)e-2μ(Bt-Bs)e-2\(\displaystyle μ^2\)t|)=e-2μ(x+Bs)E(|e-2μ(Bt-Bs)|)E(|e-2\(\displaystyle μ^2\)t|).

Since B is a typical brownian motion, is the mean value E(|e-2μ(Bt-Bs)|) equal to e0=1???
 

girdav

Member
Feb 1, 2012
96
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
I put \(\displaystyle e^{-2μ(x+B_s)}\) 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??? :confused:
 

girdav

Member
Feb 1, 2012
96
I put \(\displaystyle e^{-2μ(x+B_s)}\) outside the expectation, because I thought that since it has no t, it's like a constant...
:confused:
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Ok.... Thank you very much!!!!!! ;)