Martingale

mathmari

Junior Member
Joined
Apr 15, 2013
Messages
75
Hi!!!
I need some help at the following exercise...
Let [FONT=MathJax_Math]B[/FONT] be a typical brownian motion with [FONT=MathJax_Math]μ[/FONT][FONT=MathJax_Main]>0[/FONT] and [FONT=MathJax_Math]x[/FONT] ε [FONT=MathJax_Math]R[/FONT]. [FONT=MathJax_Math]X[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]:=[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]μ[/FONT][FONT=MathJax_Math]t[/FONT], for each [FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]>=[/FONT][FONT=MathJax_Main]0[/FONT], a brownian motion with velocity [FONT=MathJax_Math]μ[/FONT] that starts at [FONT=MathJax_Math]x[/FONT]. For [FONT=MathJax_Math]r[/FONT] ε [FONT=MathJax_Math]R[/FONT], [FONT=MathJax_Math]Tr[/FONT]:=inf{[FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main]>=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main]:[/FONT][FONT=MathJax_Math]Xs[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]r[/FONT]} and [FONT=MathJax_Math]φ[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]:=[/FONT][FONT=MathJax_Math]e[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]μ[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main])[/FONT]. Show that [FONT=MathJax_Math]M[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]:=[/FONT][FONT=MathJax_Math]φ[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]Xt[/FONT][FONT=MathJax_Main])[/FONT] for t>=0 is martingale.

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???

To find
E(|Mt|) and E(Mt|Fs) do I have to use the property E(Bt-Bs)=0?

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[FONT=MathJax_Math]μ^[/FONT][FONT=MathJax_Main]2[/FONT]t|)=e-2μ(x+Bs)E(|e-2μ(Bt-Bs)|)E(|e-2[FONT=MathJax_Math]μ^[/FONT][FONT=MathJax_Main]2[/FONT]t|).

Is the mean value E(|e-2μ(Bt-Bs)|) equal to e0=1???
 
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