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Let $B$ be a standard Brownian motion on $(\Omega, \mathcal{F}, P, ({\mathcal{F}_t})_{t\ge0})$, where the filtration is the one generated by $B$. Fix a time interval $[0,T]$. Define the process $X$ as the solution to the SDE $$ \mathrm dX_t = \sigma X_t\,\mathrm dB_t,\quad X_0 = 1. $$

Define, for each real number $\alpha$, a measure $P_{\alpha}$, such that $X$ under $P_{\alpha}$ solves the equation $$ \mathrm dX_t = \alpha X_t\,\mathrm dt + \sigma X_t\,\mathrm dB^{\alpha}, $$

where $B^{\alpha}$ is a Brownian motion under $P_{\alpha}$. Give an explicit expression for the Radon-Nikodym derivative (likelihood process) $$ L^{\alpha} = \frac{\mathrm dP_{\alpha}}{\mathrm dP_0}, $$ on $\mathcal{F}_t$.

So this is the question. And I guess you're supposed to use the Itô formula. But I've had a hard time grasping the question. Some guidance on how I could think and where I should begin would be more than appreciated!

(This is my first post on this site and also the first time i use TeX so might not look very good, hopefully you'll understand anyway!)

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Yep, looked like my attempt of using TeX was not very good. –  Good guy Mike Nov 29 '12 at 20:21
    
I've edited the TeX. Most of it was fine, except that you're missing \$'s (see the edit). –  Stefan Hansen Nov 29 '12 at 20:25
    
For a first time question this is well asked with a nice bit of context +1. +1 for also being relevant to my own current assignment :) –  Simon Hayward Nov 29 '12 at 20:42
    
Oh you needed to place those before and after, tried using them before the expressions but didn't make much of a difference. Thank you for editing! –  Good guy Mike Nov 29 '12 at 20:49

1 Answer 1

up vote 0 down vote accepted

You have the explicit solution

$$X_t=\exp{(\sigma B_t -\frac{1}{2}\sigma^2 t)}$$

which is a martingale (important!). I assume that it is meant that $B^0=B$, and $P^0$ is the original measure? What you are actually doing is going "backwards" in the BS-model. There you start with something like $dX_t=\alpha X_tdt+\sigma X_t dB^\alpha_t$ and you want to get rid of the drift term.

Define $$L^\alpha=\frac{dP^\alpha}{dP^0}:=\mathcal{E}(-\int\frac{\alpha}{\sigma}dB)$$ where $\mathcal{E}(X):=\exp{(X_t-\frac{1}{2}\langle X \rangle_t)}$ denotes the stochastic exponential. So we have to verify that this density does the job.

First note, that we have $L^\alpha >0$, $L^\alpha$ is a martingale and $E[L^\alpha_T]=1$, so we can indeed define an equivalent probability measure. What we know is: $B$ is a $P^0=P$ Brownian Motion, we have $P^\alpha$ is equivalent to $P$ and the density is of the form $\mathcal{E}(\int b_s dW_s)$, where $b_s=\frac{\alpha}{\sigma}$, which does not depend on $s$. Then Girsanov's Theorem tells us that $B=B^\alpha + \int\frac{\alpha}{\sigma}$ for a Brownian Motion $B^\alpha$ under $P^\alpha$, i.e. $dB=dB^\alpha +\frac{\alpha}{\sigma}dt$. Plug this into the original SDE, yields under $P^\alpha$:

$$dX_t=\sigma X_t d(B^\alpha_t+\frac{\alpha}{\sigma}dt)=\alpha X_tdt+\sigma X_t dB^\alpha_t$$

which is exactly the desired result.

Note:I'm just learning about this stuff myself, so please be critical! cheers

math

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Thank you very much, well presented solution. Yes I think you can assume that $B^0 = B$ (if by this you mean that B is the BM under $P^0$) and that $P^0$ is the original measure. I have a few questions though that I don't completely understand. At first, what makes it important that $dX_t$ is a martingale? –  Good guy Mike Dec 1 '12 at 13:43
    
@GoodguyMike Just let me know, which things are not clear. I try to clear up all upcoming questions –  math Dec 1 '12 at 13:44
    
@GoodguyMike Let's continue the discussion in chat chat.stackexchange.com/rooms/6577/… –  math Dec 1 '12 at 13:50
    
Ok, I'm quite new to this as I said before so still trying to figure out how everything works here :D –  Good guy Mike Dec 1 '12 at 13:52
    
Okey, I think I understood what you explained yesterday, but then it got me a bit confused since we had different notations e.t.c. So first we have these two SDEs. And I solve the first one for X using Itô's formula, giving me $X_t=\exp{(\sigma B_t -\frac{1}{2}\sigma^2 t)}$. Then I define a new measure $P^\alpha$ by $P^\alpha(A) = \int(Z_t1_A)dP$ Where $Z_t = exp(\theta B_t - frac{(\theta)^2}{2}t)$ Then by Girsanov's Thm we know that the Brownian Motion under this new measure is $B^\alpha_t = B_t - \theta t$ which gives us $B_t = B^\alpha_t + \theta t$ so we differentiate this and get –  Good guy Mike Dec 2 '12 at 11:13

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