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Suppose $B(t)$ is a standard Brownian motion, and $B_{1}(t)$ is given by $dB_{1}(t)=\mu dt+dB(t)$. Suppose $P$ is the Wiener measure induced by $B(t)$ on the $C[0,\infty)$, and $P_{1}$ is the Law induced by $B_{1}(t)$ on $C[0,\infty)$. Here we follow the definitions of law is referred to

http://math.stackexchange.com/questions/90268/how-is-the-law-of-a-stochastic-process-defined/557519#557519

According to Girsanov theorem ( for example, P155. Thereom 8.6.3 in Fifth Edtion, Stochastic Differential Equations: An introduction with Application), there exists a Law $Q$ such that $B_{1}(t)$ is a standard Brownian motion under $Q$.

Is $Q$ equal to $P_{1}$ ?

I thought they are not equal to each other. The reason is that for fixed time $t$ the expectation of $B_{1}(t)$ under $Q$ is 0, and under $P_{1}$ its expectation should be $\mu t$. If my derivation is wrong, please point out where is my mistake.

If I am correct, a new question is how to compute $\frac{d P_{1}}{dP}$?

Recall that $\frac{d Q}{dP}$ is given by Girsanov theorem. Any references are very appreciated.

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Answered on math.stackexchange.com: math.stackexchange.com/questions/231385/… –  user1157 Feb 7 at 8:31

2 Answers 2

No, that's the point of Girsanov's theorem. If $Q$ is equal to $P_1$, then nothing has changed. In order to make $B_1(t)$ a standard BM we need to transition to a new Law. Namely, $Q$.

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Your answer is very appreciated. I just added some more question. If $P_{1}\neq Q$, how to compute the radon nikodym derivative $\frac{d P_{1}}{d P}$? I am very confused with this computations. Some references said it can be obtained by Girsanov theorem. –  user2781712 Nov 8 '13 at 0:06

$$\dfrac{dP_1}{dP}=\dfrac{f(B_1)}{f(B)}=\dfrac{\frac{1}{\sqrt{2\pi}\sigma_1} e^{ -\frac{(x-\mu_1)^2}{2\sigma_1^2} })}{\frac{1}{\sqrt{2\pi}\sigma} e^{ -\frac{(x-\mu)^2}{2\sigma^2} }}=\dfrac{\sigma_1}{\sigma}e^{\dfrac{(x-\mu)^2}{2\sigma^2} -\dfrac{(x-\mu_1)^2}{2\sigma_1^2}}=\dfrac{t}{t}e^{\dfrac{(x-0)^2}{2t^2} -\dfrac{(x-\mu t)^2}{2t^2}}=e^\dfrac{x^2+x^2-2x\mu t+\mu^2t^2}{2t^2}=e^\dfrac{2x^2-\mu t(-2x+\mu t)}{2t^2}$$

because $B_1\sim N(\mu t, t),B\sim N(0,t)$.

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It seems that your computation is not correct. Actually, it only uses the concept of Brownian motion under different measure. –  user2781712 Aug 18 at 19:43
    
@user2781712 You are asking to compute $dP_1/dP$, or do you mean $dQ/dP_1$ such that $B_1$ becomes martingale? –  emcor Aug 18 at 20:32

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