# Martingale property of Brownian motion after change of measure

Let $(\Omega,\mathcal{F},\mathbb{F},P)$ be a filtered probability space and $X=(X_t)_{t\geq 0}$ be a Brownian motion with respect to the natural filtration $(\mathcal{F}_t)_{t\geq 0}$ generated by $X$.

Define $$Y_t=X_t+\mu t$$ for $t\geq 0$ and a measure $Q_t$ on $\mathcal{F}_t$ by $$Q_t(A)=E[exp(\color{red}{-}\mu X_t-\mu^2\frac{t}{2}),A]$$ for $t\geq 0$ and $A\in\mathcal{F}_t$. How can we prove that $Y$ is a $Q$-martingale elementary (see here)?

There is also this theorem called Girsanov-Thoerem:

Assume that $P' \stackrel{loc}{\ll}P$ and let $Z$ be the density process. Let $M$ be a $P$-local martingale such that the $P$-quadratic covariation $[M,Z]$ has $P$-locally integrable variation, and denote by $<M,Z>$ its $P$-compensator. Then the process $$M'=M-\frac{1}{Z}\dot\ <M,Z>$$ is $P'$-as well defined and is a $P'$-local martingale.

How do I compute $<X,Z>$ with $Z=(Z_t)_{t\geq 0}$ defined by $Z_t=exp(\color{red}{-}\mu X_t-\mu^2\frac{t}{2})$ and will M' coincide with $(X_t+\mu t)_{t\geq 0}$?

To show the martinaglity, consider $0\le s < t$. Then \begin{align*} E_Q(X_t+\mu t \mid \mathcal{F}_s) &=E_P\left((X_t + \mu t)e^{-\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right)\\ &=E_P\left((X_s +X_t-X_s + \mu t)e^{-\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right)\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(X_s+\mu t + x\sqrt{t-s})e^{-\mu\sqrt{t-s}x -\frac{\mu^2}{2}(t-s)}e^{-\frac{x^2}{2}}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(X_s+\mu t + x\sqrt{t-s})e^{-\frac{(x+\mu\sqrt{t-s})^2}{2}}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(X_s + \mu s+ y\sqrt{t-s})e^{-\frac{y^2}{2}}dy\\ &= X_s + \mu s. \end{align*}

• How do you get the third equality? You used the disintegration-theorem and that $X_t-X_s\sim \mathcal{N}(0,t-s)$. But i get $\frac{1}{\sqrt{2\pi(t-s)}}\int_{-\infty}^{\infty}(X_s+x)e^{-\frac{x^2}{2(t-s)}-\frac{\mu^2}{2}(t-s)+\mu x}dx$
– peer
Oct 26, 2016 at 15:47
• For simplicity, you can also use the fact that $X_t-X_s=\sqrt{t-s}\xi$, where $\xi\sim N(0, 1)$. But, you can get the same result either way. Oct 26, 2016 at 16:13
• Starting at the fourth term, i get $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(X_s+\mu t + x\sqrt{t-s})e^{-\frac{(x-\mu\sqrt{t-s})^2}{2}}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(X_s+\mu t)e^{-\frac{(x-\mu\sqrt{t-s})^2}{2}}dx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}( x\sqrt{t-s})e^{-\frac{(x-\mu\sqrt{t-s})^2}{2}}dx=X_s+\mu t +\sqrt{t-s}\mu\sqrt{t-s}=X_s+2\mu t -\mu s$. What is wrong here?
– peer
Oct 26, 2016 at 17:40
• Some mistakes are made. Can you make the substitution $y=x-\mu\sqrt{t-s}$? Oct 26, 2016 at 17:48
• Yes, then it is clear to me. I can probably check it on my own what i did wrong.
– peer
Oct 26, 2016 at 17:52