[Math] Girsanov: Change of drift, that depends on the process

measure-theoryprobability theorystochastic-analysisstochastic-calculus

Known:

If I am looking at an SDE like:

$dX_t = b(t,\omega) dt + dW_t$ with $W_t$ a Brownian motion under a measure $P$.

I know that I can change the drift by using Girsanov to

$dX_t = (b(t,\omega)+c(t,\omega)) dt + d\bar{W}_t$ with $\bar{W}_t$ a Brownian motion under a new measure $Q$.

if $c$ satisfies some condition such that $Z_t= \exp(-\int_{0}^{t} c(s,\omega) dW_s – 1/2 \int_{0}^{t} c(s,\omega)^2 ds)$ is a $P$-Martingale.

(Please correct me if I am wrong so far)

QUESTION

Now I am interested in a SDE with a drift that depends also on the current position $X_t$ and I want to change its drift:

$dX_t = b(t,X_t) dt + dW_t$ with $W_t$ a Brownian motion under a measure $P$ to
$dX_t = (b(t,X_t)+c(t,X_t)) dt + d\bar{W}_t$

Again I assume that
$Z_t= \exp(-\int_{0}^{t} c(s,X_s) dW_s – 1/2 \int_{0}^{t} c(s,X_s)^2 ds)$ is a $P$-Martingale.

Now my questions:

  1. Am I allowed to do this?

  2. If yes: Under which conditions is the last $Z_t$ a Martingale?
    (Someone told me, that he thinks if $c(t,x)=c(x) \leq C (1+|x|)$ then $Z_t$ is a Martingale, is this correct? (why?) Can I also use this Novikov condition here?)

  3. Do you have a reference where to find more information about this non-typical drift change by Girsanov?

Best Answer

I will use $B:=\bar{W}$ for simplicity. Looking at what you want, we see that we would like to have

$$W_t=\int c(s,X_s)ds + B_t\iff B_t=W_t-\int c(s,X_s)ds$$

for a given process $X$. Let $Y_t:=c(t,X_s)$. Let $Z:=\mathcal{E}(Y\circ W)$, where $Y\circ W=\int Y_sdW_s$. Hence you need some structural assumptions on $c$ such that you are allowed to write $Y\circ W$. In general $Z$ is a local martingale and positive on $[0,\infty)$ hence a supermartingale (Fatou). Therefore $Z_t$ converges to $Z_\infty$ $P$-a.s. It may happen that $Z_\infty = 0$ and or $E[Z_\infty] < 1$. The condition $E[Z_\infty]=1$ is equivalent to the property that $Z$ is a (uniformly integrable) martingale on $[0,\infty]$. Here you can use Novikov's condition. So if

$$E[\exp{(\frac{1}{2}\langle Y\circ W\rangle_\infty)}]<\infty$$ then $Z$ is a uniformly integrable martingale on $[0,\infty]$. We must have $\langle Y\circ W\rangle_\infty <\infty$ to guarantee that $Z_\infty$ becomes not $0$ with positive probability. Suppose it is true that $Z_\infty >0$. Then you can define a probability measure $Q$ which is equivalent to $P$ by $\frac{dQ}{dP}=Z_\infty$. The general Girsanov tells you that for a continuous local martingale $M$ w.r.t $P$ and a density process $Z$ we have

$$\tilde{M}=M-\int\frac{1}{Z}d\langle Z,M\rangle=M-\langle L, M\rangle$$ is a continuous local martingale w.r.t $Q$, where $Z$ is of the form $Z=\mathcal{E}(L)$ for a continuous local martingale $L$. Take $W=M$ and $L=Y\circ W$ to conclude

$$\tilde{M}=:B=M-\langle L,M\rangle=W-\int Y_s ds$$

It is easy to verify that $B$ is a $Q$ Brownian Motion (Lévy).

After all you are allowed to do that on some assumption on $c$:

  1. $Y\in L^2_{loc}(W)$
  2. $E[\exp{(\frac{1}{2}\langle Y\circ W\rangle_\infty)}]<\infty$
  3. $\langle Y\circ W\rangle_\infty <\infty$

Keep in mind that $2$ and $3$ are "just" sufficient condition in this case. So they are a good points to start with.

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