# Girsanov: Change of drift, that depends on the process

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?

## 2 Answers

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.

1. It does not make one iota of difference whether the coefficient $c(t,\omega)$ has an explicit dependence on $X$ or not. $X$ itself depends on $t$ and $\omega$ so just change notation from $c(t,X_t)$ to $\tilde{c}(t,\omega) := c(t,X(t,\omega))$ and you are back in business. Whatever martingale criteria you find, you will apply them to this new coefficient, as you did in the first part of your post. For martingale criteria other than Novikov see Ruf, Johannes, A new proof for the conditions of Novikov and Kazamaki, Stochastic Processes Appl. 123, No. 2, 404-421 (2013). arXiv.1111.5583.

2. The linear growth condition is related to existence of a non-explosive solution in an SDE; nothing to do with uniform integrability criteria like Novikov or Kazamaki which is what you need to turn a local martingale into a martingale.

3. This change of measure does not need a special reference.