I want to apply Girsanov's theorem to

$B_t = X_t - \int_0^t b(X_s)ds$

where $X_t$ is a Brownian motion and $b$ is Lipschitz. Therefore I have to show that

$Z_t = \exp(\int\limits_0^t b(X_s)dX_s - \frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)$

is a martingale. In order to do so I want to apply Novikov's condition, which gives me the martingale property for $Z$, if

$\mathbb{E}[\exp(\frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)] < +\infty$

Jensen and the Lipschitz property give me

$\mathbb{E}[\exp(\frac{1}{2} \int\limits_0^t|b(X_s)|^2ds)] \: \leq \: ...\leq \mathbb{E} [\int\limits_0^t\exp(\frac{C}{2}|X_s|^2)ds]$

But how can I continue to find the finiteness?


1 Answer 1


Here is a partial result for small $t$:

\begin{eqnarray*} E\left[\int_0^t exp\left(\frac{C}{2}X_s^2\right)ds\right] &=& \int_0^t E\left[exp\left(\frac{C}{2}X_s^2\right)\right]ds\\ &=& \int_0^t E\left[exp\left(\frac{Cs}{2}\Big(\frac{X_s}{\sqrt{s}}\Big)^2\right)\right]ds\\ &=& \int_0^t E\left[exp\left(\frac{Cs}{2}\chi_1^2\right)\right]ds\\ &=& \int_0^t M_{\chi_1^2}\Big(\frac{Cs}{2}\Big)ds\\ &=& \int_0^t (1-Cs)^{-\frac{1}{2}} ds\\ \end{eqnarray*} which is finite for $t\in [0,\frac{1}{C})$. I have used the following:

(1) Fubini's theorem to switch the order of integration: $E\left[\int\cdots\right]=\int E\left[\cdots\right]$

(2) $X_s\sim N(0,s)$ so $\frac{X_s}{\sqrt{s}}\sim N(0,1)$ and hence $\Big(\frac{X_s}{\sqrt{s}}\Big)^2\sim \chi_1^2$

(3) $M_{\chi_1^2}(t)=E[exp(t\chi_1^2)]=(1-2t)^{-\frac{1}{2}}$ is the MGF of $\chi_1^2$.

I'll try to find a better upper bound :)

  • $\begingroup$ Thank you very much, thats perfect. To apply Girsanov this is already enough: Novikov gives us that $Z_t$ is a martingale on $[0,T]$. This gives us $\mathbb{E}(Z_t)=1$ on that interval, which is also a sufficient condition for $Z_t$ to be a martingale in that setting. We can now prove that $Z_{Kt}$ has expectation 1 for any $K \in \mathbb{N}$ by telescoping the integrals. $\endgroup$
    – durst
    Apr 15, 2020 at 12:29

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