To confirm the Novikov's condition

Question

I have a question about Novikov's condition.

Let $L$ be a local martingale such that either $\exp \left(\frac{1}{2}L \right)$ is a submartingale or $E[\exp\left(\frac{1}{2} \langle L,L \rangle_{t} \right)]<\infty$ for every $t>0$.

Then $\mathcal{E}(L)_{t}:=\exp \left\{L_{t}-\frac{1}{2} \langle L,L\rangle_{t} \right\}$ is a martingale.

Question

Let $X$ be a Brownian motion. We can define $L_{t}:=\int_{0}^{t}b(X_{s})\,dX_{s}$ for a bounded function $b$ and it is easy to check \begin{align*} E \left[\exp\left(\frac{1}{2} \langle L,L \rangle_{t} \right) \right]=E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty \end{align*} for every $t>0$ since $b$ is bounded.

I want to know how to show \begin{align*} E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty\quad \mbox{or} \quad E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]=\infty \end{align*} when $b$ is unbounded and integrable

If you know a related research, please let me know.

Thank you for your consideration.

Answer 1

This is a partial answer, just too long for a comment.

There is a nice lemma from Dellacherie and Meyer:

Lemma If $\{A_s,s\in [0,t]\}$ is a non-decreasing adapted process such that $E[A_t - A_s|\mathcal F_s]\le K$ for all $s\in[0,t]$, then for any $\lambda < 1/K$, $$ E[e^{\lambda A_t}]< (1-\lambda K)^{-1}. $$

Now take $A_s = \int_0^t b(X_s)^2 ds$. Then $$ E[A_t - A_s|\mathcal F_s] = E\left[\int_s^t b(X_u)^2 du\,\Big|\,\mathcal F_s\right] = \int_s^t E\left[ b(X_u)^2 \,|\,\mathcal F_s\right] du \\ = \int_s^t \int_{\mathbb{R}} p(t-u,y-X_s) b(y)^2 dy\, du\le \int_{s}^t \frac{1}{\sqrt{2\pi(t-u)}}\int_{\mathbb{R}}b(y)^2 dy\,du \le \sqrt{t-s}\int_{\mathbb{R}}b(y)^2 dy, $$ where $p(t,x) = e^{-x^2/2t}/\sqrt{2\pi t}$ is the transition density of $X$.

Therefore, using Lemma, $$ E \left[\exp\left(\frac{1}{2}\int_{0}^{t}|b(X_{s})|^{2}\,ds\right) \right]<\infty $$ for $t$ small enough provided that $b\in L^2(\mathbb{R})$.

Some remarks:

1. I think that this is true for all $t>0$.

2. Having this for $t$ small enough should be sufficient to get the martingale property.

3. I think that moreover this is true for $b$ such that $\sup_{x\in\mathbb{R}}\int_{x-1}^{x+1} b(y)^2 dy<\infty$.

4. There is also a useful Bene sufficient condition: if $|b(x)|\le C(1+|x|)$, then $\mathcal{E}(L)_t$ is a martingale. (This is also true when $b$ depends on the whole path in a sublinear way.)

5. 3 and 4 suggest that the martingale property should be true if $\int_{x-1}^{x+1} b(y)^2 dy\le C(1+|x|^2)$. I have no proofs neither references at the moment.