Is M defined as below a martingale?

Question

Let $f\in L^2$. Let $$Z_t:=\int_0^t f(s)dB_s.$$

We know that $$ L=\exp\left(\int_0^t f(s)dB_s-\frac{1}{2}\int_0^t f(s)^2 ds\right), $$ is a martingale.

Now, let $\sigma$ be another random variable independent of $B_t$. Assume that the characteristic function of $\sigma$: $$ E[e^{it\sigma}]=\frac{B_1(2t)}{t} $$ where $B_1(\cdot)$ is the Bessel function (see https://en.wikipedia.org/wiki/Bessel_function).

Can we say that $$ M=\exp\left(\int_0^t \sigma f(s)dB_s-\frac{1}{2}\int_0^t \sigma^2 f(s)^2 ds\right) $$ is still a martingale?

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I try to check that with the definition of martingale...

By the comment below, by Novikov's condition (https://en.wikipedia.org/wiki/Novikov%27s_condition) we need to check $$ E\left(\exp\left(\frac{1}{2}\int_0^T \sigma^2 f^2(s)ds\right)\right)<\infty $$

But I am stuck on this step...

Answer 1

Which Bessel function do you have in mind? (There are several, and your notation $B_1$ is non-standard.)

A more appropriate condition for testing whether you have a martingale might be Kazamaki's: If for each $t>0$ you have $$ \sup_{T\le t} E[\exp(\sigma Z_T)]<\infty, $$ then $(M_t)$ is a martingale. The supremum here is taken over all stopping times $T$ that are bounded above by $t$. For this you need to work with the mgf of $\sigma$.