# Is M defined as below a martingale?

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?

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...

• You need $\sigma^{2}$ in the second integral. The proof follows by just conditioning on $\sigma$. Mar 3 at 23:16
• Make sure to verify whether the second integral is $dt$ or $dB_t$. There's a difference. Mar 3 at 23:52
• @Hamdiken Thanks! Yes, that is $dt$. Mar 4 at 6:04
• @geetha290krm What do you mean? I try to prove $E[M(t)|F_s]=M(t)$. Mar 4 at 6:05
• @Hermi what means is that the second $ds$ part is the variance of the first $dB_s$ part. By definition, the variance can be obtained through the Itô isometry, which will get you a $\sigma^2$ in the second part. Finally, with some conditions on $\sigma$ you can deduce it's a martingale. Mar 4 at 6:19

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 $$\sigma$$.