Proving integrability of the martingale $\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$

Question

In trying to prove that $M_t:=\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$, I have hit a few roadblocks. I know how to show the martingale property as has been done many times in literature; but I haven't been able to properly justify that $\mathbb{E}|M_t|<+\infty$ so that I can finally conclude that it is a true martingale. Clearly $\textrm{e}^{-\frac12\lambda^2t}$ is bounded but the $\cosh$ isn't. How do I deal with the $\cosh$ and prove the integrability? Thanks!

Answer 1

The link here (Proof that the exponential martingale is a Brownian Motion) shows that for any $\delta\in\mathbb{R}$ that $$M_{\delta,t}= e^{\delta B_t - \frac{1}{2}\delta^2 t}$$ is a Martingale. Notice that for any $\lambda \in\mathbb{R}$ that $$\frac{1}{2}(M_{\lambda,t} + M_{-\lambda,t}) = \frac{(e^{\lambda B_t}+e^{-\lambda B_t})}{2}e^{-\frac{1}{2}\lambda^2t} = \cosh(\lambda B_t)e^{-\frac{1}{2}\lambda^2 t}$$ Since linear combinations of Martingales are martingales, then $\cosh(\lambda B_t)e^{-\frac{1}{2}\lambda^2 t}$ is in fact a martingale.

Answer 2

If $X \sim N(0,\sigma^{2})$ then $Ee^{cX} =e^{c^{2}\sigma^{2} /2} $ for any real number $c$. [In fact for any complex number $c$!]. Since $\cosh x=\frac {e^{x}+e^{-x}} 2$ we are done. Ref:

1. https://en.wikipedia.org/wiki/Moment-generating_function

2. https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjM8pqt_6btAhV1yDgGHZfpDi8QFjABegQIAhAC&url=https%3A%2F%2Fwww.le.ac.uk%2Fusers%2Fdsgp1%2FCOURSES%2FMATHSTAT%2F6normgf.pdf&usg=AOvVaw3QHSFjpCFrBgTFBxRwGAnK