0
$\begingroup$

Find values of $a$ and $\lambda$ for which $Z(t)=Z_{0}e^{at+bW_{t}}-\lambda t$ is a martingale. In here $W_{t}$ is a Brownian motion and $a,b\in\mathbb{R}$ can be positive as well as negative, since $b$ is derived by substracting one variance from another.

I cannot find a form for which this is a martingale except the form of the exponential martingale $$C\cdot e^{\alpha W_{t}-\frac{\alpha^{2}t}{2}}$$ with $C$ a constant. This would imply that only for the values $$a=\frac{-b^{2}}{2}$$ and $$\lambda=0$$ $Z(t)$ is a martingale. Is this correct or is there another form for which $Z(t)$ is a martingale, since I have to find a solution for $\lambda\in\mathbb{R}^{+}$, but this is quite an ambiguous expression.

$\endgroup$
1
$\begingroup$

The mean of $Z(t)$ is $E[Z_0]\cdot e^{at+b^2t/2}-\lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=\lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.

$\endgroup$
  • $\begingroup$ How does constant expectation imply that the martingale property holds? Now $\mathbb{E}[Z(t)|\mathcal{F}_{s}]=z_{0}\neq Z(s)$ for all $0\leq s\leq t$, thus the martingale property is not necessairly satisfied. $\endgroup$ – rs4rs35 Apr 19 at 16:27
  • 1
    $\begingroup$ I asserted that if $Z(t)$ is a martingale then its mean is a constant function of time, forcing $a+b^2/2=0=\lambda$. As the OP had already observed, this condition on $a,b,\lambda$ also implies that $Z(t)$ is a martingale. $\endgroup$ – John Dawkins Apr 19 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.