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.


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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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