# Find values of $a$ and $\lambda$ for which $Z_{0}e^{at+bW_{t}}-\lambda t$ is a martingale

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.

## 1 Answer

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.

• 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. – rs4rs35 Apr 19 at 16:27
• 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. – John Dawkins Apr 19 at 20:22