Controlling the Occupation Time of an Exponential Martingale

Abstract

We consider the problem of maximizing the expected amount of time an exponential martingale spends above a constant threshold up to a finite time horizon. We assume that at any time the volatility of the martingale can be chosen to take any value between $\sigma _1$ and $\sigma _2$, where $0 < \sigma _1 < \sigma _2$. The optimal control consists in choosing the minimal volatility $\sigma _1$ when the process is above the threshold, and the maximal volatility if it is below. We give a rigorous proof using classical verification and provide integral formulas for the maximal expected occupation time above the threshold.

Appendix

We give here the proof that the function ${\hat{g}}$ defined by (18) is causal. We remark that the proof uses standard arguments that can be found e.g. in [2].

Lemma 6.1

For $t<0$, ${\hat{g}}(t)=0$.

Proof

Let $t<0$. Observe that the function $z\mapsto e^{zt}F(z)$ is holomorphic on D. Let $C_R$ be the line $[-Ri, Ri]$ added by the right-hand side semicircle $\Gamma _R$ from Ri to $-Ri$. Then

$$\begin{aligned} \int _{C_R} e^{zt}F(z)dz=0 \end{aligned}$$

To prove that ${\hat{g}}$ is causal, it is enough to prove that $\lim _{R \rightarrow \infty } \int _{\Gamma _R} e^{zt}F(z)dz=0.$ On $\Gamma _R$, one has $|e^{zt}|=e^{tR \cos \theta }$. Moreover, for all $\epsilon >0$, there exists R large enough such that $|F(z)|<\epsilon$, for $z \in \Gamma _R$. Then

$$\begin{aligned} \int _{\Gamma _R} e^{zt}F(z)dz\le 2R\epsilon \int _0^{\frac{\pi }{2}}\exp (Rt \cos \theta )d\theta \end{aligned}$$

As $t<0$ and $\sin (\theta ) \ge 2\theta /\pi$, for $\theta \in [0, \pi /2]$, we obtain

$$\begin{aligned} \int _0^{\frac{\pi }{2}}\exp (Rt \cos \theta )d\theta = \int _0^{\frac{\pi }{2}}\exp (Rt \sin \theta )d\theta \le \frac{\pi }{2R|t|}(1-\exp (Rt)), \end{aligned}$$

and hence the result.