4
$\begingroup$

Problem

I am considering an American call option which gives a domestic investor the right to buy a unit of foreign currency at a strike of $K$ units of domestic currency. I have an exchange rate $S_t$ in units of domestic to foreign currency which follows $$\frac{\mathrm{d}S_t}{S_t}=\mu\mathrm{d}t+\sigma\mathrm{d}W_t,$$ where $\mu$ and $\sigma$ are positive constants. I am trying to prove the following relations:

  1. If $r_D>0$ and $r_F\leq0$, the option should not be exercised early. I have managed to prove it, see below.

  2. If both $r_D$ and $r_F>0$, the option can be exercised early, and for a given exercise boundary $S^{\textrm{Ex}}_t$, I need to also describe the holding region and derive a PDE for the price of the call option in the holding region. I have questions about this part.

i) If $r_F>0$, I know from the inequality I prove in the next section that it is possible for $\max\{S_t-K,0\}$ to be greater than the option price. However, since the inequality that arises does not explicitly involve $r_F$, is there a way to obtain a closed form inequality (if it exists) involving $r_F$ and $r_D$ to cases for optimal exercise? Practically, when would it make sense to exercise?

ii) How would one form the PDE here? Any help is greatly appreciated!

Proof for Part 1.

Under the risk-neutral measure $\mathbb{Q}$, $S_t\mathrm{e}^{-(r_D-r_F)t}$ is a martingale, so the discounted $S_t$ (when investing domestically) satisfies $$\mathbb{E}^\mathbb{Q}\left[S_T\mathrm{e}^{-(r_D-r_F)T}\Big|\mathcal{F}_t\right]=S_t\mathrm{e}^{-r_F(T-t)}.$$

Consider a non-negative, convex function $f(\cdot)$ satisfying $f(0)=0$. Then, 

$$\mathbb{E}^\mathbb{Q}\left[\mathrm{e}^{-r_D(T-t)}f(S_T)\Big|\mathcal{F}_t\right]\geq\mathbb{E}^\mathbb{Q}\left[f(S_T\mathrm{e}^{-r_D(T-t)})\Big|\mathcal{F}_t\right].$$

By the conditional Jensen's inequality, we have

$$\mathbb{E}^\mathbb{Q}\left[f(S_T\mathrm{e}^{-r_D(T-t)})\Big|\mathcal{F}_t\right]\geq f\left(\mathbb{E}^\mathbb{Q}\left[S_T\mathrm{e}^{-r_D(T-t)}\Big|\mathcal{F}_t\right]\right)=f\left[S_t\mathrm{e}^{-r_F(T-t)}\right].$$

$f:x\mapsto\max\{x-K,0\}$ is non-negative, convex, and $f(0)=0$, so

$$\mathrm{e}^{-r_D(T-t)}\mathbb{E}^\mathbb{Q}\left[\max\{S_T-K,0\}\Big|\mathcal{F}_t\right]\geq\max\{S_t\mathrm{e}^{-r_F(T-t)}-K,0\}.$$

When exercised at $t$, the payoff is $\max\{S_t-K\}$, which is $<\max\{S_t\mathrm{e}^{-r_F(T-t)}-K,0\}$ is $r_F\leq0$, proving part 1.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.