The PDE of caplet and floors

Question

I know following PDE is the continuous payment case, but a caplet pays as rate: $\max(r - r^*,0),$ use the hedge portfolio $\Pi = V- \Delta Z$ $$d\Pi = dV- \Delta dZ +\max(r - r^*,0)dt = r\Pi dt $$ then the result PDE should be $$\dfrac{\partial V}{\partial t} + LV -rV + \max(r - r^*,0) = 0$$ Here $L$ is Black-Scholes operator, and $Z$ is zero=coupon bond respect to interest rate $r.$

But, why in the following book, the constant term is $\min(r,r^*)$ I can't understand that.

Answer 1

It must be a typo for the equation in the book. That is, the equation for a caplet is of the form \begin{align*} \frac{\partial V}{\partial t} + LV - r_t V +\max(r_t-r^*, 0) = 0, \end{align*} which can also be derived using the martingale approach.

Specifically, note that the accumulated payments from time $t$ up to maturity $T$ is given by \begin{align*} \int_t^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds. \end{align*} Let $B_t=e^{\int_0^t r_udu}$ be the money market account value at time $t$. Then, the option value at time $t$ is given by \begin{align*} V_t &= B_tE\left(\frac{\int_t^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\ &=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds - \int_0^t \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\ &=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t\right) \\ &\qquad- B_tE\left(\frac{\int_0^t \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t \right)\\ &=B_tE\left(\frac{\int_0^T \max(r_s-r^*, 0)e^{\int_s^T r_u du} ds}{B_T} \mid \mathcal{F}_t\right) -B_t\int_0^t \max(r_s-r^*, 0)e^{-\int_0^s r_u du} ds. \end{align*} That is, \begin{align*} M_t \equiv e^{-\int_0^t r_udu} V_t + \int_0^t \max(r_s-r^*, 0)e^{-\int_0^s r_u du} ds \end{align*} is a martingale. We assume that \begin{align*} dr_t = \mu(t, r_t) dt + \sigma(t, r_t) dW_t, \end{align*} where $\{W_t, t \ge 0\}$ is a standard Brownian motion. Then \begin{align*} dM_t &= -r_t e^{-\int_0^t r_udu} V dt + e^{-\int_0^t r_udu}\left(\frac{\partial V}{\partial t} + LV\right)dt\\ &\qquad + e^{-\int_0^t r_udu}\frac{\partial V}{\partial r}\sigma(t, r_t) dW_t + \max(r_t-r^*, 0)e^{-\int_0^t r_u du} dt. \end{align*} Consequently, \begin{align*} \frac{\partial V}{\partial t} + LV - r_t V +\max(r_t-r^*, 0) = 0. \end{align*}

Answer 2

How exactly did you find the first equation:

$$ d\Pi = dV-\Delta dZ + max(r-r^*,0) dt = r \Pi dt\,? $$

I mean how does the term $max(r-r^*,0) dt$ pops up?

Then, it is worth mentioning that:

$$ min(r,r^*) = - max(-r^*,-r) = -(max(r-r^*,0) - r) = r - max(r-r^*,0) $$

Maybe the latter can help.