Deriving the Black–Scholes PDE for a Dividend-Yielding Stock

# How to derive Black-Scholes equation with dividend?


# How to derive Black-Scholes equation with dividend?












> Question: The Black-Scholes equation without dividend is given by $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} -rV = 0.$$ (I attempted to derive the equation in my previous post.) If we assume that 'with dividend rate $D$', then the Black-Scholes equation becomes $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + (r-D)S \frac{\partial V}{\partial S} -rV = 0.$$ How to derive this?

By working backwards and assuming derivation of my previous post, we should have $$d\Pi = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt - \Delta S - D\Delta Sdt.$$ But I do not understand why can we add the term in $d\Pi.$

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/48867

We assume that the stock price process $\{S_t,\,t>0\}$ satisfies, under the real-world probability measure $P$, an SDE of the form \begin{align*} dS_t=S_t\big((\mu-q)dt+\sigma dW_t\big), \end{align*} where $\{W_t, \, t >0\}$ is a standard Brownian motion. Here, we need to consider the total return asset $e^{qt}S_t$, that is, the asset with the dividend payments invested in the same underlying stock. We consider a locally risk-free self-financing portfolio of the form \begin{align*} \pi_t =\Delta_t^1 \big(e^{qt}S_t\big) + \Delta_t^2 V_t, \end{align*} where $V_t$ is the option price. Then, \begin{align*} d\pi_t &= \Delta_t^1 d\big(e^{qt}S_t\big) + \Delta_t^2 dV_t\\ &= \Delta_t^1 e^{qt}\big(q S_t dt + dS_t \big) + \Delta_t^2\left(\frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S}dS_t + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 dt\right)\\ &=\left[\mu\Delta_t^1 e^{qt} S_t + \Delta_t^2\left(\frac{\partial V}{\partial t} + (\mu-q) S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 \right)\right]dt \\ &\qquad\qquad\qquad\qquad\qquad\quad + \left(\sigma\Delta_t^1 e^{qt}S_t + \sigma \Delta_t^2 S_t \frac{\partial V}{\partial S}\right)dW_t. \end{align*} Since $\pi_t$ is locally risk-free, we assume that $\pi_t$ earns the risk-free interest rate $r$, that is, \begin{align*} d\pi_t = r \pi_t dt, \end{align*} Then, \begin{align*} &\left[\mu \Delta_t^1 e^{qt} S_t + \Delta_t^2\left(\frac{\partial V}{\partial t} + (\mu-q) S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 \right)\right]dt \\ &\qquad\qquad\qquad\qquad\qquad + \left(\sigma\Delta_t^1 e^{qt} S_t + \sigma \Delta_t^2 S_t \frac{\partial V}{\partial S}\right)dW_t= r \pi_t dt. \end{align*} Consequently, \begin{align*} \sigma\Delta_t^1 e^{qt}S_t + \sigma \Delta_t^2 S_t \frac{\partial V}{\partial S}=0, \tag{1} \end{align*} and \begin{align*} \mu e^{qt} \Delta_t^1 S_t + \Delta_t^2\left(\frac{\partial V}{\partial t} + (\mu-q) S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 \right) = r(\Delta_t^1 e^{qt}S_t + \Delta_t^2 V_t). \end{align*} From $(1)$, \begin{align*} \Delta_t^1 = -e^{-qt} \Delta_t^2 \frac{\partial V}{\partial S}. \end{align*} Then, \begin{align*} -\mu \Delta_t^2 S_t \frac{\partial V}{\partial S}+ \Delta_t^2\left(\frac{\partial V}{\partial t} + (\mu-q) S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 \right) = r\Big(-\Delta_t^2 S_t\frac{\partial V}{\partial S} + \Delta_t^2 V_t\Big), \end{align*} or \begin{align*} \Delta_t^2\left(\frac{\partial V}{\partial t} -q S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2\right) &= r\Delta_t^2\Big(-\frac{\partial V}{\partial S} S_t + V_t\Big). \tag{2} \end{align*} Canceling the term $\Delta_t^2$ from both sides of $(2)$, we obtain the Black–Scholes equation of the form \begin{align*} \frac{\partial V}{\partial t} + (r-q) S_t \frac{\partial V}{\partial S} + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2S_t^2 -rV = 0. \end{align*}

## Answer by Turquoise Hexagon Sun (score 2)

https://quant.stackexchange.com/a/48861

The only difference in the derivation when you have a dividend-yield paying stock lies in the value of the Riskless Portfolio $\Pi_t$.

The financial meaning here is the key: to delta-hedge your option you buy a quantity $\Delta$ of the stock $S$, and only the stock is paying you the dividend, so you have to add this contribution in time to your hedge. The change in value of the stock is therefore: $$ dS=(\mu-q)Sdt+\sigma S dW. $$ We short a quantity $\Delta$ of the stock \begin{equation} \Pi=V-\Delta S. \end{equation} In the interval $dt$ the portfolio variation is therefore given by: \begin{equation} d\Pi= dV - \Delta dS - q \Delta S dt. \end{equation} The last term $ qS\Delta dt$ denotes the value added to the portfolio due to the dividend yield.

Now, for the Ito's Lemma the value of $dV$ is: $$ dV= \left( \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} \right) dt + \frac{\partial V}{\partial S} dS. $$ Finally you use the no arbitrage condition which states that a riskless portfolio has the same value as a riskless zero-coupon bond, i.e. $d\Pi=r\Pi dt$. Substituting you finally obtain: $$ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + (r-q)S \frac{\partial V}{\partial S} -rV = 0 $$ where you can see directly the additional $q$ term to the hedging due to the dividend yield.