# 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 $$d\Pi.$$

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*}

• Grossly oversimplifying, if the risk-free interest rate is 5% and the dividend is 3%, it seems like the underlying would have to increase by only 2% (roughly) for the instrument holder to breakeven with a risk-free portfolio? Or am I missing something? Apr 25, 2023 at 2:15
• Over an infinitesimal interval, I agree with your observation. Apr 25, 2023 at 13:42
• But not over a full year? Apr 25, 2023 at 16:24
• no, only locally. Apr 25, 2023 at 18:04
• Why not over a full year? Apr 26, 2023 at 12:14

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 $$$$\Pi=V-\Delta S.$$$$ In the interval $$dt$$ the portfolio variation is therefore given by: $$$$d\Pi= dV - \Delta dS - q \Delta S dt.$$$$ The last term $$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.

• I do not understand why we minus dividend in $dS.$ Sep 23, 2019 at 10:13
• When a company pays dividends its value decrease, then the same appens to its stock value.
– ab94
Sep 23, 2019 at 10:34
• first: $$r \Pi dt = r \left( V - \Delta S \right) dt = (\Theta + \frac{1}{2} \left( \sigma S \right)^2 \Gamma ) dt + \frac{\partial V}{\partial S} dS - \Delta dS - q \Delta S dt.$$ and then, delta hedging ( i.e. $\Delta = \frac{\partial V}{\partial S}$) and substituting, you find the correct answer. Financially you can think exactly as @gk94 suggested. If S is the price of my stock and it pays a dividend (think of it as a discrete dividend) then this value drops of the dividend amount. Sep 23, 2019 at 11:19
• See this question. The portfolio $\Pi = V-\frac{\partial V}{\partial S} S$ may not be self-financing. Sep 23, 2019 at 19:49