Deriving Black-Scholes Partial Differential Equation

Black-Scholes Partial Differential Equation

Black-Scholes PDE or Black-Scholes equation is an equation that governs the dynamics of European option value under the Black-Scholes model. The equation is given by,

$$\frac{\partial v}{\partial t}+\frac{1}{2}{\sigma }^2{s}^2\frac{{\partial }^{2}v}{\partial s^2}+rs\frac{\partial v}{\partial s}-rv=0$$

Black-Scholes Model

$d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$

$d{B}_t=r{B}_{t}dt$

where $(W_t{)}_{0\le t\le T}$ is a Brownian motion.


I will use the following known facts without proof.

For infinitesimally small $dt$,

$d{W}_t\sim O(\sqrt{dt})=O(d{t}^{1/2})$

$(d{W}_t{)}^2=dt$

Pricing by replication

Consider a European option with payoff $\phi (S_T)$ at maturity $T$ that has its value of $V_t$ at time $t$.

Construct a portfolio $X$ that replicates the payoff of this European option. i.e., $X_t=V_t$

Let $X_t={\mathrm{\Delta }}_t{S}_t+\frac{X_t-{\mathrm{\Delta }}_t{S}_t}{B_t}{B}_t$.

where ${\mathrm{\Delta }}_t$ indicates the number of shares of stocks in the portfolio.

If the value of $X_t$ only depends on the value of $S$ and $B$ then

$$\begin{array}{rl}d{X}_t& ={\mathrm{\Delta }}_{t}d{S}_t+\frac{X_t-{\mathrm{\Delta }}_t{S}_t}{B_t}d{B}_t\\ & ={\mathrm{\Delta }}_t(\mu S_{t}dt+\sigma S_{t}d{W}_t)+\frac{X_t-{\mathrm{\Delta }}_t{S}_t}{B_t}r{B}_{t}dt\\ & ={\mathrm{\Delta }}_t(\mu S_{t}dt+\sigma S_{t}d{W}_t)+(X_t-{\mathrm{\Delta }}_t{S}_t)rdt\\ & ={\mathrm{\Delta }}_t{S}_t(\mu -r)dt+{\mathrm{\Delta }}_t{S}_t\sigma d{W}_t+r{X}_{t}dt\end{array}$$

$$d\left(\frac{X_t}{B_t}\right)=d(X_t\cdot \frac{1}{B_t})=d{X}_t\cdot \frac{1}{B_t}+X_t\cdot d\left(\frac{1}{B_t}\right)+d{X}_{t}d\left(\frac{1}{B_t}\right)$$

Note that by letting $f(B_t)=\frac{1}{B_t}$ , for an infinitesimally small increment $dt$, we have

$$\begin{array}{rl}d\left(\frac{1}{B_t}\right)=df(B_t)& =f^{\prime }(B_t)d{B}_t+\frac{1}{2}{f}^{″}(B_t)(d{B}_t{)}^2\\ & =-\frac{1}{B_t^2}\cdot d{B}_t+O(d{t}^2)\\ & =-\frac{r{B}_{t}dt}{B_t^2}\\ & =-\frac{r}{B_t}dt\end{array}$$

$$\begin{array}{rl}d\left(\frac{X_t}{B_t}\right)& =d{X}_t\cdot \frac{1}{B_t}+X_t\cdot d\left(\frac{1}{B_t}\right)+d{X}_{t}d\left(\frac{1}{B_t}\right)\\ & =\frac{d{X}_t}{B_t}+X_t(-\frac{r}{B_t}dt)+d{X}_t(-\frac{r}{B_t}dt)\\ & =\frac{{\mathrm{\Delta }}_t{S}_t(\mu -r)dt+{\mathrm{\Delta }}_t{S}_t\sigma d{W}_t+r{X}_{t}dt}{B_t}-\frac{r{X}_t}{B_t}dt+O(d{t}^2+d{t}^{3/2})\\ & =\frac{{\mathrm{\Delta }}_t{S}_t(\mu -r)dt+{\mathrm{\Delta }}_t{S}_t\sigma d{W}_t+r{X}_{t}dt}{B_t}-\frac{r{X}_t}{B_t}dt\\ & =\frac{{\mathrm{\Delta }}_t{S}_t}{B_t}(\mu -r)dt+\frac{{\mathrm{\Delta }}_t{S}_t}{B_t}\sigma d{W}_t\end{array}$$

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Now consider the value of the European option at time $t$, $V_t.$

It is reasonable to assume that the value of a European option will only depend on time $t$ and the stock price $S_t$.

That is, assume $V_t$$=v(t,S_t)$.

By using the multi-dimensional Ito’s lemma,

$$\begin{array}{rl}d{V}_t& =dv(t,S_t)\\ & =\frac{\partial v}{\partial t}dt+\frac{\partial v}{\partial s}d{S}_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}(d{S}_t{)}^2\\ & =\frac{\partial v}{\partial t}dt+\frac{\partial v}{\partial s}(\mu S_{t}dt+\sigma S_{t}d{W}_t)+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}(\mu S_{t}dt+\sigma S_{t}d{W}_t{)}^2\\ & =\frac{\partial v}{\partial t}dt+\frac{\partial v}{\partial s}(\mu S_{t}dt+\sigma S_{t}d{W}_t)+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}({\mu }^2{S}_t^2(dt{)}^2+{\sigma }^2{S}_t^2(d{W}_t{)}^2+2\mu \sigma S_t^{2}dtd{W}_t)\\ & =\frac{\partial v}{\partial t}dt+\frac{\partial v}{\partial s}(\mu S_{t}dt+\sigma S_{t}d{W}_t)+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}(O(d{t}^2)+{\sigma }^2{S}_t^{2}dt+O(d{t}^{3/2}))\\ & =\frac{\partial v}{\partial t}dt+\frac{\partial v}{\partial s}\mu S_{t}dt+\frac{\partial v}{\partial s}\sigma S_{t}d{W}_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^{2}dt\\ & =[\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2]dt+\frac{\partial v}{\partial s}\sigma S_{t}d{W}_t\end{array}$$

$$\begin{array}{rl}d\left(\frac{V_t}{B_t}\right)& =d(V_t\cdot \frac{1}{B_t})=d{V}_t\cdot \frac{1}{B_t}+V_t\cdot d\left(\frac{1}{B_t}\right)+d{V}_{t}d\left(\frac{1}{B_t}\right)\\ & =\frac{d{V}_t}{B_t}+V_t(-\frac{r}{B_t}dt)+d{V}_{t}d\left(\frac{1}{B_t}\right)\\ & =\frac{d{V}_t}{B_t}-\frac{r{V}_t}{B_t}dt+O(d{t}^2+d{t}^{3/2})\\ & =\frac{d{V}_t}{B_t}-\frac{r{V}_t}{B_t}dt+O(d{t}^2+d{t}^{3/2})\\ & =\frac{1}{B_t}(d{V}_t-r{V}_{t}dt)\\ & =\frac{1}{B_t}([\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2]dt+\frac{\partial v}{\partial s}\sigma S_{t}d{W}_t-r{V}_{t}dt)\\ & =\frac{1}{B_t}([\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t]dt+\frac{\partial v}{\partial s}\sigma S_{t}d{W}_t)\end{array}$$

Suppose $X_0=V_0$ and $d\left(\frac{V_t}{B_t}\right)=d\left(\frac{X_t}{B_t}\right)$, then we have

$\frac{X_t}{B_t}=\frac{X_0}{B_0}+{\int }_0^{t}d\left(\frac{X_t}{B_t}\right)=\frac{V_0}{B_0}+{\int }_0^{t}d\left(\frac{V_t}{B_t}\right)=\frac{V_t}{B_t}$.

From $d\left(\frac{V_t}{B_t}\right)=d\left(\frac{X_t}{B_t}\right)$,

$$\begin{array}{c}d\left(\frac{V_t}{B_t}\right)-d\left(\frac{X_t}{B_t}\right)=0\\ \frac{1}{B_t}([\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t]dt+\frac{\partial v}{\partial s}\sigma S_{t}d{W}_t)-(\frac{{\mathrm{\Delta }}_t{S}_t}{B_t}(\mu -r)dt+\frac{{\mathrm{\Delta }}_t{S}_t}{B_t}\sigma d{W}_t)=0\\ \frac{1}{B_t}([\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t-{\mathrm{\Delta }}_t{S}_t(\mu -r)]dt+[\frac{\partial v}{\partial s}\sigma S_t-{\mathrm{\Delta }}_t{S}_t\sigma ]d{W}_t)=0\\ [\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t-{\mathrm{\Delta }}_t{S}_t(\mu -r)]dt+[\frac{\partial v}{\partial s}-{\mathrm{\Delta }}_t]\sigma S_{t}d{W}_t=0\end{array}$$

Choose ${\mathrm{\Delta }}_t$ s.t. $[\frac{\partial v}{\partial s}-{\mathrm{\Delta }}_t]\sigma S_{t}d{W}_t=0$, i.e., ${\mathrm{\Delta }}_t=\frac{\partial v}{\partial s}$.

$$\begin{array}{c}[\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t-\frac{\partial v}{\partial s}{S}_t(\mu -r)]dt=0\\ [\frac{\partial v}{\partial t}+\frac{\partial v}{\partial s}\mu S_t+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t-\frac{\partial v}{\partial s}{S}_t\mu +\frac{\partial v}{\partial s}{S}_{t}r]dt=0\\ [\frac{\partial v}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{S}_t^2-r{V}_t+\frac{\partial v}{\partial s}{S}_{t}r]dt=0\\ \frac{\partial v}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}v}{\partial s^2}{\sigma }^2{s}^2-rv+\frac{\partial v}{\partial s}rs=0\end{array}$$

By rearranging the expression, we have the Black-Scholes Partial Differential Equation.

$$\frac{\partial v}{\partial t}+\frac{1}{2}{\sigma }^2{s}^2\frac{{\partial }^{2}v}{\partial s^2}+rs\frac{\partial v}{\partial s}-rv=0$$