# Is my derivation of Black-Scholes equation correct or am I missing something (eg assumption)?

Question: The following is my derivation of the Black-Scholes equation. Is it correct or am I missing some details (eg assumption)?

Let $$V$$ be value of an option.

Suppose value $$\Pi$$ of a portfolio consisting of a long option position and short position with quantity $$\Delta$$ of underlying: $$\Pi = V - \Delta S.$$ Then we have $$d\Pi = dV-\Delta dS.$$ By assuming the underlying follow a lognormal random walk, we have $$dS = \mu S dt + \sigma S dW$$ where $$W$$ is the standard Brownian motion. By Ito's lemma on $$V$$, we have $$dV = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt.$$ Therefore, $$d\Pi = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt - \Delta S.$$ To hedge it, we need to remove the stochastic component $$(\frac{\partial V}{\partial S} - \Delta) dS$$ by setting $$\Delta = \frac{\partial V}{\partial S}.$$ So we have $$d\Pi = \frac{\partial V}{\partial t} dt + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt.$$ Since the portfolio is riskless, it must follow $$d\Pi = r \Pi dt$$ where $$r$$ is the riskless interest rate. So we have $$\frac{\partial V}{\partial t} dt + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt = r (V - \frac{\partial V}{\partial S} S)dt.$$ Dividing $$dt$$ on both side and rearrangement lead to the well-known Black-Scholes equation $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} = rV.$$

• See this question. The portfolio $\Pi = V-\frac{\partial V}{\partial S} S$ may not be self-financing. – Gordon Sep 23 '19 at 19:48