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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 $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + rS \frac{\partial V}{\partial S} = rV.$$

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

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