本科水平讲 BS 推导

面试版思路：

1. 假设股票 $S$ 服从几何布朗：$dS=\mu Sdt+\sigma SdW$。
2. 用 Itô 写出 $dV$，其中随机项为 $\sigma SV_S dW$。
3. 取组合 $\Pi=V-V_S S$ 消掉随机项，使组合无风险。
4. 无套利要求无风险组合收益率等于 $r$，推出 BS PDE。
5. 配上到期 payoff 解 PDE（或用风险中性期望）得到 BS 定价公式。

---

英文解析

What sort of undergrad are we dealing with here? Obviously there is a large difference between a student directly out of high school and one nearing the end of their studies in probability theory or financial mathematics. The best interpretation of this question is to give an explanation which is as simple as possible.

One unavoidable, and somewhat technical, statement is that in the BlackScholes world the arbitrage- free stock price evolves according to the stochastic differential equation

$$dS_{t} = rS_{t}dt + \sigma S_{t}dW_{t},$$

where $r$ is the risk- free rate of return (whether the undergrad understands much stochastic calculus is questionable, but short of giving a brief explanation of what the above equation represents there is little we can do to avoid using this). Here you should mention that 'arbitrage- free' essentially implies that there does not exist opportunities to make money for nothing without any risk in the market. One could also give an elementary explanation of what this equation represents; see the extension questions below. We require one other asset to use in the derivation, the risk- free bank account. This grows at the continuously compounding rate $r$ and hence its value at time $t$ , $B_{t}$ ,

is given by

$$B_{t} = e^{r t}\Rightarrow d B_{t} = r B_{t}d t,$$

which is a result from ordinary calculus.

The final necessary piece of technical mathematics we require is Itô's formula: the stochastic differential equation of a function $f\left(t,S_{t}\right)$ is given by

$$d f\left(t,S_{t}\right) = \frac{\partial f}{\partial t}\left(t,S_{t}\right)d t + \frac{\partial f}{\partial S}\left(t,S_{t}\right)d S_{t} + \frac{1}{2}\frac{\partial^{2}f}{\partial S^{2}}\left(t,S_{t}\right)\left(d S_{t}\right)^{2}.$$

Evaluating this requires the relations $(d t)^{2} = (d W_{t})\left(d t\right) = 0,\left(d W_{t}\right)^{2} = d t$ . Here we can compare this result to those from ordinary calculus, noting the extra term as a consequence of differentiation using stochastic processes.

Next we state that the price of a derivative is a function of the current time $t$ and the current stock price $S_{t}$ (this can be proved, but is beyond the scope of the question). We therefore denote such a price by $C\left(t,S_{t}\right)$

Finally we need that $C\left(t,S_{t}\right)B_{t}^{- 1}$ is a martingale. How do we justify this, and what does it mean? A simple explanation of its meaning is that we expect it to have zero growth: our option price is expected to grow at the same rate as the bank account and hence the growth of each cancels out in the given process. This is what it means to be a martingale, we do not expect change over time so we have zero expected growth. We perhaps overused the word 'expected' here, but it should be emphasized that there will be changes in the discounted price, we just expect it to be zero on average. This translates to the discounted price having a zero drift term. We apply Itô's formula to calculate the drift (see Question 2.1), equate to zero and get

$$\frac{\partial C}{\partial t} +\frac{\partial C}{\partial S_{t}} r S_{t} + \frac{1}{2}\frac{\partial^{2}C}{\partial S_{t}^{2}}\sigma^{2}S_{t}^{2} - r C = 0,$$

the Black- Scholes equation.

Here are some possible related questions:

• Give a non-technical explanation of the stochastic differential equation describing the evolution of the stock price.

• What is the mathematical definition of 'arbitrage free'. Explain this in everyday language.