The Deriving of Black-Scholes Equation

Basic Wiener Process

$$\mathrm{d}z = \varepsilon \sqrt{\mathrm{d}t}$$

where $\varepsilon$ has a standard normal distribution $\phi(0, 1)$.

Generalized Wiener Process

A generalized Wiener process for a variable $x$ can be defined in terms of $\mathrm{d} x$

$$\mathrm{d} x = a \mathrm{d} t + b\mathrm{d}z$$

where a and b are constants.

Ito Process

An Ito process for a variable $\mathrm{d}x$ can be written as

$$\mathrm{d}x = a( x, t) \mathrm{d}t + b( x, t )\mathrm{d}z$$

where $a(x,t)$ and $b(x, t)$ are functions of the value of the underlying variable x and time t, and dz is a Wiener process.

Ito's Lemma

Suppose that the value of a variable $\mathrm{x}$ follows the Ito process, Ito's lemma shows that a function G of x and t follows the process

$$dG = \left ( \frac{\partial{G}}{\partial{x}}a(x,t)+\frac{\partial{G}}{\partial{t}}+\frac{1}{2} \frac{\partial^2{G}}{\partial{x^2}} b(x,t)^2 \right )\mathrm{d}t+\frac{\partial{G}}{\partial{x}}b(x,t)\mathrm{d}z$$

Note, G also follows an Ito process.

Proof

First of all, let's solve the mean and variance of $Y$given $Y=X^2$where $X \sim \phi(0, \sigma^2)$.

Since $\sigma^2 = E(X^2) - E(X)^2$, and $E(X) = 0$, we get

$$E(X^2) = \sigma^2$$

And becuase$VarX^2 = EX^4 - (EX^2)^2$, and

$$\begin{aligned} E(X^4) &= \int \frac{x^4}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx \\ &=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}d-\frac{x^2}{2\sigma^2} \\&=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}de^{-\frac{x^2}{2\sigma^2}} \\&= \frac{1}{\sqrt{2\pi}\sigma}\left ( -\sigma^2x^3e^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^2\int x^2e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= \frac{1}{\sqrt{2\pi}\sigma} \left ( 0 - 3\sigma^4xe^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^4\int e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= 3\sigma^4 \end{aligned}$$

Actually, the generalized form of $E[X^{2n}]$is,

$E\left [ X^{2n}\right ] = (2n - 1)!!\sigma^{2n}$,

where$!!$ is the double factorial.

Thus,

$$\begin{aligned} VarX^2 &= EX^4 - (EX^2)^2 \\ &= 3\sigma^4 - \sigma^4 \\ &= 2\sigma^4 \end{aligned}$$

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Consider a continuous and differentiable function G of two variables x and t, a Taylor series expansion of $\Delta G$ is

$$\Delta{G}=\frac{\partial{G}}{\partial{x}}\Delta{x} + \frac{\partial{G}}{\partial{t}}\Delta{t} + \frac{1}{2}\frac{\partial^2G}{\partial{x}^2}\Delta{x}^2 + \frac{\partial^2{G}}{\partial{x}\partial{t}}\Delta{x}\Delta{t}+\frac{1}{2}\frac{\partial^2{G}}{\partial{t}^2}\Delta{t}^2+\dots$$

discretize the Ito process to

$$\begin{aligned} \mathrm{\Delta}x &= a( x, t) \mathrm{\Delta}t + b( x, t )\mathrm{\Delta}z \\ &= a(x, t)\Delta{t}+b(x,t)\varepsilon\sqrt{\Delta{t}} \end{aligned}$$

then we have

$$\Delta{x}^2 = b^2(x, t)\varepsilon^2\Delta{t} + \text{terms of higher order in } \Delta{t}$$

It shows that the term involving $\Delta{x}^2$ in the Taylor series expansion cannot be ignored, and

$$E(X^2) = \sigma^2 \text{ and }Var{(X^2)} = 2\sigma^4$$

Since the variance of a standard normal distribution is 1, $E(\varepsilon^2\Delta{t}) = \Delta{t}$ and $Var(\varepsilon^2\Delta{t}) = 2\Delta{t}^2.$

We know that the variance of the change in a stochastic variable in time $\Delta{t}$ is proportional to $\Delta{t}$, not $\Delta{t}^2$. The variance of $\varepsilon^2\Delta{t}$ is therefore too small for it to have a stochastic component. As a result, we can treat it as nonstochastic and equal to its expected value, $\Delta{t}$, as $\Delta{t}$ tends to zero.

Taking limits as $\Delta{x}$ and $\Delta{t}$ tend to zero, and using $dx^2 = b^2(x,t)dt$, we obtain

$$dG = \frac{\partial{G}}{\partial{x}}dx + \frac{\partial{G}}{\partial{t}}dt + \frac{1}{2}\frac{\partial^2{G}}{\partial{x}^2}b^2(x,t)dt$$

if we substitute the dx with $\mathrm{d}x = a( x, t) \mathrm{d}t + b( x, t )\mathrm{d}z$, equation becomes

$$dG = \left ( \frac{\partial{G}}{\partial{x}}a(x,t) + \frac{\partial{G}}{\partial{t}} + \frac{1}{2}\frac{\partial^2G}{\partial{x}^2} b^2(x,t)\right )dt + \frac{\partial{G}}{\partial{x}}b(x,t)dz$$

Property of Stock Prices

The Process for A Stock Return

The most widely used model of stock return is

$$\frac{dS}{S} = \mu \mathrm{d}t + \sigma \mathrm{d}z \text{ and } \frac{\Delta{S}}{S} \sim \phi(\mu\Delta{t}, \sigma^2\Delta{t})$$

where $\mu$ is the stock's expected rate of return, and $\sigma$ is the volatility of the stock price. It represents the stock return process in the real world. In a risk-neutral world, $\mu$ equals the risk-free rate $r$.

The Log Return

The Taylor series expansion of $\ln{(1+x)}$ is

$$\ln{(1+x)} = 0 + \frac{1}{1+0}\times \Delta{x} - \frac{1}{2}\frac{1}{(1+0)^2} \times \Delta{x}^2 + \dots \approx x$$

for small $\Delta{x}$ and ignore the terms of higher order in $\Delta{x}$.

Thus, $\ln{\frac{S_t}{S_0}} = \ln{(1 + \frac{S_t-S_0}{S_0})} \approx \frac{\Delta{S}}{S_0} \text{ ONLY for small } \Delta{S}$, also note that $S_t$ here is instantaneous price.

The log rate of return is also called the continuously compounded rate of return.

The Lognormal Property of Stock Prices

Define

$$G = \ln{S}$$

Where S follows $\frac{dS}{S} = \mu \mathrm{d}t + \sigma \mathrm{d}z$.

Since

$$\frac{\partial{G}}{\partial{S}} = \frac{1}{S}, \frac{\partial^2G}{\partial{S^2}} = - \frac{1}{S^2}, \frac{\partial{G}}{\partial{t}} = 0$$

Using Ito's Lemma, we can get

$$dG = \left ( \mu - \frac{\sigma^2}{2}\right)dt + \sigma dz$$

Since $\mu$and $\sigma$are constant, this equation indicates that $G = \ln{S}$follows a generalized Wiener process. It has constant drift rate $\mu - \sigma^2/2$and constant variance rate $\sigma^2$. The change in $\ln{S}$between time 0 and some future time $T$is therefore normally distributed, with mean $\left( \mu-\sigma^2/2\right)$and variance $\sigma^2T$. This means that

$$\ln\frac{S_T}{S_0} \sim \phi{\left [\left ( u - \sigma^2/2 \right )T, \sigma^2T \right ]}$$

Where $S_T$is the stock price at time T, $S_0$is the stock price at time 0.

Black-Scholes Equation

$$c = S_0N(d_1) - Ke^{-rT}N(d_2)$$

$$p = Ke^{-rT}N(-d_2)-S_0N(-d_1)$$

Where

$$d_1 = \frac{\ln{\left ( S_0 / K \right )} + \left ( r + \sigma^2/2\right )T}{\sigma \sqrt{T}}$$

$$d_2 = \frac{\ln{\left ( S_0/K\right )} +\left ( r - \sigma^2/2\right )T}{\sigma \sqrt T} = d_1 - \sigma \sqrt{T}$$

Proof

Key Result

If V is lognormally distributed, and the mean of $\ln{V}$is $m$ and the standard deviation of $\ln{V}$is $w$, then

$$E( \max(V - K, 0)) = E(V)N(d_1) - KN(d_2)$$

Where

$$d_1 = \frac{\ln{\left [ E(V)/K\right ]} + w^2/2}{w}$$

$$d_2 = \frac{\ln \left [ E(V) / K\right ] - w^2/2}{w}$$

Note that as to the call option,

• $N(d_2)$is the probability of exercise,

• $S_0e^{rT}N(d_1)/N(d_2)$is the expected stock price in a risk-neutral world if option is exercised

Similarly, as to the put option

• $N(-d_2)$is the probability of exercise,

• $S_0N(-d_1)/N(-d_2)$is the expected stock price in a risk-neutral world if option is exercised

Proof of Key Result

Define $g(V)$as the probability density function of V. It follows that

$$E\left [ \max(V-K, 0) \right ] = \int_{K}^{\infty} (V-K)g(V)dV$$

From the properties of the lognormal distribution,

$$E(V)=e^{m+\frac{1}{2}\sigma^2}$$

$$m = \ln{E(V)} - w^2/2$$

Define a new variable

$$Q = f(V) = \frac{\ln{V} - m}{w}$$

It follows standard normal distribution,

Denote the density function of $Q$by $h(Q)$so that$h(Q) = \frac{1}{\sqrt{2\pi}}e^{-Q^2/2}$,

And since $f(V)$is strictly increasing, we can use the formula for the density of a strictly increasing function.

Support $R_{Q} = \left \{ q = f(v) : v \in R_{V}\right \}$,

$$\begin{aligned} h(q) = \left \{ \begin{aligned} &g(f^{-1}(q))\frac{df^{-1}(q)}{dq}, && \text{if}\ q \in R_{Q}\\ &0, && \text{if} q \notin R_{Q} \end{aligned} \right. \end{aligned}$$

$$\begin{aligned} E\left [ \max(V-K, 0) \right ] &= \int_{K}^{\infty} (V-K)g(V)dv \\ &= \int_{f(K)}^{\infty} \left [f^{-1}(Q) - K\right ]g(f^{-1}(Q))\frac{d(f^{-1}(Q))}{dQ}dQ \\ &= \int_{(\ln{K}-m)/w}^{\infty}\left( e^{Qw+m} - K \right )h(Q)dQ \\ &= \int_{(\ln{K} -m)/w}^{\infty}e^{Qw+m}h(Q)dQ-K\int_{(\ln{K} - m)/2}^{\infty}h(Q)d(Q) \end{aligned}$$

Now

$$\begin{aligned} e^{Qw+m}h(Q) &= \frac{1}{\sqrt{2\pi}}e^{(-Q^2+2Qw+2m)/2} \\ &=\frac{1}{\sqrt{2\pi}}e^{\left [ -(Q-w)^2+2m+w^2 \right ]/2} \\ &=\frac{e^{m+w^2/2}}{\sqrt{2\pi}}e^{\left [ -(Q-w)^2\right ]/2} \\ &= e^{m+w^2/2}h(Q-w) \end{aligned}$$

This means that

$$\begin{aligned} E(max(V-K, 0)) &= e^{m+w^2/2}\int_{(\ln{K}-m)/w}^{\infty}h(Q-w)dQ - K\int_{(\ln{K}-m)/w}^{\infty}h(Q)dQ \\ &= e^{m+w^2/2}\int_{(\ln{K}-m)/w-w}^{\infty}h(Q-w)d(Q-w) - K\int_{(\ln{K}-m)/w}^{\infty}h(Q)dQ \\ &= e^{m+w^2/2}\left[1 - N\left (\frac{\ln{K}-m}{w}-w\right )\right] - K\left[1 - N\left(\frac{\ln{K-m}}{w}\right) \right] \\ &=e^{m+w^2/2} N\left (\frac{-\ln{K}+m}{w}+w \right) - K N\left ( \frac{-\ln{K}+m}{w}\right ) \\ &= E(V) N\left (\frac{-\ln{K}+\ln{E(V)}-w^2/2}{w}+w \right) - K N\left ( \frac{-\ln{K}+\ln{E(V)}-w^2/2}{w}\right ) \\ &= E(V) N\left (\frac{\ln{\frac{E(V)}{K}}+w^2/2}{w} \right) - K N\left ( \frac{\ln{\frac{E(V)}{K}}-w^2/2}{w}\right ) \\ &= E(V) N\left (d_1\right) - K N\left ( d_2\right ) \end{aligned}$$

Where

$$d_1 = \frac{\ln{\frac{E(V)}{K}}+w^2/2}{w}$$

$$d_2 = \frac{\ln{\frac{E(V)}{K}}-w^2/2}{w}$$

The Black-Scholes Result

We now consider a call option on a non-dividend-paying stock maturing at time T. The strike price is K, the risk-free rate is r, the current stock price is $S_0$, the stock price at time T is $S_{T}$ and the volatility is $\sigma$. The call price is given by

$$c = e^{-rT}\^{E}(\max(S_{T}-K, 0))$$

Where $\^{E}$denotes the expectation in a risk-neutral world. As mentioned above, under the stochastic process assumed by Black-Scholes, $S_{T}$is lognormal. Based on the deduction above, this means that

$$\ln\frac{S_T}{S_0} \sim \phi{\left [\left ( r - \sigma^2/2 \right )T, \sigma^2T \right ]}$$

Based on the property of lognormal, $\^{E}(\frac{S_T}{S_0})=e^{rT}$, and $Var\left( \frac{S_T}{S_0}\right)=e^{2rT}\left(e^{\sigma^2T}-1 \right)$, from the key result just proved, it implies

$$c = e^{-rT}\left [ S_{0}e^{rT}N(d_1)-KN(d_2) \right ] = S_0N(d_1)-Ke^{-rT}N(d_2)$$

Where

$$d_1 = \frac{\ln{\left [ \^{E}(S_T)/K\right ]+\sigma^2T/2}}{\sigma \sqrt{T}} = \frac{\ln(S_0/K)+(r+\sigma^2/2)T}{\sigma \sqrt{T}}$$

$$d_2 = \frac{\ln{\left [ \^{E}(S_T)/K\right ]-\sigma^2T/2}}{\sigma \sqrt{T}} = \frac{\ln(S_0/K)+(r-\sigma^2/2)T}{\sigma \sqrt{T}}$$

Proof of the Put Option

We can do it in the same way as above. However, here we will do it in another way.

We know that as put option, $N(-d_2)$ is the probability of exercise,

$$\begin{aligned} \^{E}(\max(K-S_T, 0))&=\int_{-\infty}^{-d_2}\left ( K - S_0e^{\left [(r-\frac{1}{2}\sigma^2)T+\sigma \sqrt{T}\varepsilon\right ]}\right )f(\varepsilon)d\varepsilon \\ &=KN(-d_2) - \int_{-\infty}^{-d_2}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d\varepsilon \\ &=KN(-d_2) - \int_{-\infty}^{-d_2-\sigma \sqrt{T}}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d(\varepsilon-\sigma \sqrt{T}) \\ &= KN(-d_2) - \int_{-\infty}^{-d_1}S_0e^{rT}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\varepsilon-\sigma \sqrt{T})^2}{2}}d(\varepsilon-\sigma \sqrt{T}) \\ &= KN(-d_2)-S_0e^{rT}N(-d_{1}) \end{aligned}$$

where

• $\varepsilon$ has a standard normal distribution $\phi(0, 1)$

• $f(\varepsilon) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{\varepsilon^2}{2}}$

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Strictly speaking, this is not a deduction of the Black-Scholes equation, because we were using the result from the equation that $N(-d_2)$is the probability of exercise. And if the function underneath the expectation has been changed, the probability of exercise could change along with that.