Black-Scholes equation

Here, I will present three independent derivations of the Black-Scholes equation, assuming the risk-neutral pricing principle without proof. First, I will show how the equation emerges in the continuum limit of a binomial tree. The next method makes use of stochastic calculus and a simple no-arbitrage argument. Finally, I present the martingale pricing method, which is the simplest but requires a lot of formal mathematical background.

Claim. Let $D(S,t)$ be a derivative whose value depends on an underlying $S$ and time $t$. Assume the underlying follows geometric Brownian motion with drift $\mu$ and volatility $\sigma$, which defines the stochastic process

$$ \frac{dS}{S} = \mu dt + \sigma dW, $$

where $W$ is a Brownian motion. Let the risk-free interest rate be $r$. Then, the unique arbitrage-free price of the derivative follows the Black-Scholes equation:

$$ \newcommand\pdv[2]{\frac{\partial#1}{\partial#2}} rD = \pdv{D}{t} + rS\pdv{D}{S} + \frac{1}{2} \sigma^2 S^2 \pdv{^2 D}{S^2}. $$

Risk-neutral pricing: we state without proof that the unique arbitrage-free price of a derivative is given by the expectation value of the derivative under the risk-neutral probability measure. Let us define this measure now.

Let $B_t$ be the money-market account continuously compounding at rate $r$, so that $B_t = e^{rt}B_0 $. We measure all values with respect to this account, which is equivalent to discounting or using present values. The risk-neutral probability measure is defined so that the expectation value of any tradable asset $A_t$, measured with respect to $B_t$, is constant:

$$ \mathbb{E}_{RN}\left(\frac{A_t}{B_t}\right) = \mathbb{E}_{RN}\left(\frac{A_{t'}}{B_{t'}}\right) \quad\Leftrightarrow\quad \mathbb{E}_{RN}(A_t) = e^{r(t-t')} \mathbb{E}_{RN}(A_{t'}). $$

Binomial tree: continuum limit

Consider a simple two-state world in which the stock evolves geometrically with each time step:

$$ S_t \to S_{t+\Delta t} \in \{ e^u S_t \;, \;e^{-d}S_t \} \quad \Leftrightarrow \quad \log S_t \to \log S_{t+\Delta t} \in \{ \log S_t + u \;,\; \log S_t - d \}, $$

where $u,d > 0$. Suppose the up move occurs with probability $p$, which in the risk-neutral measure is fixed to

$$ p = \frac{e^{r\Delta t} - e^{-d}}{e^{u} - e^{-d}}. $$

First, we must determine a well-defined scaling limit as $\Delta t \to 0$, clearly $u, d$ must scale with $\Delta t $ in a particular way. So, let us define a collection of random variables $ X_t $ taking values $u$ and $-d$ with probabilities $p$ and $1 -p $. Next, consider the $N$-step evolution:

$$ \log S_{t+\Delta t} = \log S_t + X_t \quad \Rightarrow\quad \log S_T = \log S_0 + \sum_{i=1}^N X_i, $$

where $ T = N / \Delta t $ and $X_i$ are iid. The central limit theorem implies

$$ \log S_T = \log S_0 + \sqrt{N}\sigma_X \mathcal{N}(0,1) + N \mu_X \quad \Rightarrow\quad S_T = S_0 \exp\bigg[ \frac{T}{\Delta t}\mu_X + \sqrt{\frac{T}{\Delta t}}\sigma_X \mathcal{N}(0,1) \bigg], $$

where $ \mathcal{N}(0,1) $ is a random variable drawn from the normal distribution with unit variance. Clearly, to have a well-defined non-trivial continuum limit the quantities $\tilde{\mu} = \mu/\Delta t $ and $\tilde{\sigma} = \sigma/\sqrt{\Delta t} $ both have to be finite. This determines implicitly the scalings of $u, d$ with $\Delta t$.

First, consider $ \mu = \mathbb{E}(X) = p u - (1-p) d $. Let us define rescaled jumps $u = \Delta t^\nu \tilde{u} $, $d = \Delta t^\nu \tilde{d} $. With some algebra, one obtains:

$$ (\tilde{u} + \tilde{d})\tilde{\mu} = r(\tilde{u} + \tilde{d}) - \frac{1}{2}\Delta t^{2\nu - 1} (\tilde{u}\tilde{d}^2 + \tilde{u}^2\tilde{d}). $$

A non-trivial relation between rescaled quantities is recovered when $\nu = 1/2$. A choice of $\nu < 1/2 $ gives $|\tilde{\mu}| \to \infty $, and so the model jumps too violentlly and blows up. A choice of $\nu > 1/2$ yields $\tilde{\mu} = r$ for all choices of $\tilde{u}, \tilde{d}$, so the model jumps too little and becomes trivial.

Hence, our ansatz is $ u = \sqrt{\Delta t} \tilde{u} $ and $d = \sqrt{\Delta t} \tilde{d}$. One can check that this choice also yields a finite $\tilde{\sigma}$:

$$ \tilde{\sigma} = \sqrt{\frac{\tilde{u}\tilde{d}^2 + \tilde{u}^2\tilde{d}}{\tilde{u} + \tilde{d}}} \quad \Rightarrow \quad \tilde{\mu} = r - \frac{1}{2} \tilde{\sigma}^2. $$

This relationship between $\tilde{\mu}$ and $\tilde{\sigma}$ is a key result. With this, the expression for $S_T$ reads:

$$ S_T = S_0 \exp\left[ rT - \frac{1}{2}\tilde{\sigma}^2 T + \sqrt{T}\tilde{\sigma}\mathcal{N}(0,1) \right]. $$

Note that this depends only on the rescaled variance, not on the rescaled drift. This is a property of the risk-neutral measure and a key element of option pricing: real-world drift of underlying assets don't affect the arbitrage-free option prices.

It is easy to recover the Black-Scholes equation at this point, simply consider the small change in $\log S $:

$$ \Delta \log S = \left(r- \frac{1}{2} \tilde{\sigma}\right)\Delta t + \sqrt{\Delta t} \tilde{\sigma}\mathcal{N}(0,1). $$

Consider the expected value of the change in $D(S,t)$ under the risk-neutral measure:

\begin{align*} \mathbb{E}[\Delta D] &= \pdv{D}{t}\Delta t + \pdv{D}{\log S} \mathbb{E}[\Delta \log S] + \frac{1}{2} \pdv{^2 D}{\log S^2} \mathbb{E}[(\Delta \log S)^2] \\ &= \pdv{D}{t}\Delta t + S\pdv{D}{S} \left(r- \frac{1}{2} \tilde{\sigma}\right)\Delta t + \frac{1}{2} \left( S\pdv{D}{S} + S^2\pdv{^2 D}{S^2} \right) \tilde{\sigma}^2 \Delta t \\ &= \pdv{D}{t}\Delta t + rS\pdv{D}{S}\Delta t + \frac{1}{2} \tilde{\sigma}^2 S^2 \pdv{^2 D}{S^2} \Delta t, \end{align*}

where I used $ \mathbb{E}[\mathcal{N}(0,1)^2] = 1 $. By risk-neutral pricing, the expected change in the value of $D$ is $r\Delta t D$, and so we recover the Black-Scholes equation.

Stochastic calculus

This method is much more straightforward, provided one is familiar with Ito's lemma.

Ito's lemma: Consider a stochastic process $X_t$, and a function $f = f(t,X)$. The process followed by $f$ is:

$$ df = \pdv{f}{t}dt + \pdv{f}{X}dX + \frac{1}{2}\pdv{^2f}{X^2} dX^2, \quad \text{where} \quad dt^2 = 0,\quad dt dW = 0, \quad dW^2 = dt. $$

Appliying Ito's lemma to a derivative $D$ on an underlying following geometric Brownian motion yields:

\begin{align*} dD &= \pdv{D}{t}dt + \pdv{D}{S} dS + \frac{1}{2} \pdv{^2 D}{S^2} dS^2 \\ &= dt\left[ \pdv{D}{t} + \pdv{D}{S} S\mu + \frac{1}{2}S^2\sigma^2\pdv{^2 D}{S^2} \right] + S\sigma \pdv{D}{S}dW. \end{align*}

Now, let us consider instead a portfolio without a stochastic component. Consider the portfolio long one unit of $D$ and $\alpha$ units of the underlying $S$. The change in the value of this portfolio due to movements of the underlying and elapsed time (fixing our holding $\alpha$ constant) is:

$$ d(D + \alpha S) = dt\left[ \pdv{D}{t} + \pdv{D}{S} S\mu + \alpha \mu S + \frac{1}{2}S^2\sigma^2\pdv{^2 D}{S^2} \right] + S\sigma \left(\pdv{D}{S} + \alpha \right)dW. $$

To cancel the stochastic component, we can fix $\alpha = -\partial D / \partial S$, obtaining a portfolio which grows at some constant rate. No-arbitrage requires this rate to equal the risk-free interest rate, hence we obtain:

$$ r(D - \pdv{D}{S} S) = \pdv{D}{t} + \frac{1}{2}S^2\sigma^2\pdv{^2 D}{S^2}, $$

which is the Black-Scholes equation.

Martingale pricing

We need a series of mathematical facts about martingales and measure changes on the space of Brownian motions.

Fact 1 (Martingales): a process $X_t$ is a martingale if for all $t,s $ with $t > s$, the conditional expectation of $X_t$ in filtration $\mathcal{F}_s$ is $X_s$:

$$ \mathbb{E}(X_t \mid \mathcal{F}_s) = X_s. $$

This is essentially the statement that information available time $s$ contains no information about the future process $ X_t - X_s $. For our purposes, we can simply state that driftless processes of the general form

$$ dX_t = \varphi(X_t) dW_t $$

are martingales.

Fact 2 (Martingale pricing): there can be no arbitrage if all tradable assets in a market are martingales. This is easy to see in zero-interest case: suppose all tradable assets are martingales. The expected of any portfolio equals today's value, and so if the portfolio has value $ P=0 $ initially, the existence of a non-zero probability of $P>0$ at a later time implies the existence of a non-zero probability of $P<0$ at that same time.

In the non-zero interest case, one denominates all assets in zero-coupon riskless bonds, effectively working with the discounted values. Therefore, denoting the riskless bond as $B_t$, one demands for all tradable assets $A_t$ that the process $ A_t/B_t $ is a martingale.

Fact 3 (Measure change): a measure change on the space of Brownian paths is equivalent to changing the drift of the Brownian motion. That is, given some Brownian motion $W_t$ under some measure, there exists an equivalent measure under which the process $\tilde{W}_t = W_t - \nu t$ is a Brownian motion, for a reasonable function $\nu$.

Combining these facts, let us consider a stock following geometric Brownian motion in the real-world measure and a riskless bond with continuously compounding interest rate $r$:

$$ dS = \mu S dt + \sigma S dW, \quad dB = r B dt. $$

The process for $S/B$ follows from Ito's lemma:

$$ d\left(\frac{S}{B}\right) = (\mu - r) \frac{S}{B} dt + \sigma \frac{S}{B} dW. $$

Now, we change to the risk-neutral measure under which the process above is driftless, which is equivalent to setting $ \mu = r $. Hence, under the risk-neutral measure the stock follows the process

$$ dS = r S dt + \sigma S dW. $$

Now, consider a derivative $D$. Under the risk-neutral measure, it follows the process

$$ dD = \left( \pdv{D}{t} + rS\pdv{D}{S} + \frac{1}{2}\sigma^2 S^2 \pdv{^2D}{S^2} \right) dt + \sigma S \pdv{D}{S}dW $$

As before we demand that $D/B $ is a martingale:

$$ d\left(\frac{D}{B}\right) = \frac{1}{B}(dD - rD dt) \quad\Rightarrow\quad dD - \sigma S \pdv{D}{S}dW = rDdt, $$

hence recovering the Black-Scholes equation.

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