Deriving Black-Scholes from Geometric Brownian Probability Density Function

Nothing revolutionary, maybe it can help with your stochastic calculus problem set :)

In this article Pricing Uniswap v3 LP Positions: Towards a New Options Paradigm?, Prof. Guillaume Lambert left the following exercise.

from Lambert’s post linked above

I will show my solution here.

First of all, let's calculate the probability of S_t > K, aka {S_t > K}.

Recall that S_t follows a Geometric Brownian Motion (GBM). The Wiener process W_t is normally distributed, aka W_t ~ N(0,t) , therefore it can be rewritten as Z*sqrt(t) where Z ~ N(0,1) .

\(\begin{align*} S_t &= S_0 \exp \left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t \right) \\ &= S_0 \exp \left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma Z\sqrt{t}\right) \end{align*}\)

Substituting this back to {S_t > K} and rearranging the equation we get the following condition.

\(\begin{align*} \{S_t > K\} &= \left\{S_0 \exp \left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma Z\sqrt{t} \right) > K\right\} \\ &= \left\{Z > \frac{1}{\sigma \sqrt{t}} \left( \ln \frac{K}{S_0} - \left(r - \frac{\sigma^2}{2}\right)t\right)\right\} \end{align*}\)

1. Equation

Now let’s start with the actual equation, we first abbreviate the GBM probability density function (pdf) as lower case phi since φ is commonly used to represent pdf.

\( \text{Call}(P,t) = \int_K^\infty f(P,\mu, \sigma, t) \cdot e^{-\mu t} \cdot (x - K) \ dx = \int_K^\infty \phi \cdot e^{-\mu t} \cdot (x - K) \ dx \)

We can take e^{-\mu t} out, manipulate the equation, and tackle each integral. We can’t take out φ cause φ depends on x(see the very first screenshot at the top of this post).

\(\begin{align*} \text{Call}(P,t) &= e^{-\mu t} \int_K^\infty \phi \cdot (x - K) \ dx & \phi = f(P,\mu, \sigma, t) \\ &= e^{-\mu t} \int_K^\infty \phi \cdot x \ dx - Ke^{-\mu t}\int_K^\infty \phi \ dx & \text{algebra} \end{align*}\)

This looks a lot like the Black-Scholes equation already.

1.1 Solving the first part

Recall that φ is a pdf, and the definition of expected value is the integral of x times its pdf. If you squint your eyes and look at the integral carefully, it is the same as below,

\(\int_K^\infty x \cdot \phi \ dx = \mathbb{E}\left[S_t \cdot \mathbf{1}_{\{S_t > K\}}\right]\)

where “1” is an indicator function. This is essentially saying that the integral equals to the expected value of S_t on the condition that S_t > K.

\(\mathbf{1}_{\{S_t > K\}} := \begin{cases} 1 & \text{if } S_t > K \\ 0 & \text{otherwise} \end{cases}\)

We can now substitute S_t, the integral, and {S_t > K} into the equation.

\(\begin{align*} e^{-\mu t} \int_K^\infty \phi \cdot x \ dx &= e^{-\mu t} \cdot \mathbb{E}\left[S_t \mathbf{1}_{\{S_t > K\}}\right] \\ &= e^{-\mu t} \cdot \mathbb{E}\left[S_0 \exp \left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma Z\sqrt{t}\right) \cdot \mathbf{1}_{\{S_T > K\}}\right] \\ &= \mathbb{E}\left[S_0 \exp \left(-\mu t + \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma Z\sqrt{t}\right) \cdot \mathbf{1}_{\{S_T > K\}}\right] \\ &= S_0 \mathbb{E}\left[\exp \left(-\frac{\sigma^2}{2}t + \sigma Z\sqrt{t}\right) \cdot \mathbf{1}_{\{S_T > K\}}\right] \\ &= S_0 \mathbb{E}\left[\exp \left(-\frac{\sigma^2}{2}t + \sigma Z\sqrt{t}\right) \cdot \mathbf{1}_{\left\{Z > \frac{1}{\sigma \sqrt{t}} \left( \ln \frac{K}{S_0} - \left(r - \frac{\sigma^2}{2}\right)t\right)\right\}}\right] \end{align*}\)

Since X is a random variable with a normal distribution of mean 0 and standard deviation 1, we can rewrite the expected value back to an integral with the pdf of a normal distribution.

\(\begin{align*} e^{-\mu t} \int_K^\infty \phi \cdot x \ dx &= S_0 \int_{\frac{1}{\sigma \sqrt{t}} \left( \ln \frac{K}{S_0} - \left(r - \frac{\sigma^2}{2}\right)t\right)}^\infty \exp\left( \sigma\sqrt{t}x - \frac{\sigma^2}{2}t \right) \cdot \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{x^2}{2} \right)\ dx \\ &= S_0 \int_{\frac{1}{\sigma \sqrt{t}} \left( \ln \frac{K}{S_0} - \left(r - \frac{\sigma^2}{2}\right)t\right)}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{\left(x - \sqrt{t}\sigma\right)^2}{2} } \ dx \\ &= S_0 \Phi^*(d_1) \end{align*}\)

where Φ* is the cumulative distribution function of N(σ\sqrt(t), 1) and d1 is the lower bound of the integral. To make Φ the CDF of N(0,1), we just need to make d1

\(d_1 = \frac{1}{\sigma \sqrt{t}} \left( \ln \frac{S_0}{K} + \left(r - \frac{\sigma^2}{2}\right)t\right) - \sqrt{t}\sigma\)

1.2 Solving the second part

This part is shorter. Following the same intuition as above, we can rewrite

\(\begin{align*} Ke^{-\mu t}\int_K^\infty \phi \ dx &= Ke^{-\mu t}\mathbb{E}[\mathbf{1}_{\{S_t > K\}}] \\ &= Ke^{-\mu t} \int_{\frac{1}{\sigma\sqrt{t}} \left(\ln\frac{K}{S_0} - \left(r-\frac{\sigma^2}{2}\right)t \right)}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \ dx & \text{pdf of } N(0,1) \\ &= Ke^{-\mu t} \Phi(-d_2) & d_2 = \frac{1}{\sigma\sqrt{t}} \left(\ln\frac{K}{S_0} - \left(r-\frac{\sigma^2}{2}\right)t \right) \\ &= Ke^{-\mu t} \Phi(d_2) & d_2 = \frac{1}{\sigma\sqrt{t}} \left(\ln\frac{S_0}{K} + \left(r-\frac{\sigma^2}{2}\right)t \right) \end{align*}\)

Notice how the expected value does not have x in it. That’s because the integrand only has the pdf.

2. Piecing them together

\(\begin{align*} \text{Call}(P,t) &= e^{-\mu t} \int_K^\infty \phi \cdot x \ dx - Ke^{-\mu t}\int_K^\infty \phi \ dx \\ &= S_0 \Phi(d_1) - Ke^{-\mu t} \Phi(d_2) \end{align*}\)

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