Stochastic Calculus
DISCLAIMER THERE IS AT TIMES LOT OF HAND WAVING ,THIS IS NOT AS MATHEMATICALLY RIGOROUS AS YOU WOULD EXPECT LIKE IN A PURE MATH TEXTBOOK.
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Mathematical modelling and computation in finance by Cornerlis Ooseterlee and Lech Grzelak (Main Text)
Stochastic CalcuIus for Finance II Continuous-Time Models by Steven Shreve (Main Reference text)
Stochastic Differential Equations By Bernt Oksendal
Preliminaries
Definition 1 (\sigma algebra) : Let \Omega be a nonempty set , and let \mathcal{F} be a set of collections of subsets of \Omega. We say that \mathcal{F} is a \sigma algebra provided that
- The empty set \emptyset belongs to \mathcal{F}
- A and A^{c} belong to \mathcal{F} whenever A \in \mathcal{F}
- whenever A_1 ,A_2 ,A_3 ..... \in \mathcal{F} then \bigcup_{i=1}^\infty A_{i} \in \mathcal{F}
Definition 2 (Probability measure) : Let \Omega be a nonempty set , and let \mathbb{F} be \sigma algebra of subsets of \Omega. A probability measure \mathbb{P} is a function that, to every set A \in \mathcal{F} , assigns a number in [0,1] called the probability of A written as \mathbb{P}(A) .We require
\mathbb{P}(\Omega) = 1 and
(Countable Additivity) Whenever A_1,A_2,A_3.... is a sequence of disjoint sets in \mathcal{F} , then \mathbb{P}(\bigcup_{n=1}^\infty A_n) = \sum_{n=1}^{\infty} \mathbb{P}(A_n)
The Triple (\Omega, \mathcal{F},\mathbb{}P) is called probability measure
From the above it follows that for finitely many disjoint sets A_1, A_2,A_3....A_n \mathbb{P}(\bigcup_{n=1}^N) = \sum_{n=1}^N \mathbb{P}(A_n)
and \mathbb{P}(A^c) = 1 - \mathbb{P}(A)
Example 1 (Uniform(Lebesgue) measure on [0,1])
\Omega = [0,1]
\mathbb{P}[a,b] = b-a ,0 \leq a \leq b \leq1
single points have zero probability and
(a,b) can be written as \bigcup_{n=1}^\infty[a+\frac{1}{n} , b -\frac{1}{n}]
The \sigma-algebra beginning with closed intervals and adding everything else necessary in order to have a \sigma- algebra is called a Borel \sigma-algebra and denoted by \mathcal{B}[0,1]. Borel \sigma-algebras are \sigma-algebra generated by open sets and equivalently closed sets over any topological space.
Definition 3 let (\Omega,\mathcal{F},\mathbb{P}) be a probability space. If a set A \in \mathcal{F} satisfies \mathbb{P}(A) = 1, we say the event A occurs almost surely
Example 2 Let \mathbb{P} be the uniform measure as defined in example 1. Define X(\omega) = \omega and Y(\omega) = 1 - \omega for all \omega \in [0,1] , then the distribution measure of X is uniform \mu_{X}[a,b] = \mathbb{P}\{\omega;a\leq X(w)\leq b\} = \mathbb{P}[a,b] = b-a, 0\leq a \leq b \leq 1 by definition of \mathbb{P}.
Its easy to see to see that X and Y have the same distribution. But under the probability measure\mathbb{\widetilde{P}} on [0,1] defined by \mathbb{\widetilde{P}[a,b]} = \int_a^b 2\omega \, d\omega = b^2-a^2 , 0 \leq a\leq b \leq1 X and Y have different distributions.
Definition 4 (cumulative distribution function and density function) F(x) = \mathbb{P}\{X \leq x\} , x \in \mathbb{R} \mu_X[a,b] = \mathbb{P}\{a \leq X \leq b \} = \int_a^b f(x) \, dx , -\infty < a \leq b < \infty f(x) is called the density funtion and F(x) is called the cummulative distribution function.
Example 3 (Standard normal random variable) Let \phi(x) = \frac {a}{\sqrt{2\pi}}e^{\frac{-x^2}{2}} be the standard normal density and definiing the cummulative normal distribution function as N(x) = \int_{-\infty}^x \phi(\xi)\,d\xi The function N(x) is strictly increasing function which is surjective onto (0,1) from \mathbb{R}, so it has a strictly increasing inverse function N^{-1}(y),y \in (0,1). Let Y be a uniformly distributed random variable defined on some probability space (\Omega,\mathcal{F},\mathbb{P}) and set X = N^{-1}(Y) then
\begin{align*} \mu_X[a,b] &= \mathbb{P}\{\omega \in \Omega ; a \leq X(\omega) \leq b\} \\ &= \mathbb{P}\{\omega \in \Omega ; a \leq N^{-1}(Y(\omega)) \leq b\}\\ &= \mathbb{P}\{\omega \in \Omega ; N(a) \leq Y(\omega) \leq N(b) \} \\ &= N(b) - N(a) \\ &= \int_a^b \phi(x) \, dx \end{align*}Any random variable that has this distribution regardless of the probability space is called standard normal distribution.
Thing to note the use of uniformly distributed random variable for generating a standard random variable is called probability integral transform and this is commonly used in Monte Carlo simulation
Stochastic Processes
Weiner process
A fundamental stochastic process, which is also commonly used in the construction of stochastic differential equations (SDEs) to describe asset price movements, is the Wiener process, also called Brownian motion. Mathematically, a Wiener process, W(t), is characterized by the following properties:
a)\quad W(t\_{0}) = 0 (technically:\mathbb{P}[W(t_0)=0] =1)
b)\quad W(t) is\quad almost \quad surely \quad continous
c)\quad W(t) \quad has \quad independent \quad incerements \quad with \quad distribution\quad W(t)- W(s) \sim \mathcal{N}(0,t-s)
Black-Scholes model
Part I: Mathematical Foundations and Definitions
This section establishes all the mathematical machinery needed for the Black-Scholes proofs. We assume knowledge of measure theory and basic stochastic calculus but will define all financial and probabilistic concepts explicitly.
Financial Market Framework
The explicit solution to geometric Brownian motion is: S_t = S_0 \exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right) This can be verified using Itô’s lemma on f(t,x) = \ln x.
Portfolio and Trading Strategy Concepts
Martingale Theory for Finance
Options and Derivatives
Fundamental Theorems of Asset Pricing
Part II: The Black-Scholes Model Setup
Model Specification
We consider a financial market on a probability space (\Omega, \mathcal{F}, \mathbb{P}) with filtration \{\mathcal{F}_t\}_{t \geq 0} and two traded assets:
Key Properties of the Model
Let Y_t = \ln S_t. By Itô’s lemma with f(x) = \ln x:
\begin{align} dY_t &= f'(S_t) dS_t + \frac{1}{2}f''(S_t)(dS_t)^2 \\ &= \frac{1}{S_t} dS_t + \frac{1}{2}\left(-\frac{1}{S_t^2}\right)(dS_t)^2 \\ &= \frac{1}{S_t}(\mu S_t \, dt + \sigma S_t \, dW_t) - \frac{1}{2S_t^2}(\sigma S_t)^2 dt \\ &= \mu \, dt + \sigma \, dW_t - \frac{\sigma^2}{2} dt \\ &= \left(\mu - \frac{\sigma^2}{2}\right) dt + \sigma \, dW_t \end{align}
Integrating from 0 to t: Y_t = Y_0 + \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t
Since Y_t = \ln S_t and Y_0 = \ln S_0: \ln S_t = \ln S_0 + \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t
Therefore: S_t = S_0 \exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right)
Part III: Proof via Martingale Approach
Step 1: Construction of Risk-Neutral Measure
Step 1: Define the market price of risk: \theta = \frac{\mu - r}{\sigma}
This is the constant that will appear in Girsanov’s theorem.
Step 2: Define the Radon-Nikodym density process: \begin{align} Z_t &= \exp\left(-\theta W_t - \frac{1}{2}\theta^2 t\right) \\ &= \exp\left(-\frac{\mu - r}{\sigma} W_t - \frac{1}{2}\left(\frac{\mu - r}{\sigma}\right)^2 t\right) \end{align}
Step 3: Verify that Z_t is a martingale and \mathbb{E}[Z_T] = 1.
Since \theta is constant, we can compute: \mathbb{E}[Z_t] = \mathbb{E}\left[\exp\left(-\theta W_t - \frac{1}{2}\theta^2 t\right)\right]
Since W_t \sim N(0,t) under \mathbb{P}: \begin{align} \mathbb{E}[Z_t] &= \int_{-\infty}^{\infty} \exp\left(-\theta w - \frac{1}{2}\theta^2 t\right) \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{w^2}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{w^2 + 2\theta tw}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{(w + \theta t)^2 - \theta^2 t^2}{2t}\right) dw \\ &= \exp\left(-\frac{1}{2}\theta^2 t\right) \exp\left(\frac{1}{2}\theta^2 t\right) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp\left(-\frac{(w + \theta t)^2}{2t}\right) dw \\ &= 1 \end{align}
The last integral equals 1 since it’s the integral of a normal density.
Step 4: Define \mathbb{Q} by \frac{d\mathbb{Q}}{d\mathbb{P}} = Z_T.
Step 5: Apply Girsanov’s theorem. Under \mathbb{Q}, the process: \tilde{W}_t = W_t + \int_0^t \theta \, ds = W_t + \theta t is a \mathbb{Q}-Brownian motion.
Step 6: Transform the stock price dynamics. Under \mathbb{Q}: \begin{align} dS_t &= \mu S_t \, dt + \sigma S_t \, dW_t \\ &= \mu S_t \, dt + \sigma S_t (d\tilde{W}_t - \theta \, dt) \\ &= \mu S_t \, dt + \sigma S_t \, d\tilde{W}_t - \sigma S_t \theta \, dt \\ &= (\mu - \sigma\theta) S_t \, dt + \sigma S_t \, d\tilde{W}_t \\ &= \left(\mu - \sigma \cdot \frac{\mu - r}{\sigma}\right) S_t \, dt + \sigma S_t \, d\tilde{W}_t \\ &= r S_t \, dt + \sigma S_t \, d\tilde{W}_t \end{align}
Step 7: Verify the martingale property. The discounted stock price is: e^{-rt}S_t
Its differential is: \begin{align} d(e^{-rt}S_t) &= -re^{-rt}S_t \, dt + e^{-rt} dS_t \\ &= -re^{-rt}S_t \, dt + e^{-rt}(rS_t \, dt + \sigma S_t \, d\tilde{W}_t) \\ &= e^{-rt}\sigma S_t \, d\tilde{W}_t \end{align}
Since this has no dt term, e^{-rt}S_t is a \mathbb{Q}-martingale.
Step 2: Stock Price Under Risk-Neutral Measure
Under \mathbb{Q}, the stock follows: dS_t = rS_t \, dt + \sigma S_t \, d\tilde{W}_t
Using the same technique as in the original measure, let Y_t = \ln S_t: \begin{align} dY_t &= \frac{1}{S_t} dS_t - \frac{1}{2S_t^2}(dS_t)^2 \\ &= \frac{1}{S_t}(rS_t \, dt + \sigma S_t \, d\tilde{W}_t) - \frac{1}{2S_t^2}(\sigma S_t)^2 dt \\ &= r \, dt + \sigma \, d\tilde{W}_t - \frac{\sigma^2}{2} dt \\ &= \left(r - \frac{\sigma^2}{2}\right) dt + \sigma \, d\tilde{W}_t \end{align}
Integrating: Y_t = Y_0 + \left(r - \frac{\sigma^2}{2}\right)t + \sigma \tilde{W}_t
Therefore: S_t = S_0 \exp\left(\left(r - \frac{\sigma^2}{2}\right)t + \sigma \tilde{W}_t\right)
Step 3: Risk-Neutral Valuation
This follows directly from the risk-neutral valuation principle. In an arbitrage-free complete market, the option price equals the discounted expected payoff under the risk-neutral measure.
Step 4: Computing the Expectation
Step 1: Express the stock price at maturity. Under \mathbb{Q}, using the strong Markov property: S_T = S_t \exp\left(\left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma (\tilde{W}_T - \tilde{W}_t)\right)
Since \tilde{W}_T - \tilde{W}_t \sim N(0, T-t) under \mathbb{Q}, let Z \sim N(0,1). Then: S_T = S_t \exp\left(\left(r - \frac{\sigma^2}{2}\right)(T-t) + \sigma\sqrt{T-t} \cdot Z\right)
Step 2: Set up the expectation. \begin{align} V(t, S_t) &= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[(S_T - K)^+ | \mathcal{F}_t] \\ &= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}\left[\left(S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z} - K\right)^+\right] \end{align}
Step 3: Find the exercise region. The option is exercised when S_T > K, i.e., when: S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z} > K
Taking logarithms: \begin{align} &(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} Z > \ln(K/S_t) \\ &Z > \frac{\ln(K/S_t) - (r-\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}} = -d_2 \end{align}
Step 4: Evaluate the integral. \begin{align} V(t, S_t) &= e^{-r(T-t)} \int_{-d_2}^{\infty} \left(S_t e^{(r-\sigma^2/2)(T-t) + \sigma\sqrt{T-t} z} - K\right) \\ &\quad \times \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &= e^{-r(T-t)} S_t e^{(r-\sigma^2/2)(T-t)} \int_{-d_2}^{\infty} e^{\sigma\sqrt{T-t} z} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &\quad - e^{-r(T-t)} K \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \end{align}
Step 5: Evaluate the first integral. For the first integral, complete the square in the exponent: \sigma\sqrt{T-t} z - \frac{z^2}{2} = -\frac{1}{2}(z - \sigma\sqrt{T-t})^2 + \frac{\sigma^2(T-t)}{2}
Therefore: \begin{align} &\int_{-d_2}^{\infty} e^{\sigma\sqrt{T-t} z} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz \\ &= e^{\sigma^2(T-t)/2} \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-(z-\sigma\sqrt{T-t})^2/2} dz \\ &= e^{\sigma^2(T-t)/2} \int_{-d_2-\sigma\sqrt{T-t}}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-u^2/2} du \\ &= e^{\sigma^2(T-t)/2} \Phi(d_2 + \sigma\sqrt{T-t}) \\ &= e^{\sigma^2(T-t)/2} \Phi(d_1) \end{align}
where we used the substitution u = z - \sigma\sqrt{T-t} and the fact that: d_2 + \sigma\sqrt{T-t} = d_1
Step 6: Evaluate the second integral. \int_{-d_2}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} dz = \Phi(d_2)
Step 7: Combine the results. \begin{align} V(t, S_t) &= e^{-r(T-t)} S_t e^{(r-\sigma^2/2)(T-t)} e^{\sigma^2(T-t)/2} \Phi(d_1) - e^{-r(T-t)} K \Phi(d_2) \\ &= S_t e^{-r(T-t)} e^{(r-\sigma^2/2)(T-t)} e^{\sigma^2(T-t)/2} \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \\ &= S_t e^{-r(T-t) + r(T-t) - \sigma^2(T-t)/2 + \sigma^2(T-t)/2} \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \\ &= S_t \Phi(d_1) - K e^{-r(T-t)} \Phi(d_2) \end{align}
Part IV: Proof via PDE Approach
Step 1: Delta Hedging Strategy
Step 2: Application of Itô’s Lemma to Option Value
Since dS_t = \mu S_t \, dt + \sigma S_t \, dW_t, we have: (dS_t)^2 = (\mu S_t \, dt + \sigma S_t \, dW_t)^2 = \sigma^2 S_t^2 (dW_t)^2 = \sigma^2 S_t^2 dt
where we used the fact that (dW_t)^2 = dt, dt \cdot dW_t = 0, and (dt)^2 = 0.
Applying Itô’s lemma: \begin{align} dV &= \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS_t + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} (dS_t)^2 \\ &= \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} (\mu S_t \, dt + \sigma S_t \, dW_t) + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} \sigma^2 S_t^2 dt \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt + \sigma S_t \frac{\partial V}{\partial S} dW_t \end{align}
Step 3: Portfolio Dynamics
We have: \begin{align} d\Pi_t &= dV - \Delta_t dS_t \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt + \sigma S_t \frac{\partial V}{\partial S} dW_t \\ &\quad - \Delta_t (\mu S_t dt + \sigma S_t dW_t) \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} - \Delta_t \mu S_t\right] dt \\ &\quad + \sigma S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) dW_t \\ &= \left[\frac{\partial V}{\partial t} + \mu S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right] dt \\ &\quad + \sigma S_t \left(\frac{\partial V}{\partial S} - \Delta_t\right) dW_t \end{align}
Step 4: Eliminating Risk
Step 5: No-Arbitrage Condition
If a risk-free portfolio earned more than the risk-free rate, we could borrow at rate r, invest in the portfolio, and earn a risk-free profit (arbitrage). If it earned less, we could short the portfolio, lend at rate r, and again earn risk-free profit.
Step 6: Deriving the Black-Scholes PDE
Equating the two expressions for d\Pi_t: \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} = r\left(V - \frac{\partial V}{\partial S} S_t\right)
Rearranging: \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} = rV - rS_t \frac{\partial V}{\partial S}
Therefore: \frac{\partial V}{\partial t} + rS_t \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2} - rV = 0
The terminal condition comes from the option payoff at maturity.
Step 7: Solving the PDE
To solve this PDE, we use a change of variables to transform it into the heat equation.
We need to compute the partial derivatives of V in terms of u. We have: V(t,S) = K e^{-r(T-t)} u(T-t, \ln(S/K))
Let \tau = T-t and x = \ln(S/K). Then:
\begin{align} \frac{\partial V}{\partial t} &= K e^{-r\tau} \left[r u - \frac{\partial u}{\partial \tau}\right] \\ \frac{\partial V}{\partial S} &= K e^{-r\tau} \frac{1}{S} \frac{\partial u}{\partial x} \\ \frac{\partial^2 V}{\partial S^2} &= K e^{-r\tau} \frac{1}{S^2} \left[\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x}\right] \end{align}
Substituting into the Black-Scholes PDE and simplifying (after dividing by K e^{-r\tau}): -\frac{\partial u}{\partial \tau} + \left(r - \frac{\sigma^2}{2}\right) \frac{\partial u}{\partial x} + \frac{1}{2}\sigma^2 \frac{\partial^2 u}{\partial x^2} = 0
Therefore: \frac{\partial u}{\partial \tau} = \frac{1}{2}\sigma^2 \frac{\partial^2 u}{\partial x^2} + \left(r - \frac{\sigma^2}{2}\right) \frac{\partial u}{\partial x}
The initial condition at \tau = 0 (i.e., t = T) is: u(0, x) = \frac{(Ke^x - K)^+}{K} = (e^x - 1)^+
Step 8: Computing the Solution
We have: V(t,S) = K e^{-r(T-t)} u(T-t, \ln(S/K))
The transformation from the u solution to the V solution follows the same integration techniques as in the martingale approach. The key insight is that both methods yield identical results, demonstrating the equivalence of risk-neutral valuation and PDE approaches.
Verification and Conclusion
Economic Interpretation
Key Insights
Greeks and Risk Management
Extensions and Limitations
Computational Implementation
This complete treatment of the Black-Scholes model provides both theoretical rigor and practical applicability, showing how mathematical finance bridges pure mathematics and real-world financial applications.