Notes on Courses QMF & MathModelling of CUHK
2023-04-24
Thanks for my professors, Dr. Willem J. Van Vliet, Dr. Keith Law and Mr. Kelvin Lam
Yibo Li
yli12@stevens.edu; yiboli@link.cuhk.edu.hk
http://eborlee.github.io
Table Of Contents
1 Probability
1.1 Introductory Background: Flipping Coins
Flipping Two fair coins. Use H and T to denote heads and tail.
P[HH] = P[HT] = P[TH] = P[TT] = 1/4
P[H on first coin] = P[HH] + P[TT] = 1/2
[Outcomes] we do not directly assign probabilities to outcomes but to events. An event is a collection of outcomes. It’s a way of grouping together outcomes.
For the flipping coins, there are 4 outcomes: HH, HT, TT, TH.
A prob measure is a function that maps events to [0,1]
16 possible events and their corresponding probs:
P(HH) makes no sense. Right expression is P({HH}).
1.2 How to define probability
Requires 3 parts:
- Sample Space Ω as the set of all possible outcomes
- Event Space as the set of events that we will assign probs to
- Prob Measure, which assigns probs to events
An event is a subset A ⊆ Ω including ∅(nothing happens) and Ω(something happens)
So the event space would be a set of subsets of Ω.
Sigma-Algebra
Def: Ω be a set. The set F of subsets of Ω is called a sigma-algebra if it satisfies:
- Ω ∈ F
- If A ∈ F, then Ac ∈ F
- F is closed under countable unions. If A1, A2 … are in F, then ∪Ai ∈ F.
Then we have the definition of Event Space:
Some examples:
The discrete sigma-algebra is the largest sigma-algebra.
Measureable Space
Prob Measure:
P(Ω) is not Ω, but much larger.
Now we only modeled “group together”. Actually we should also consider conditional prob:
2 Random Variables
10 Derivation of Black-Scholes-Merton Fomula
Method 1: Long Way
Firstly, we could derive the SDE of the European Call:
\[\begin{align*} &\text{Let}~r_{t} = r, \text{ constant}\\ &\Rightarrow dB_{t}=rB_{t}\mathrm{d}_{t}\to \frac{dB_{t}}{B_{t}}=r\mathrm{d}_{t}\\ &\text{Let}~u=\ln B_{t} \to \frac{\mathrm{d}u}{\mathrm{d}B_{t}}=\frac{1}{B_{t}}\to \mathrm{d}u=\frac{\mathrm{d}B_{t}}{B_{t}}\\ &\int_{0}^{t}\frac{\mathrm{d}B_{s}}{B_s} = \int_{0}^{t}r\mathrm{d}_{s}=rt\\ &\text{The left side also equals to: }\int_{\ln B_{0}}^{\ln B_{t}}\mathrm{d}u=\ln B_{t}-\ln B_{0}=\ln\frac{B_{t}}{B_{0}}\\ &\ln\frac{B_{t}}{B_{0}}=rt\\ &B_t=e^{rt}B_{0}\\ &\text{Assume the underlying stock return follows: }\\ &\frac{\mathrm{d}S_t}{S_t}=\mu \mathrm{d}_t+\sigma \mathrm{d}W_t\\ &\text{And the payoff of a European Call is: }C_T=\max(S_T-K,0)=(S_T-K)^+\\ &\text{From Markov,}C_T\text{ only depends on }S_T\to C_t=C(t,S_t)\\ &\text{From Ito Lemma: }\mathrm{d}C=\frac{\partial C}{\partial t}\mathrm{d}_t + \frac{\partial C}{\partial S}\mathrm{d}_S+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\mathrm{d}_S)^2\\ &\Rightarrow \mathrm{d}C_t=\left(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}\sigma^2S_t^2 + \frac{\partial C}{\partial S}\mu S_t\right)\mathrm{d}_t+\frac{\partial C}{\partial S}\sigma S_t\mathrm{d}W_t\\ \end{align*}\]Then we discount the Ct to Time 0: \(\begin{align*} &e^{-rt}C_t=e^{-rt}C(t,S_t)\\ &\Rightarrow\\ &\text{Let} ~f(t,C)=e^{-rt}C_t\\ &\mathrm{d}(e^{-rt}C_t)= \mathrm{d}f=-re^{-rt}C_t\mathrm{d}t+e^{-rt}\mathrm{d}C_t\\ &=-re^{-rt}C_t\mathrm{d}t+e^{-rt}\left(\left(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}\sigma^2S_t^2 + \frac{\partial C}{\partial S}\mu S_t\right)\mathrm{d}t+\frac{\partial C}{\partial S}\sigma S_t\mathrm{d}W_t\right)\\ &=e^{-rt}\left(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}\sigma^2S_t^2 + \frac{\partial C}{\partial S}\mu S_t-rC\right)\mathrm{d}t+e^{-rt}\frac{\partial C}{\partial S}\sigma S_t\mathrm{d}W_t \end{align*}\)
According to self-financing strategy, we want to construct a portfolio to replicate the call option exposing to money market fund and the stock.
\[\begin{aligned} &\text{Let}~X_t=\theta_{B,t}B_t+\theta_{S,t}St~\text{denote the of the portfolio, }\theta_{B,t}~\text{and}~\theta_{S,t}~\text{are the exposures} \\ &\mathrm{d}X_t=\theta_{B,t}\mathrm{d}B_t+\theta_{S,t}\mathrm{d}S_t \\ &=\theta_{B,t}rB_t\mathrm{d}t+\theta_{S,t}(\mu S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t) \end{aligned}\]Then we consider:
\[\begin{aligned} f(t,X_t)&=e^{-rt}X_t\\ \Rightarrow\\ \mathrm{d}(e^{-rt}X_t)&=\frac{\partial f}{\partial t}\mathrm{d}t+\frac{\partial f}{\partial X_t}\mathrm{d}X_t\\ &=-re^{-rt}X_t\mathrm{d}t+e^{-rt}\mathrm{d}X_t\\ &=-re^{-rt}(\theta_{B,t}B_t+\theta_{S,t}S_t)\mathrm{d}t+e^{-rt}\mathrm{d}X_t\\ &=-re^{-rt}(\theta_{B,t}B_t+\theta_{S,t}S_t)\mathrm{d}t+e^{-rt}(\theta_{B,t}\mathrm{d}B_t+\theta_{S,t}\mathrm{d}S_t)\\ &=-re^{-rt}(\theta_{B,t}B_t+\theta_{S,t}S_t)\mathrm{d}t+e^{-rt}(\theta_{B,t}\mathrm{d}B_t+\theta_{S,t}(\mu S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t))\\ \text{Simplify:}\\ &=-re^{-rt}\theta_{S,t}S_t\mathrm{d}t+e^{-rt}\theta_{S,t}(\mu S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t)\\ &=(\mu -r)e^{-rt}\theta_{S,t}S_t\mathrm{d}t+e^{-rt}\theta_{S,t}\sigma S_t\mathrm{d}W_t \end{aligned}\]Now we combine (2) and (4), we could get the BSM PDE:
\[\begin{align*} &\text{Since}~C_t=X_t,~\text{we have: }d(e^{-rt}C_t)=d(e^{-rt}X_t)\\ &\begin{cases} e^{-rt}\left(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}\sigma^2S_t^2 + \frac{\partial C}{\partial S}\mu S_t-rC\right)\mathrm{d}_t = (\mu -r)e^{-rt}\theta_{S,t}S_t\mathrm{d}_t \\ e^{-rt}\frac{\partial C}{\partial S}\sigma S_t\mathrm{d}W_t = e^{-rt}\theta_{S,t}\sigma S_t\mathrm{d}W_t~\to~\frac{\partial C}{\partial S}=\theta_{S,t} \end{cases}\\ &\text{Simplify:}\\ &\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2\frac{\partial ^2C}{\partial S^2}S_t^2+rS_t\frac{\partial C}{\partial S}-rC=0 \end{align*}\]Convert the variable coefficients into constant coefficients:
\[\begin{aligned} &Let~V_t=ln \frac{S_t}{K}~\\ &\text{Then we have:}\\ &\frac{\partial V}{\partial t}=0,~\frac{\partial V}{\partial S}=\frac{1}{S_t},~\frac{\partial V}{\partial t}=-\frac{1}{S_t^2}\\ &dV_t=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}dS_t+\frac{1}{2}\frac{\partial ^2V}{\partial S^2}dS_t^2\\ &=\frac{1}{S_t}(\mu S_tdt+\sigma S_tdW_t)-\frac{1}{2}\frac{1}{S_t^2}dS_t^2\\ &=\mu dt+\sigma dW_t-\frac{1}{2}\frac{1}{S_t^2}(\sigma^2S_t^2dt)\\ &=(\mu -\frac{1}{2}\sigma^2)dt+\sigma dW_t\\ \\ &\text{For the previous dC(St, t), now change it to dC(Vt, t):}\\ &dC=\frac{\partial C}{\partial t}d_t + \frac{\partial C}{\partial V}dV+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}(dV)^2\\ &\to dC_t=\frac{\partial C}{\partial t}d_t + \frac{\partial C}{\partial V}((\mu -\frac{1}{2}\sigma^2)dt+\sigma dW_t)+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}(\sigma^2d_t) \\ &\to dC_t=(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}\sigma^2 + \frac{\partial C}{\partial V}(\mu-\frac{1}{2}\sigma^2) )d_t+\frac{\partial C}{\partial V}\sigma dW_t\\ \\ &\text{Consider}f(t,C_t)=e^{-rt}C_t \\ &d(e^{-rt}C_t)= df=-re^{-rt}C_tdt+e^{-rt}dC_t \\ &= -re^{-rt}C_tdt+e^{-rt}((\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}\sigma^2 + \frac{\partial C}{\partial V}(\mu-\frac{1}{2}\sigma^2) )d_t+\frac{\partial C}{\partial V}\sigma dW_t) \\ &=e^{-rt}(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}\sigma^2 + \frac{\partial C}{\partial V}(\mu-\frac{1}{2}\sigma^2) -rC)d_t+e^{-rt}\frac{\partial C}{\partial V}\sigma dW_t\\ \\ &\text{Now, combine the } d(e^{-rt}Xt)\text{ and new }d(e^{-rt}C_t):\\ &\begin{cases} &\frac{\partial C}{\partial V}\sigma dW_t = \theta_{S,t}\sigma S_tdW_t~\to~\frac{\partial C}{\partial V}=\theta_{S,t}S_t~\to~\theta_{S,t}=\frac{1}{S_t}\frac{\partial C}{\partial V}\\ &(\frac{\partial C}{\partial t}+\frac{1}{2}\frac{\partial^2 C}{\partial V^2}\sigma^2 + \frac{\partial C}{\partial V}(\mu-\frac{1}{2}\sigma^2)-rC )d_t = (\mu -r)\theta_{S,t}S_tdt=(\mu-r)\frac{\partial C}{\partial V}dt \\ &\end{cases}\\ &\text{Simplify:}\\ &\frac{\partial C}{\partial t}+\frac{1}{2}\sigma^2\frac{\partial ^2C}{\partial V^2}+(r-\frac{1}{2}\sigma^2)\frac{\partial C}{\partial V}-rC=0 \end{aligned}\]Now we have a new PDE with constant coefficients. No more St in it. Then we could solve this PDE with one-dimensional Poisson formula for heat conduction equation.
