title: “📖 Notes on MSFM - Option Theory”
date: 2019-12-04
tags: notes
mathjax: true

home_only: true

This is a study note on the fundamental theory of the pricing of a financial derivative, whose payoff is defined in terms of an underlying asset. We hereby try to compute a consistent price of the derivative in relative terms to the market price of the underlying asset.

Option Pricing Theory

We make our first assumption that the market is frictionless, by which we mean that:

no transaction cost (commission, bid-ask spread, taxes)
can hold negative asset (shortting) and there is no margin constraint
can hold fractional asset
no market impact from trading

Arbitrage (Static Portfolio)

We assume that the market lives in a probability space $\mathbb{P}$ and it includes $N$ tradable assets with non-random time- $0$ prices and random time- $T$ prices:

$\textbf{X}_t := (X^1_t, \dots, X^N_t)$

A static portfolio is a vector of quantities, where each $\theta$ is non-random and constant in time:

$\Theta := (\theta^1, \dots, \theta^N)$

Thus the time- $t$ value of the static portfolio $\Theta$ is;

$V_t := \Theta \cdot \textbf{X}_t$

A static portfolio $\Theta$ is an arbitrage if its value $V_t$ satisfies that:

$V_0 = 0 \text{, and both } \mathbb{P}[V_T \geq 0] = 1 \text{ and } \mathbb{P}[V_T > 0] > 0$

Suppose portfolio $\Theta^a$ super-replicates portfolio $\Theta^b$ , which means that $\mathbb{P}[V^a_T > V^b_T] = 1$ . Then $V^a_0 \geq V^b_0$ , otherwise arbitrage exists. Same goes if it is a sub-replication. Therefore, if $\Theta^a$ replicate $\Theta^b$ , which menas that $\mathbb{P}[V^a_T = V^b_T] = 1$ , then $V^a_0 = V^b_0$ . This is called the law of one price.

Assets

Discount Bond

A discount bond $Z$ pays $1$ at maturity $T$ . Given non-random interest rate $r_t$ , the no-arbitrage price of the discount bond is:

$Z_0 = 1/B_T = e^{-\int_0^Tr_tdt} \\ = e^{-rT} \text{ if r is constant}$

Forward Contract

A forward contract on $S_T$ with non-random delivery price $K$ obligates its holder to pay $K$ and receive $S_T$ at time $T$ . The time- $0$ value of the forward contract is $S_0 - KZ_0$ .

A forward price $F_0$ is delivery price such that the value of forward contract at time- $0$ is zero.

$F_0 = S_0/Z_0\\ = S_0e^{-rT} \text{ if r is constant}$

European Call Option

An European call option gives its holder the right at time $T$ to pay $K$ and receive $S_T$ . A call has payoff $(S_T-K)^+$ , and it is in the money if $S_t>K$ at time $t\leq T$ .

The time- $0$ price $C_0$ of a call option satisfies:

$(S_0-KZ_0)^+\leq C_0 \leq S_0$

For strike $K_1<K_2$ :

$0\leq C_0(K_1) - C_0(K_2) \leq (K_2 - K_1)Z_0 \\ \text{this can be proved by bull call spread}$

European Put Option

An European put option gives its holder the right at time $T$ to pay $S_T$ and receive $K$ . A put has payoff $(K-S_T)^+$ , and it is in the money if $S_t<K$ at time $t\leq T$ .

The time- $0$ price $P_0$ of a put option satisfies:

$(KZ_0 - S_0)^+\leq P_0 \leq KZ_0$

For strike $K_1<K_2$ :

$0\leq C_0(K_2) - C_0(K_1) \leq (K_2 - K_1)Z_0$

In addition,

$P_0(K_1)\leq \dfrac{K_1}{K_2}P_0(K_2) \leq P_0(K_2)$

Put-Call Parity

$C_0 - P_0 = S_0 - KZ_0$

Binomial Tree

We can create a replicating portfolio to calculate the value of a call option under a simple binomial tree:

$C_0 = \alpha + \beta S_0$

Where,

$\begin{align} \alpha e^{rT} + \beta s_u &= c_u\\ \alpha e^{rT} + \beta s_d &= c_d \end{align}$

And,

$\beta = \dfrac{c_u - c_d}{s_u - s_d} \\ \text{and, } \alpha=e^{-rT}(c_d - \beta s_d)$

Plugging in $\alpha$ and $\beta$ :

$C_0 = e^{-rT}(p_uc_u + p_dc_d) \\ \text{where } p_u := \dfrac{S_0e^{rT} - s_d}{s_u - s_d} \text{, and } p_d := \dfrac{s_u - s_0e^{rT}}{s_u - s_d}$

We can interpret $p_u$ and $p_d$ as probabilities that construct a risk-neutal measure $\mathbb{P}$ and that:

$C_0 = e^{-rT}\mathbb{E}C_T$

The Fundamental Theorem

The fundamental theorem of asset pricing states that:

no arbitrage

if and only if:

there exists a probability measure $\mathbb{P}$ equivalent to P such that the discounted prices of all tradable assets are martingales w.r.t. $\mathbb{P}$

The proof can be summarized as two ideas:

$\exists\ M.G. \mathbb{P} \rightarrow \text{no arb}$ :

a martingale is the cumulative P&L from betting on zero- $\mathbb{E}$ games, which is always zero no matter how you vary your bet size across games and time. you cannot riskless make something from nothing.
$\text{no arb} \rightarrow \exists\ M.G. \mathbb{P}$ :

the $\mathbb{P}$ probability of an event is simply the price of an asset that pays 1 unit of B iff that event happen

Risk-Neutral Measure

The physical probability is not accurate in evaluating a payoff’s true market price. Considering a 50/50 coin flip worth $1M$ or nothing. Using physical probability the price will be $500K$ .

However, the actual market price would be different. If the market is risk-adverse, the price would be lower, say $300K$ . We can view it as this market represents a risk-neutral measure where the down move has higher risk-neutral probabilities than up move.

We can see that the risk-neutral probability is price, that the risk-neutral probability of an event is the price of one-unit payout contingent on the event. Taking a risk-neutral expectation is the same as pricing by replication.

Radon-Nikodym derivative

In a discrete settings with outcomes $\{\omega_1\,\dots ,\omega_n\}$ , the relatioship between the risk-neutral measure and physical measure $P$ can be expressed by the Radon-Nikodym Derivative, or liklehood ratio:

$\mathbb{P}(\omega) = LR * P(\omega)$

The LR is typically larger in bad states than good states, reflecting the price margin on adverse events.

The Second Fundamental Theorem

A market is said to be complete if every random variable $Y_T$ can be replicated by a static portfolio $\Theta$ .

The second fundamental theorem of asset pricing states that:

a no arbitrage market is complete

if and only if:

there exists a unqiue measure $\mathbb{P}$ equivalent to P such that the discounted prices of all tradable assets are martingales w.r.t. $\mathbb{P}$

Trading Strategy

A filtration $\{\mathcal{F}_t\}$ represents all information revealed at or before time $t$ . A stochastic process $X$ is adapted to $\{\mathcal{F}_t\}$ if $X_t$ is $\mathcal{F}_t$ -measurable for each $t$ , meaning that the value of $X_t$ is determined by the information in $\mathcal{F}_t$ .

A trading strategy is a sequence of static strategy $\Theta_t$ adapted to $\mathcal{F}_t$ . A trading strategy is self-financing if for all $t>0$ :

$\Theta_{t-1}\cdot\textbf{X}_t = \Theta_t\cdot\textbf{X}_t \text{ with probability 1}$

This implies that the change in the portfolio value is fully attributable to gains and losses in asset prices:

$V_{t+1} - V_t = \Theta_t\cdot(\textbf{X}_{t+1} - \textbf{X}_t) \\ \text{or equivalently, } dV_t = \Theta_t\cdot d\textbf{X}_t$

Therefore,

$V_T - V_0 = \sum\Theta_t\cdot (\textbf{X}_{t+1} - \textbf{X_t}) \\ \text{or equivalently, } V_T - V_0 = \int_0^T\Theta_t\cdot d\textbf{X}_t$

We define that a trading strategy $\Theta$ replicates a time-T payoff $X_T$ if it is self-financing and the value $V_T=X_T$ . By the law of one price, at any time $t$ , the no-arbitrage price of an asset paying $X_T$ must have the same value of the replicating portfolio.

Arbitrage (Trading Strategy)

We now expand on the previous definition of arbitrage, that an arbitrage is a self-finance trading strategy $\Theta_t$ whose value $V_t$ satisfies:

$V_0 = 0 \text{, and both } \mathbb{P}[V_T \geq 0] = 1 \text{ and } \mathbb{P}[V_T > 0] > 0$

Ito integral

Let $\mu_t$ and $\sigma_t$ satisfies integrability, adapted to $\mathcal{F}_t^W$ , and be continuous on $t$ . We define the Reimann integrals:

$\int_0^T \mu_tdt := \lim_{N\rightarrow\infty} \Sigma_{n=0}^{N-1}\mu_{t_n}\Delta t$

And define the Ito integral of $\sigma$ with respect to $W$ by:

$\int_0^T \sigma_tdW_t := \lim_{N\rightarrow\infty} \Sigma_{n=0}^{N-1}\sigma_{t_n}\Delta W_{t_n}$

Note: $W$ flutuates so much that the limit does not necessarily exist in pathwise sense - it does exist in $L^2$ and probability.

Ito integrals are martingales. $\mathbb{E}$ of any Ito integral is zero.

Ito Process

We define an Ito process to be a stochastic process $X$ of the form of the sum of an initial value, a Riemann integral and an Ito integral:

$X_t = X_0+\int_0^t\mu_sds + \int_0^t\sigma_sdW_s \\ \text{equivalently, } dX_t = \mu_tdt + \sigma_tdW_t$

Note that $X_t$ is continuous in $t$ (b.c. $W$ is), is adapted to $\mathcal{F}_t^W$ , and is a martingale iff $\mu_t=0$ for all $t>0$ , with probability $1$ .

Ito’s Rule

Given an Ito process $X_t$ with $dX_t=\mu_tdt+\sigma_tdW_t$ , and a sufficiently smooth, real-value function $f(X_t)$ :, then $f(X_t, t)$ is an Ito process with:

$df(X_t) = \dfrac{\partial f}{\partial x}dX_t + \dfrac{1}{2}\dfrac{\partial^2f}{\partial x^2}(dX_t)^2$

With two processes $X_t$ and $Y_t$ , and $f(X_t, Y_t)$ :

$df(X_t, Y_t) = \dfrac{\partial f}{\partial x}dX_t + \dfrac{\partial f}{\partial y}dY_t + \dfrac{1}{2}\dfrac{\partial^2f}{\partial x^2}(dX_t)^2 + \dfrac{1}{2}\dfrac{\partial^2f}{\partial y^2}(dY_t)^2 + \dfrac{\partial^2f}{\partial x\partial y}(dX_t)(dY_t)$

In a special case where $Y_t = t$ , the formula becomes:

$df(X_t, t) = \dfrac{\partial f}{\partial x}dX_t + \dfrac{\partial f}{\partial t}dt + \dfrac{1}{2}\dfrac{\partial^2f}{\partial x^2}(dX_t)^2$

Note that the Ito’s Rule applies under any probability measure, it is purely math.

Black-Scholes Model

Assumptions Consider two basic assets $B_t$ and $S_t$ in continuous time, where:

$dB_t = rB_tdt \text{ , }\\ \text{ where }B_0=1$

And $S_t$ follows GBM dynamics,

$dS_t = \mu S_tdt + \sigma S_tdW_t \\ \text{ where }S_0>0 \text{ , }\sigma>0\text{ ,and W is BM under physical measure}$

Think of volatility $\sigma$ as $\sqrt{\text{variance of log-returns per unit time}}$

Conclusion Then by no-arbitrage and Ito's rule, the time- $t$ price $C_t$ of a call option with payoff $(S_T-K)^{+}$ satisfies the Black-Scholes PDE for $(S, t)\in [0, \infty]\times (0, T)$

$\dfrac{\partial C}{\partial t} + rS\dfrac{\partial C}{\partial S} + \dfrac{1}{2}\sigma^2S^2\dfrac{\partial^2C}{\partial S^2} = rC \\ \text{ with terminal condition: } C(S, T)=(S-K)^{+}$

We can solve the call price analytically with the Black-Scholes formula:

$C^{BS}(S_t, t) := S_tN(d_1) - Ke^{-r(T-t)}N(d_2) \\ \text{where } d_{1,\ 2} := \dfrac{log(S_t/K) + (r \pm \sigma^2/2)(T-t))}{\sigma\sqrt(T-t)}$

Here we plotted the BS call price $C^{BS}$ , the and the against the current underlying price $S_t$ , with paramters $K=100$ , $T-t=1$ , $\sigma=0.2$ and $r=0.05$

Girsanov’s Theorem

Theorem If $W$ is a Brownian motion under P, and if $\mathbb{P}$ is a probability measure on $\mathcal{F}_T^W$ that is equivalent to P, then there exists an adapted process $\lambda$ such that for all $t \in [0,T]$ :

$\bar{W_t} := W_t + \int_0^t\lambda_sds \text{ is a Brownian motion under }\mathbb{P}$

In other words, $X_t$ plus drift under $\mathbb{P}$ has the same distribution as $X_t$ under P

$dW_t = \lambda_tdt + d\bar{W_t}$

Therefore given:

$dB_t = rB_tdt \\ dS_t = \mu S_tdt + \sigma S_tdW_t$

No arbitrage implies that $\exists$ $\mathbb{P}$ such that $S/B$ is a $\mathbb{P}$ -MG. By Girsanov, $\exists$ $\lambda$ such that $\bar{W_t}:=W_t+\int_0^t\lambda_sds$ is $\mathbb{P}$ -BM. Therefore:

$d_s = (\mu - \lambda_t\sigma)S_tdt + \sigma S_td\bar{W_t}$

Also because $S/B$ has not drift, $d(S_t/B_t)=d(e^{-rt}S_t)=e^{-rt}(dS_t-rS_tdt)$ has no drift term, so the drift of $dS_t$ under $\mathbb{P}$ -BM must be $rS_tdt$ :

$dS_t = rS_tdt + \sigma S_td\bar{W_t}$

Under $\mathbb{P}$ :

$logS_T \sim \mathcal{N}(logS_t + (r- \sigma^2/2)(T-t), \sigma^2(T-t))$

Computing Call Price

By fundamental theorem, the call price $C_t$ satisfies MG: $C_t/B_t=\mathbb{E}_t(C_T/B_T)$ . Therefore,

$\begin{align} C_t &= \mathbb{E}[e^{-r(T-t)}C_T] \\ &= \mathbb{E}_t[e^{-r(T-t)}(S_T-K)^{+}] \\ &= \mathbb{E}_t[e^{-r(T-t)}S_T \textbf{1}_{S_T>K}] + \mathbb{E}_t[e^{-r(T-t)}K \textbf{1}_{S_T>K}] \\ &= S_t\mathbb{E}_t[\dfrac{S_Te^{rt}}{S_te^{rT}}\textbf{1}_{S_T>K}] - Ke^{-r(T-t)}\mathbb{P}_t(S_T>K) \end{align}$

Since $\mathbb{P}_t(S_T>K)$ is equivalent to $\mathbb{P}_t(log(S_T) > logK)$ , and under $\mathbb{P}$ , $log(S_T)\sim\mathcal{N}(logS_t+(r-\sigma^2/2)(T-t), \sigma(T-t))$ :

$\begin{align} \mathbb{P}_t(S_T>K) &= \mathbb{P}_t(logS_T > logK) \\ &= \mathbb{P}_t(\dfrac{logS_T - (logS_t+(r-\sigma^2/2)(T-t)}{\sigma\sqrt(T-t)} > \dfrac{logK - (logS_t + r-\sigma^2/2)(T-t)}{\sigma\sqrt(T-t)}) \\ &= N(-\dfrac{log(K/S_t) - (r-\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}) \\ &= N(d_2) \end{align}$

To calculate the first term, let $F_t=S_te^{r(T-t)}$ :

$\begin{align} \mathbb{E}_t[\dfrac{S_Te^{rt}}{S_te^{rT}}\textbf{1}_{S_T>K}] &= \mathbb{E}_t[e^{log(S_T/F_t)}\textbf{1}_{log(S_T/F_t)>log(K/F_t}] \\ &= \int_{log(K/F_t)}^{\infty} e^y\frac{1}{\sqrt{2\pi \sigma^2(T-t)}} e^{[-\dfrac{(y+(\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}]} dy \\ &= \int_{log(K/F_t)}^{\infty} \frac{1}{\sqrt{2\pi \sigma^2(T-t)}} e^{[-\dfrac{(y-(\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}]} dy \\ &= \int_{\dfrac{log(K/F_t)-(\sigma^2/2)(T-t)}{\sigma\sqrt(T-t)}}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{(-z^2/2)} dz \\ &= N(d_1) \end{align}$

Note that as $z=\dfrac{y-(\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}$ , therefore $dz=\dfrac{dy}{\sigma\sqrt{T-t}}$

Numeraire

Definition A numeraire is any tradeable asset whose price process is always positive.

Theorem For any numeraire $N$ , No arbitrage iff $\exists$ $\mathbb{P}^N$ , equivalent to P, under which all tradeable assets divided by $N_t$ are martingales.

Definition The martingale measure w.r.t. $S$ is called share measure, under which:

$\dfrac{C_t}{S_t} = \mathbb{E}_t^S\dfrac{C_T}{S_T}$

Under $\mathbb{P}^S$ , we derive similarly by Girsanov theorem that:

$dS_t = (r+\sigma^2)S_tdt + \sigma S_tdW_t^S$

And so:

$logS_T\sim \mathcal{N}^S(logS_t + (r+\sigma^2/2)(T-t), \sigma^2T)$

we can see that:

$\begin{align} \mathbb{E}^S\textbf{1}_{S_T>K} &= \mathbb{P}^S(logS_T>logK) \\ &= N(-\dfrac{logK - (logS_t + (r+\sigma^2/2)(T-t))}{\sigma^2\sqrt{T-t}}) \\ &= N(d1) \end{align}$

To price a vanilla call, we can then use both share measure and forward measure:

$\begin{align} C_T &= (S_T - K)^{+} = S_T\textbf{1}_{S_T>K} - K\textbf{1}_{S_T>K} \\ C_t &= S_t\mathbb{P}_t^S(S_T>K) - KZ_0\mathbb{P}_t^Z(S_t>K) \\ &= S_0N(d_1) - KZ_0N(d_2) \end{align}$

In the B.S. case, $N(d_2)$ is the probability under risk-neutral/forward measure that the option expires ITM. The $N(d_1)$ is the probability under share measure that the option expires ITM.

The Greeks

Delta

Suppose an asset has a time t value $V_t(S_t, t)$ , then its Delta at time $t$ is $\partial V_t(S_t, t)/\partial S_t$ . Delta can be interpreted as:

the slope of the asset value $V_t$ , plotted as a function of S_t.
how much the asset value movies per unit move in $S_t$
humber of $S_t$ needed to replicate this asset.

If the asset is a call option on $S_t$ and we assumes the Black-Scholes assumptions on $S_t$ , then:

$\begin{align} \text{Delta} &:= \dfrac{\partial C^{BS}}{\partial S_t} \\ &=N(d_1) + S_tN'(d_1)\dfrac{\partial d_1}{\partial S_t} - Ke^{-(T-t)}N'(d_2)\dfrac{\partial d_2}{\partial S_t} \\ &=N(d_1) \end{align}$

The Delta of a call option is strictly between 0 and 1. As the time-to-maturity decreases, the Delta increases faster the the option becomes more ITM. Here we plotted the BS Delta for $T-t$ equals $1$ and against the current underlying price $S_t$ .

Gamma

For a call option in a B-S model,

$\text{Gamma} = \dfrac{\partial^2 C^{BS}}{\partial S_t^2} = N'(d_1)\dfrac{1}{S_t\sigma\sqrt{T-t}}$

In this case, the Gamma can be interpreted as:

the convextity of $C^{BS}$ w.r.t. S_t
how much the Delta moves, per unit move in $S_t$
how much rebalancing of the replicating portfolio is needed, per unit move in $S_t$

The Gamma of a call option is strictly positive. As the time-to-maturity decreases, the Gamma increases for ATM options. Here we plotted the BS Delta for $T-t$ equals $1$ and against the current underlying price $S_t$ .

Theta

For a call in B-S model,

$\text{Theta} = \dfrac{\partial C^{BS}}{\partial t}$

The Theta of a call option is strictly negative. As the time-to-maturity decreases, the Theta decreases for ATM options (faster time-decay). Here we plotted the BS Theta for $T-t$ equals $1$ and against the current underlying price $S_t$ .

Discrete Delta Hedge and Gamma Scalping

A discretely Delta-hedged portfolio could buy $C$ and short $\text{Delta} \cdot S$ . In this case it is a Delta neutral and long Gamma/Gamma scalping portfolio:

Delta of the portfolio is $0$
Gamma of the portfolio is positive
achieve net profit only if the realized volatility of $S$ is high enough to overcome time decay, otherwise portfolio loss happens. This is the opposite from a short Gamma position, e.g. sell $C$ and long Delta $S$

We can visualize the P&L of a long Gamma portfolio in the following graph, where the area indicate profits and the area indicate losses. The curved line is $C_{t+\Delta t}$ the straight line is $\text{Delta}_t \cdot S_{t+\Delta t}$ . As $\Delta t$ increases, $C_{t+\Delta t}$ shifts downwards due to time-decay.

In addition, we can show that the P&L of such portfolio $dV = dC - C_S dS$ does not depend on the drift $\mu$ of the stock:

$\begin{align} dV &= dC - C_S dS \\ &= C_tdt + C_SdS^{\sigma_{implied}} + C_{SS}dS^2 - C_SdS^{\sigma_{realized}} \\ &= (C_t + C_{SS}\sigma_{implied}^2S^2/2)dt + C_S(\sigma_{implied} - \sigma_{realized})SdW \end{align}$

Continue on L5

Numerical Methods

The Taylor series of a real or complex value function $f(x)$ that is differentiable at $a$ is:

$f(x) = f(a) + \dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(a)}{2!}(x-a)^2 ...$

Implied Volatility

Given the time- $t$ price of a European call option on a non-dividend stock $S$ , the time- $t$ Black Scholes implied volatility $\sigma(t)$ is the unique solution to $C_t = C^{BS}(\sigma(t))$ .

Uniqueness is because $C^{BS}$ is strictly increasing in $\sigma$ and Existence is because $C^{BS}$ covers the full range of arbitrage-free prices of the European option $[S_0-Ke^{-rT}, S_0]$

If $S_t$ follows the SDE dynamic $dS_t = rS_tdt + \sigma(t)S_tdW_t$ , where $\sigma(t)$ a non-random function of $t$ , then we can first find the implied volatility $\bar{\sigma}_T$ given call prices with different maturity $T$ , and use the equation below to find (not uniquely) the true function $\sigma(t)$ :

$\begin{align} logS_T &= logS_0 + (r-\bar{\sigma}^2_T/2)T + \int_0^T \sigma(t)dW_t \\ &\sim \mathcal{N}(logS_0 + (r-\bar{\sigma}^2_T/2)T, \;\bar{\sigma}^2_TT) \\ &\text{where } \bar{\sigma}_T := \sqrt{\dfrac{1}{T} \int_0^T \sigma^2(t)dt} \end{align}$

Volatility Smile, Skew and Surface

If $S_t$ truely follows GBM with constant volatility $\sigma$ , then $\sigma_{imp}(K, T) = \sigma, \;\forall\;K, T$ . However, empirically the $\sigma_{imp}$ is lower when $K \approx S_t$ (volatility smile), possibly because

the market price option using a risk-neutral distribution of log-returns with fatter tails than Normal

Note that $\sigma_{imp}$ is also higher when $K < S_t$ (volatility skew), possibly due to:

instantaneous volatility increases as price decreases
possibility of severe crash fuels demand for downside protection

In addition, the $\sigma_{imp}$ has a term structure and varies for different $T$ . The function $\sigma_{imp}(K, T)$ is call the implied volatility surface

Tree Model

Binomial Tree

European Option

Given option price at the $j$ -th node $C_T^j = f(S_T^j)$ , we can induct backward to find $C_t^j$ :

$C_t^j = e^{-r\tau}[p_n^jC_{t+1}^{j+1} + (1-p_n^j)C_{t+1}^{j-1}] \\ \text{where } p_n^j = \dfrac{S_t^je^{r\tau} - S_{t+1}^{j-1}\;}{S_{t+1}^{j+1} - S_{t+1}^{j-1}\;}$

American Option - Put

Given option price at the $j$ -th node $C_T^j = (K - S_T^j)^+$ , we can induct backward to find $C_t^j$ :

$C_t^j = max[(K - S_t^j)^+,\;e^{-r\tau}[p_n^jC_{t+1}^{j+1} + (1-p_n^j)C_{t+1}^{j-1}]] \\ \text{where } p_n^j = \dfrac{S_t^je^{r\tau} - S_{t+1}^{j-1}\;}{S_{t+1}^{j+1} - S_{t+1}^{j-1}\;}$

American Option - Call

Given option price at the $j$ -th node $C_T^j = (S_T^j - K)^+$ . If $r>0$ and stock dividend $\delta=0$ , then it is never optimal to exercise early on an American call option. Therefore $C^{American} = C^{European}$

Argument 1 At all $t>0$ , the American call is worth more than the exercise payoff $S_t - K$ :

$\text{American Call} \geq \text{European Call} \geq S_t - KZ_t > S_t - K$

Argument 2 If $C^{American} > C^{European}$ then construct portfolio $V = [-C^{American}, C^{European}]$ . Then V is an arbitrage as $V_0 > 0$ and $V_T \geq 0$ .

Trinomial Tree

Let $\Delta t:= T/N$ and choose $\Delta x \approx \sigma\sqrt{3\Delta t}$ to improve accruacy.

Finite Difference Model

Explicit Scheme

Inducting backward from $t=T$ to $0$ :

$\begin{align} \dfrac{\partial C}{\partial t} &\approx \dfrac{C^j_{t+1} - C^j_t}{\Delta t} \\ \dfrac{\partial C}{\partial x} &\approx \dfrac{C^{j+1}_{t+1} - C^{j-1}_{t+1}\;}{2\Delta x} \\ \dfrac{\partial^2 C}{\partial x^2} &\approx \dfrac{1}{\Delta x} (\dfrac{C^{j+1}_{t+1} - C^{j}_{t+1}\;}{\Delta x} - \dfrac{C^{j}_{t+1} - C^{j-1}_{t+1}\;}{\Delta x}) = \dfrac{C^{j+1}_{t+1} -2C^{j}_{t+1} + C^{j-1}_{t+1}\;}{(\Delta x)^2} \end{align}$

Solving for the B-S PDE: $rC = C_t + vC_S + 0.5\sigma^2C_{SS}$ where $v = (r-\sigma^2/2)$ , we get:

$C^j_t = \dfrac{1}{1 + r\Delta t}(q_uC^{j+1}_{t+1} + q_mC^{j}_{t+1} + q_dC^{j-1}_{t+1})$

Where:

$\begin{align} q_u &= \dfrac{1}{2}[\dfrac{\sigma^2\Delta t}{(\Delta x)^2} + \dfrac{v\Delta t}{\Delta x}] &=p_u - \dfrac{v^2(\Delta t)^2}{2(\Delta x)^2}\\ q_m &= 1 - \dfrac{\sigma^2\Delta t}{(\Delta x)^2} &=p_m + \dfrac{v^2(\Delta t)^2}{(\Delta x)^2}\\ q_d &= \dfrac{1}{2}[\dfrac{\sigma^2\Delta t}{(\Delta x)^2} - \dfrac{v\Delta t}{\Delta x}] &=p_d - \dfrac{v^2(\Delta t)^2}{2(\Delta x)^2}\\ \end{align}$

Note that $p_u, p_m, p_d$ are trinomial tree probabilities.

Implicit Scheme

Inducting backward from $t=T$ to $0$ :

$(-\alpha C^{j+1}_{t} + (1+2\alpha)C^{j}_{t} + \alpha C^{j-1}_{t}) = C^j_{t+1}$

Solving the $LHS$ requires solutions of a system of $2J-1$ equation with $2J-1$ unknowns.

Crank-Nicolson Scheme

Inducting backward from $t=T$ to $0$ :

$-F^j_tC^{j+1}_t + (1+G^j_t)C^j_t - H^j_tC^{j-1}_t = F^j_{t+1}C^{j+1}_{t+1} + (1-G^j_{t+1})C^j_{t+1} + H^j_{t+1}C^{j-1}_{t+1}$

If given terminal conditions, then we know $C_{t+1}$ ‘s and can solve for $C_t$ .

Monte Carlo Model

Given $Y$ be a discounted payoff and the time- $0$ price of the payoff $C=\mathbb{E}Y$ . The Monte Carlo estimator $\hat{C}_M$ of $C$ :

$\hat{C}_M = \dfrac{1}{M}\sum Y \\ \text{where } \mathbb{E}\hat{C}_M = C \text{, and } Var(\hat{C}_M) = Var(Y)/M$

By the strong law of large numbers, the sample average $\hat{C}_M$ converges almost surely to the expected value $C$ as $M \rightarrow \infty$ . By the central limit theorem:

$\dfrac{\hat{C}_M - C}{Var(Y)/M} \rightarrow \mathcal{N}(0, 1) \text{, as } M \rightarrow \infty$

Often times we need to estimate $\sigma$ with sample estimator for the variance of $Y$ :

$\hat\sigma^2_M := \dfrac{\sum (Y - \hat{C}_M)^2}{M-1} \\ \text{so then } \dfrac{\hat{C}_M - C}{\hat{\sigma}_M^2/M} \rightarrow \mathcal{N}(0, 1) \text{, as } M \rightarrow \infty$

The standard error $SE = \hat{\sigma}_M/\sqrt{M}$ , and a $95\%$ confident interval for $C$ is $\hat{C}_M \pm 1.96SE$

Variance Reduction Techniques

Antithetic Variate

Let $\tilde{Y}:=Y_{Z = -z\;}$ . The antithetic variate estimoator $\hat{C}^{av}_M$ :

$\hat{C}^{av}_M = \dfrac{1}{M}\sum \dfrac{Y + \tilde{Y}\;}{2} \\ \text{where } Var(\hat{C}^{av}_M) = \dfrac{Var(Y) + Cov(Y, \tilde{Y})}{M}$

Control Variate

A control variate $Y^\ast$ is a random variable, correlated to $Y$ such that $C^\ast := \mathbb{E}Y^\ast$ has an explicit formula.

Example Let $Y$ be the discounted payoff on a call on $S_t$ where $dS_t = \sigma(t)S_tdW_t$ . We can choose $Y^\ast$ to be the discounted payoff on a call on $S_t^{\ast}$ where $dS_t^{\ast} = \sigma S^{\ast}_tdW_t$ , in which case $C^{\ast}$ can be calculated explicitely through B-S formula given constant $\sigma$ close to $\sigma(t)$ .

The control variate estimator $\hat{C}^{cv, \beta}_M$ estimates $C$ by simulating $Y - \beta Y^{\ast}$ .

$Y^{cv, \beta}_M = \beta C^{\ast} + (Y - \beta dsY^{\ast}) \\ \hat{C}^{cv, \beta}_M = \mathbb{E}Y^{cv, \beta}_M = \beta C^{\ast} + \dfrac{1}{M}\sum (Y - \beta Y^{\ast}) \\ \text{and }Var(\hat{C}^{cv}_M) = \dfrac{1}{M}[Var(Y) - 2\beta Cov(Y, Y^{\ast}) + \beta^2Var(Y^{\ast})]$

Choose $\beta$ to minimize $Var(\hat{C}^{cv}_M)$ , we get:

$\beta^{\ast} = \dfrac{Cov(Y, Y^{\ast})}{Var(Y^{\ast})} \rightarrow \hat{\beta^{\ast}\;} = \dfrac{\sum(Y - \bar{Y})(Y^{\ast} - \bar{Y}^{\ast})}{\sum(Y^{\ast} - \bar{Y}^{\ast})^2} \\ \text{so then } Var(\hat{C}^{cv, \beta^{\ast}\;}_M) = Var(\hat{C}_M)[1 - Corr^2(Y, Y^{\ast})]$

Note that when using sample estimate $\hat{\beta^{\ast}\;}$ , the estimated $\hat{C}^{cv, \hat{\beta}^{\ast}\;}_M$ is biased, only when $M$ is small.

Importance Sampling

Suppose $X$ are IID draws from density $f$ , and $C := \mathbb{E}h(X)$ . Ordinary Monte Carlo estimator provides:

$\hat{C}_M = \dfrac{1}{M}\sum h(X)$

With importance sampling, find $g$ s.t. $g(x) > 0$ iff $f(x)g(x) \neq 0$ . Then re-draw $X$ from density $g$ and the importance sampling estimator $\hat{C}^{is}_M$ is:

$\hat{C}^{is}_M = \dfrac{1}{M}\sum h(X)\dfrac{f(X)}{g(X)} \\ \text{where } Var(\hat{C}^{is}_M) = \dfrac{1}{M}Var(h(X)\dfrac{f(X)}{g(X)})$

Conditional Monte Carlo

Given a random variable $X$ :

$C = \mathbb{E}Y = \mathbb{E}[\mathbb{E}(Y|X)] = f(X)$

The condintional Monte Carlo estimator:

$\hat{C}^{cmc}_M = \dfrac{1}{M}\sum f(X) \\ \text{where } Var(\hat{C}^{cmc}_M) = \dfrac{1}{M}Varf(X) < Var(\hat{C}_M) \\ \text{because } Var(Y) = Var[\mathbb{E}(Y|X)] + \mathbb{E}Var(Y|X)$

Fourier Transform Model

Given $f:\mathbb{R}\rightarrow\mathbb{R}$ be integrable, meaning $\int|f(x)|dx < \infty$ . The Fourier transform of $f$ is the function $\hat{f}: \mathbb{R}\rightarrow\mathbb{C}$ defined by:

$\hat{f}(z) = \int_{-\infty}^{\infty} \hat{f}(x)e^{izx}dx$

Theorem If $\hat{f}$ is also integrable, then the inversion formula holds:

$f(x) = \dfrac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(z)e^{-izx}dz$

Characteristic Function

The complex conjugate of a complex number $z = x + yi$ is given by $\bar{z} = x - yi$ . so $\text{Re}(z) = \text{Re}(\bar{z})$ .

The characteristic function of any random variable $X$ is the function $F_X(z)$ defined by:

$F_X(z) := \mathbb{E}e^{izX}$

Therefore if $X$ has density $f$ , then $F_X(z) = \hat{f}(z)$ . A characteristic function uniquely identifies a distribution. For example, $F_X(z) = e^{-z^2/2}$ , if $X\sim\mathcal{N}(0,1)$

To calculate the moments of $X$ using CF, take the $n$ -derivatives of $F_X(z)$ w.r.t. $z$ :

$\mathbb{E}X^n = (-i)^nF_X^{(n)}(0)$

To calculate the CDF of $X$ using CF:

$\mathbb{P}[X < k] = 0.5 - \dfrac{1}{\pi}\int_0^{\infty} \text{Re}[\dfrac{F_X(z)}{iz}e^{-izk}]\;dz$

To calculate asset-or-nothing call price using CF, given $e^X$ be the asset share price, define the share measure $\mathbb{P}^{\ast}$ with likelihood ratio $e^X/\mathbb{E}e^X$ .

$F^{\ast}_X(z) = \mathbb{E}^{\ast}e^{izX} = F_X(z - i)/F_X(-i)$

Therefore for any $k\in\mathbb{R}$ , the asset-or-nothing call price:

$\begin{align} e^{-rT}\mathbb{E}e^X\textbf{1}_{X>k} &= e^{-rT}\mathbb{E}e^X\mathbb{P}^{\ast}(X>K) \\ &= e^{-rT}[\dfrac{F_X(-i)}{2} + \dfrac{1}{\pi}\int_0^{\infty} \text{Re}[\dfrac{F_X(z - i)}{iz}e^{-izk}]\;dz] \end{align}$

To calculate a vanilla European call price on $e^X$ struck at $K$ with $k := log\;K$ :

$C_0 = e^{-rT}[\mathbb{E}e^X\textbf{1}_{X>k} - K\mathbb{P}[X > k]]$

Heston Model

Provided that:

$\begin{align} dS_t &= rS_tdt + \sqrt{V_t}S_tdW_t^S \text{ , and let} X = logS_t\\ \text{so that } dX &= (r - 0.5V_t)dt + \sqrt{V_t}dW_t^S \\ dV_t &= \kappa(\theta - V_t)dt + \eta\sqrt{V_t}dW_t^V \end{align}$

Where $W^S$ and $W^V$ are $\mathbb{P}$ BM with correlation $\rho$ , $\kappa$ is the rate of mean-reversion, $\theta$ is the long-term mean, and $\eta$ is the volatility of volatility.

We want to find the CF of $X$ in order to price options on $S_T$ . The time- $t$ conditional Heston CF provides an answer:

$F_X(z) = e^{A + izX_t + BV_t}$