Introduction

We want to buy oil in 6 months time. We can buy it now for $30/barrel or in 6 months time for eg. $36 with probability p, or $18 with probably 1-p. (Assuming 0 interest rates, storage costs etc.)

A cap contract compensates the holder for any increase in price, eg. it would pay off $6 if the price rose to $36, or $0 if the price fell. This reduces risk – the pain of the worst case scenario is lessened. Is an example of a financial derivative, or contigent claim. Where do caps come from? – insurance companies (risk transfer) – oil company (reduces risk for both parties).

How do we value the cap? One (incorrect) way would be to take the expected value = $6p + 0(1-p) = 6p$ .

Replicating strategies

Is it possible to reproduce this contract with some strategy of buying and selling existing assets? How about: buy 1/3 barrel now for $10 (using $6 borrowed money) then sell the oil in 6 months time – has the same payoffs as the cap, so the cap should cost the same as implementing this strategy, $4.

More generally, if a contract payoff can be replicated by a trading strategy with existing assets, then the contract has the same cost as the strategy.

Hidden assumption: no arbitrage – no strategy can have zero cost and non-negative payoffs, unless the payoffs are all 0. Why should be believe arbitrage is impossible? – exploiting arbitrage opportunities tends to diminish them.

More general single period models

A strategy is a vector $\phi$ of length $d+1$ giving the amount of each asset to hold.

Assumption: If prices are consistent then no arbitrage strategy should exist.

Theorem: No arbitrage strategies $\iff$ there exists a vector $q$ with $S q = (1+r)s_0$ , all $q \gt 0, \sum q_i = 1$ . $q$ is called an equivalent martingale measure. Can be rewritten as $E_q ( \frac{S_i(1)}{1+r}) = S_1(0)$ , ie. according to the probabilities $q_1, ..., q_N$ every asset is a fair bet.

Proof: $\Rightarrow$ suppose $q$ exists, then for any strategy $\phi$ we have $\phi^T Sq = (1+r)\phi^T s_0$ . If $\phi^T S \ge 0$ then $\phi S q \ge 0$ , unless $\phi^T S = 0$ , hence $\phi^t s_0 \gt 0$ .

How do you value a new asset?

Suppose the payoff the new asset is a ramdon variable $X_1$ with possible value scenarios $\omega_1, \omega_2, ..., \omega_N$ given by a vector $x$ of length $N$ . Find a replicating strategy $\phi$ , ie. $\phi^T S = x^T$ . Then the value of the new asset is $\phi^T s_0$ .

Possible to calculate the value without explicitly finding $\phi$ :

\[\phi^T s_0 = \phi^T \frac{Sq}{1+r} = x^T \frac{q}{1+r} = E_q [ \frac{X}{1+r}] \]

ie. take the expected discounted value with q-probabilities, called risk neutral valuation.

But what if $q$ isn’t unique?

Theorem: A vector $x$ can be replicated $\iff x^T q$ has the same value for all EMMs $q$ . In particular, if there is a unique EMM then all payoff vectors can be replicated. Markets with a unique EMM are called complete.

A model with few assets and many scenarios will be incomplete – a problem for realistic models. What to do? Find a better model!

An introduction to probability theory

Sample space = set of possible outcomes $\Omega$

Probability:

$\sigma$ -fields

A non-empty set $\mathcal{F}$ of subsets of $\Omega$ closed under complementation and countable unions. For example, with $\Omega = \{1,2,3\}$ .

Purpose

Represent partial information about the outcome of a random experiment (the events in $\mathcal{F}$ that can be determined with the available information). eg. $\mathcal{F}_0$ represents knowing nothing, $\mathcal{F}_1$ represents knowing if the outcome was 1.

$\sigma$ -fields are useful for experiments with a time structure, where information is revealed in a series of stages. $\sigma$ -fields also have another use: in large sample spaces (eg. $[0,1]$ ) trying to assign a probability to every $\omega \in \Omega$ leads to paradoxes. To avoid this, the full $\sigma$ field is taken to be something smaller than all subsets. For example, with $\mathbb{R}$ , we use the Borel $\sigma$ -field, which contains all intervals, open sets and others. The can be used to represent full info for $\mathbb{R}$ -valued experiments.

A filtration, $\mathcal{F}_0 \subset \mathcal{F}_1 \subset ...$ , is an increasing sequence of $\sigma$ -fields (information revealed progressively)

$\sigma$ -fields for finite $\Omega$

A $\sigma$ -field $\mathcal{F}$ will consist of unions of sets from some partition $B_1, ..., B_k$ of $\Omega$ . Trivial $\sigma$ -field generated by $\{\Omega\}$ , full $\sigma$ -field by $\{ \{\omega_1\}, ..., \{\omega_N\} \}$ .

Random variables

A random variable is numerical measure of the outcome of an experiment, $X: \Omega \to \mathbb{R}$ . Can be used to describe events and $\sigma$ -fields. $X \in \mathcal{F}$ , means $\{ X \in B \} \in \mathcal{F}$ , for any Borel set $B$ – the value of $X$ is determined by the info represented by $\mathcal{F}$ .

NOTE: if $X \in \mathcal{F}$ then $f(X) \in \mathcal{F}$ .

If $X$ is a random variable then $\sigma (X)$ is the smallest $\sigma$ -field $\mathcal{F}$ with $X \in \mathcal{F}$ .

Distributions

A random variable $X$ has a distribution, the probabilities $P(X) \in B$ , for all Borel sets $B$ . This can be thought of as a probability measure on the reals.

Independence

Equivalent probability measures

$P, Q$ are probabilities measures of a sample space $\Omega$ . If $P ~ Q$ (P equivalent to Q) then $P$ and $Q$ have the same null sets. $P(A) = 0 \iff Q(A) = 0$ .

It can be shown that equivalent probability measures can be used to calculate each other’s expectations:

If $P ~ Q$ there is a random variable, T, $T: \Omega \to \mathbb{R}$ , s.t. $E_q [X] = E_p [XT]$ and $E_p [X] = E_q [X/T]$ , for all random variables X. $T$ is written $\frac{dQ}{dp}$ and called the Radon-Nikodym derivative.

Example: Let $T = exp(\frac{1}{2\sigma^2} (2 \mu X - \mu^2))$ and define $Q$ by $Q(A) = E[T 1_A]$ , then $Q$ is an equivalent probablity measure under which $X ~ N(\mu, \sigma^2)$ .

More generally, if $X$ has density $f$ under $P$ , let $T = g(X) / f(X)$ and define $Q$ by $Q(A) = E[T 1_A]$ . Then $Q$ is an equivalent probability measure under which $X$ has density $g$ . This works provided $g$ is a density, with the same support as $f$ (eg. $f(x) = 0 \iff g(x) = 0$ ). Similarly for collections of random variables.

Conditional expectations

For a $\sigma$ -field $\mathcal{F}$ and rv $X$ , $E[{X | \mathcal{F}}]$ is the unique random variable $Y \in \mathcal{F}$ with $E[Y 1_B] = E[X 1_B]$ for all events $B \in \mathcal{F}$ .

In the finite case, $\mathcal{F}$ is given by a partition $B_1, B_2, ..., B_m$ of $\sigma$ . $E[X | \mathcal{F}]$ is consistent over each $B_j$ , and has the value $\sum X(\omega) P(\{\omega\}) / P(B_j)$ .

For random variables, $X, Z$ , $E[X | Z] = E[X | \sigma (Z)]$ . Similarly $E[X | Z_1, Z_2, ...] = E[X | \sigma (Z_1, Z_2, ...)]$ . $P(A | \mathcal{F}) = E[ 1_A | \mathcal{F}$ .

More facts about conditional expectation:

As inner product

Let $\mathcal{L}_2 =$ {random variables $X$ on $\omega$ with $E[X^2] \lt \infinity$ }, a vector space. $S_\mathcal{F} = \{ X \in \mathcal{L}_2 : X$ , measure wrt to $X\}$ is a vector subspace. If we use the inner product $\lt X,Y \gt = E[X,Y]$ then $E[X | \mathcal{F}]$ is the orthogonal projection of $X$ onto $S_\mathcal{F}$ .

This means for any $Z \in S_\mathcal{F}$ , $E[ Z(X - E[X | mathcal{F}])] = 0$ .

Martingales

Wrt a filtration $( \mathcal{F}_n )^\infinity_{n=0}$ , a martingale is a sequence of random variables $( X_n )^\infinity_{n=0}$ with:

Properties:

Submartingales have $E[X_{n+1} | \mathcal{F}_n] \ge X_n$ . Supermartingales have $E[X_{n+1} | \mathcal{F}_n] \le X_n$ .

Regular martingales

If $( \mathcal{F}_n )^\infinity_{n=0}$ is a filtration and $X$ is a rv with $E[|X|] \lt \infinity$ then $M_n = E[X | \mathcal{F}_n]$ makes a martingale wrt $( \mathcal{F}_0)$ . Many martingales have this form.

Martingale transforms

A process (collection of rv’s) $(C_n)^\infinity_{n=1}$ is predictable or previsible wrt $(\mathcal{F}_n)^\infinity_{n=1}$ if $C_n \in F_{n-1}$ .

If $C_n$ is predictable and $(M_n)$ a martingale, then let $Y_n = \sum_{k=\phi}^n C_k (M_k = M_{k-1})$ , then $(Y_n)$ is a martingale.

Proof: $E[Y_{n+1} | \mathcal{F}_n] = E[Y_n + C_{n+1} (M_{n+1} = M_n) | \mathcal{F}] = Y_n + C_n+1 E[M_{n+1} - M_n | \mathcal{F}_n] = Y_n$

Multiperiod models

A trading strategy is a sequence of random vectors $(\phi_0(t), \phi_1(t), ..., \phi_d(t))$ for $t= 1, ..., T$ where $\phi_j(t) $ = amount of asset $j$ to hold from $t-1$ to $t$ , which is predictable ( $\phi_j(t) \in \mathcal{F}_{t-1}$ . This gives the portfolio to hold at all times.

Discounted values are $\tilde{V}_\phi (t) = \frac{V_\phi (t)}{S_0 (t)}$ , and similarly for prices and gains.

Lemma: A strategy $\phi$ is self-financing wrt prices $\iff$ it is self-financing with respect to discounted prices.

Lemma: A strategy $\phi$ is self-financing $\iff V_\phi (t) = V_\phi (0) + G_\phi (t) \iff \tilde{V}_\phi (t) = \tilde{V}_\phi (0) + \tilde{G}_\phi (t)$

Artbitrage

A trading strategy is an arbitrage opportunity if:

(All arbitrage opportunities are local – use same techniques as single period models)

Martingale measures

A probability measure $Q$ is a martingale measure if all discounted asset prices $\tilde{S}_i (t)$ are martingales when $Q$ is used.

Theorem: The model is arbitrage free $\iff$ there is a martingale measure $Q$ equivalent to the original $P$ .

Proof:

$\Leftarrow$ : Let $\phi$ be a self-financing strategy. Then $\tilde{V}_\phi (t)$ is a martingale when $Q$ is used. \[ \tilde{V}_\phi (t) = \tilde{V}_\phi (0) + \sum^t_{s=1} \phi(s) \cdot \left( \tilde{S}(s) - \tilde{S}(s-1) \right) \] \[ = \tilde{V}_\phi (0) + \sum^d_{r=1} \sum^t_{s=1} \phi(s) \cdot \left( \tilde{S}_i(s) - \tilde{S}_d(s-1) \right) \] (the sum of a martingale is a martingale) so $E_Q [\tilde{V}_\phi (t+1) | \mathcal{F}_t] = \tilde{V}_\phi (t)$ and even $E_Q [\tilde{V}_\phi (T)] = \tilde{V}_\phi (0)$ .

For an arbitrage strategry $\tilde{V}_\phi (0) =0$ but $\tilde{V}_\phi (T) \ge 0$ with $P \gt 0$ and so $E_Q [\tilde{V}_\phi (T)] \gt 0$ , so if $Q$ exists then there can not be an arbitrage strategy.

$\Rightarrow$ : Suppose no arbitrage. Let $V = \{ \tilde{V}_\phi (T) : \phi $ is self-financing $V_\phi (0) = 0 \}$ (can think of $V$ as a linear subspace of $\mathcal{R}^N$ . Define $Q$ s.t $Q(\{ \omega_i \} ) \gt 0$ and $\sum^N_{i=1} Q(\{ \omega_i \} ) Y(\omega_i) =0$ , $\forall Y \in V$ . ie. $E_Q [Y] = 0$ , $\forall Y \in V$ , ie. $E_Q [\tilde{V}_\phi (T) ] = 0 \forall$ self-financing $\phi$

Now show every discounted asset price is a martingale under $Q$ . ie. $E_Q [ \tilde(S)_i (t+1) - \tilde{S}_i (t) | \mathcal{F}_t] = 0$ , $\forall i,t$ $E_Q [( \tilde(S)_i (t+1) - \tilde{S}_i (t)) \cdot 1_B ] = 0$ , $\forall B \in \mathcal{F}_t$

We need to show $(( \tilde(S)_i (t+1) - \tilde{S}_i (t)) \cdot 1_B = \tilde{V}_\phi (t)$ for some $\phi$

Take: $\phi(s) = 0$ , for $s = 1, ... , t$ $\phi(t) = ( - S_i(t), 0,0,...,0,1_i,0,0,...,0) \cdot 1_B$ $\phi(s) = ( S_i(t+1) - S_i(t), 0, ..., 0) \cdot 1_B$ for $s \gt t$

Replicating contingent claims

A contingent claim is a random variable on $\Omega$ . A contingent claim $X$ is attainable if there is some self-financing strategy $\phi$ with $V_\phi(T)= X$ .

Theorem: Suppose we have an arbitrage-free multi-period model. Let $X$ be a contingent claim. Then any strategy $\phi$ replicating $X$ has the same $V_\phi (t) \forall t$ , ie if $\phi, \psi$ are two strategies replicating $X$ , then $V_\phi (t) = V_\psi (t) \forall t$ .

Proof: Let $Q$ be an equivalent martingale measure. Then $\tilde{V}_\phi (t)$ and $\tilde{V}_\psi (t)$ are $Q$ -martingales.

The arbitrage price process of an attainable contigent claim $X$ is $(V_t(t))^T_{t=0}$ for any $\phi$ replicating $X$ . It is denoted $(\pi_X (t))^T_{t=0}$ .

The risk-neutral valuation formula gives the value of claim $X$ at time $t$ ,

\[\pi_X (t) = S_0 (t) \cdot E_Q {[\frac{X}{S_0 (T)} | \mathcal{F}_t ]} \] In particular $\pi_X (0) = E {[ \frac{X}{S_0 (T)}]}$ . $\tilde{\pi}_X$ is a martingale under $Q$ , if X is attainable. This only works in $X$ is attainable, must check if any doubt.

Complete markets

A market is complete if any contingent claim $X$ is attainable.

Theorem: An arbitrage free multi-period model is complete $\iff$ it has a unique equivalent martingale measure. For such markets, can use risk-neutral valuation formula without checking that the claim is attainable.

Fundamental theorem of asset pricing

For scenario trees, having a unique solution implies $d+1$ branches at each point.

Cox-Ross-Rubinstein model

This is a $binomial$ model. Numeraire is bank account with $S_0 (t) = (1+r)^t$ . Other account is stock, with:

\[ S_1 (t+1) = { \{ \array{ u s_1(t) & prob p \\

Number of periods, $T$ , gives size of $\Omega$ , as $2^T$ . Number of possible values of $S_1(T) = T + 1$ . Let $Z_T = \frac{S_1 (t)}{S_1(t-1)}$ for $t = 1,..., T$

Arbitrage-free and complete

If $d \lt 1+r \lt u$ , the CRR model is arbitrage free and complete with EMM given by:

Otherwise the model contains arbitrage.

Proof:

Look at one branch-point, compute EMM:

If:

Pricing some contingent claims

An important class of claims are given by $X = f( S_1(T))$ for some function $f$ (payoff is some function of final value).

European call

: $f(S) = (S - K)_T$ for some constant “strike price” $K$ (Want to buy at T for no more than K, payoff is difference in price above K)

European put

: $f(S) = (K - S)_T$ for some constant “strike price” $K$ (Want to sell at T for no less than K, payoff is difference in price below K)

Collar

Cash or nothing

Path dependent options (‘exotic options’)

Barrier option

eg. “knockout option”, “down and out” (similarly “up and out”, “up and in”, “down and in”

Lookback option

Calims where the payoff involves $\min_t S_i(t)$ or $\max_t S_i(t)$ . Eg. $[ \max_t S_i (t) ] - S_i(T)$ , eliminates regret at having sold at a bad time

Asian options

Based on average price over time: $X = -\frac{1}{T} \sum_{t=0}^{T} S_i(t) + S_i(T)$ . Gives the efects of having sold at the average prices if held in conjunction with 1 share to be sold at time $T$ .

American put

Holder must make a decision when to claim, eg. American put. $X_\tau = [K - S_i(\tau)]_t (1+r)^{T-\tau}$ . Like Euro, excpe that holder must claim the payoff at time $\tau$ (which the holder decides) called the exercise time. $\tau$ may be any rv in $\{ 0, 1, ..., T\}$ satisfying $\{ \tau \gt t\} \in \mathcal{F}_t$ (non-anticipation).

For a given strategy $\tau$ being followed the payoff $X_\tau$ is a random varibale so is attainable. Hence it has arbitrage price process $\pi_{X_\tau} (t) = (1+r)^{t-\tau} E_Q [ X_\tau | \mathcal{F}_t]$ .

Market value assumes the holder is using the bst strategy $\tau$ (ie. the one which maximises the value of the put). So we have an optimisation problem:

\[\max _{X_\tau} (0) s.t. \tau : \Omega \to \{0, 1, ..., T\}$ is a rv with $\{ \tau \gt t \} \in \mathcal{F}_t \forall t \]

Only finitely many possible $\tau$ so an optimal strategy exists.

Consider a node $\nu$ at time $t$ . Should we exercise at $\nu$ ? (can assume no exercise on any path leading to $\nu$ ).

Let $\lambda(\nu) =$ value of $\tilde{\pi}_{X_\tau}$ at $\nu$ when $\tau$ is don’t exercise before time $t$ , then folllow a strategy to maximise $\tilde{\pi}_{X_\tau}$ (if $t=T$ , $\lambda(\nu) = [K - S_1(\nu)]_t[1+r]^-T)$ .

If $t \lt T$ and we exercise now, then we get $[K - S_1 (\nu)]_t (1+r)^-t$ If $t \lt T$ and we wait, then we get $q \lambda(\nu^+) + (1-q)\lambda(\nu^-)$

So, $\lambda(\nu) = max \{ [K- S_1(\nu)]_t (1+r)^-t, q \lambda{\nu^+} + (1-q)\lambda(\nu^-)\}$ .

If $t=0$ then $\lambda(\nu) = \tilde{\pi}_{X_\tau} (0)$ for the optimal strategy that achieves it.

So this calculation lets us find the arbitrage value of the options and the optimal strategy that achieves it.

American call

Like Euro, but payoff can be claimed at time $\tau$ . $X_\tau = [S_i(\tau) - K]_t (1+r)^{T-\tau}$

CRR in real-life

Represent a time $T$ by $n$ discrete steps of length $h$ . Use $t \in [0, T]$ for real-time, the time at step $k$ is $t = kh$ .

Interest

Usually continuously compounded interest, $e^rt$ after time $t$ . If $r_0$ is the interest rate per step in the CRR model, then $1 + r_0 = e^rh$ .

Asset prices

Many asset prices behave (approximately) like CRR model:

Calculating $u$ and $d$

Under the martingale measure we want:

$S_1(t) = S_1(0) exp(rt - \frac{1}{2} \sigma^2_t + \sigma_t Z)$

What is $\sigma_t$ ?

Note that after $k$ periods of length $t$ :

So $\sigma_kt = \sqrt{k} \sigma_t$ , ie $\sigma_k \propto \sqrt{t}$ . $\sigma_t = \sigma \sqrt{t}$ .

$\sigma$ is called the *volatility parameter of the asset. Typical values for $\sigma$ are $0.1 - 0.4 year^{-1/2}$ .

CRR model

We need to approximate the distribuition of $S_1(0) exp( (r-\frac{1}{2}\sigma^2_t)h + \sigma \sqrt{t} Z)$ by the discrete distribution:

Choose $u, d$ so that means and variances match:

or since $T= nh$ .

To satisfy first equation, let:

And note $q = \frac{1+ r - d}{u-d} = \frac{e^rt -d}{u-d}$ , so $\alpha$ must satisfy:

\[ \frac{{( e^{\sigma^2 \frac{h}{2}} - e^{-\alpha} )} {( e^{-\alpha} - e^{\sigma^2 \frac{h}{2}} )}}{ {( e^\alpha - e^{-\alpha} )}^2} 4 \alpha^2 = \sigma^2 h \]

This can be solved numerically, or use CRR approximation, $\alpha = \sigma \sqrt{h}$ , which is very good when $\sigma$ is small (which it usually is).

Stochastic calculus

Suppose we have a sample space $\Omega$ with prob $P$ and a random variable $Z ~ N(0,1)$ .

Example 1.

Let $\gamma \in \mathbb{R}$ be a constant. We can find equivalent probability measure $Q$ under which $Z ~ N(\gamma, 1)$ with

\[\frac{dQ}{dP} = \frac{f_\gamma (Z)}{f_0(Z)} = \frac{\frac{1}{\sqrt{2\pi}} exp(-\frac{1}{2}(Z - \gamma)^2)}{\frac{1}{\sqrt{2\pi}} exp(-\frac{1}{2}Z^2)} = exp(Z \gamma - \gamma^2 /2) \]

Example 2

Suppose $Z_1, ..., Z_n$ are iid N(0,1) under P. Let $\gamma_1, ..., \gamma_n$ be constants. We can definie a new probability measure Q by:

\[ \frac{dP}{dQ} = \frac{ f_\gamma(Z_1, ..., Z_n)}{f_0(Z_1, ..., Z_n)} = exp(\sum \gamma_i Z_i - \frac{1}{2} \sum \gamma_i^2) \]

Under $Q$ , $Z_1, ..., Z_n$ , are indepdendent with distributions $Z_i ~ N(\gamma_i , 1)$

Example 3

Let $\Omega$ be a sample space with prob $P$ and a Brownian motion $W$ . Let $(\gamma_s)$ be a process with $\gamma_t \in \mathcal{F}_t$ and $E[ exp{(\frac{1}{2} \int^T_0 \gamma_t^2 dt )}] \lt \infinity$ . Define a new prob measure $Q$ by:

\[ \frac{dQ}{dP} = exp {( \int^T_0 \gamma_t dW_t - \frac{1}{2} \int^T_0 \gamma_t^2 dt )} \]

Under $Q$ the process $\tilde{W}_t = W_t - \int^T_0 \gamma^2_t dt$ is a Brownian motion. ie. $\tilde{W}_t$ is a Ito diffusion with $dW_t = \gamma_t dt + d \tilde{W}_t$ (drift $\gamma_t$ , speed1).

If $(X_t)$ is an Ito diffusion under $P$ then under $Q$ :

\[ dX_t = U_t d_t + V_t (\gamma + d\tilde{W}_t) dX_t = (U_t + \gamma_t V_t) dt + V_t d\tilde{W}_t) \]

In particular if $\gamma_t = -\frac{U_t}{V_t}$ then $(X_t)$ is a martingale

Continuous time financial models

Suppose a sample space $\Omega$ probability measure $P$ and filtration $(\mathcal{F}_t)$ .

Have $d+1$ assets, with prices $(S_i(t))_{0 \le t \le T}$ , $i = 0, ..., d$ , and $S_i(t) \in \mathcal{F}_t$ . These prices are Ito diffusions: $dS_i (t) = U_i (t) dt + V_i (t) dW_i (t)$ . The numeraire $S_0 (t)$ is strictly positive, and usually $= e^rt$ .

Main example

Geometric brownian model (or Black-Scholes model)

Or equivalently:

An arbitrage opportunity is a strategy $\phi$ with $V_\phi(0) = 0$ , $V_\phi (T) \gt 0$ , $P( V_\phi (T) \gt 0) \gt 0$ . Should not exist in model with consistent prices.

Martingale measures

A martingale measure is a probability measure under which all discounted asset prices, $\tilde{S}_i$ , are martingales. (sometimes called strong marginale measure).

Proposition: Under a martingale measure, the discounted value $\tilde{V}_\phi$ of any self-financing strategy is a martingale. Proof: $d\tilde{V}_\phi (t) = \sum \phi_i (t) d\tilde{S}_i (t)$ $ = \sum \phi_i V_i (t) dW_i (t)$ , so $\tilde{V}_\phi$ is also a martingale.

Theorem: If there is a martingale measure $Q$ equivalent to the orginal $P$ , then the model contains no arbitrage opportunities.

Proof: For any self-financing strategy $\phi$ , $\tilde{V}_\phi$ is a martingale under Q. So:

For an arbitrage opportunity, $V_\phi(0) = 0$ , but $V_\phi (T) \ge 0$ and $P(V_\phi (T) \gt 0) \gt 0$ so $P(V_\phi (T) \gt 0) \gt 0$ so $E_Q{[\frac{V_\phi (T)}{S_0 (T)}]} \gt 0$ .

Note: Reverse is not true – there are arbitrage-free models in continuous time with not EMM.

Pricing contingent claims

Definition: A contingent claim is a random variable $X \in \mathcal{F}$ , thought of as amount of money to paid to the holder at time $T$ . The claim is attainable if there is a self-financing strategy $\phi$ with $V_\phi (T) = X$ . The model is complete if all contingent claims are attainable.

Risk-neutral pricing formula

Theorem: If $X$ is an attainable contingent claim and $Q$ an EMM, then any strategy that replicates $X$ will have price process $\pi_X (t) = V_\phi(t) = S_0 (t) E_Q {[ \frac{X}{S_0 (T)} | \mathcal{F} ]} $

Doesn’t depend on $\phi$ so can find $\pi_X$ without knowing $\phi$ (but $X$ must be attainable). If there is more than one EMM any can be used – they will all give the same results.

Proposition: If there is a unique EMM $Q$ then any contingent claim $X$ with $E {[ {| \frac{X}{S_0 (T)} |} ]} \lt \infinity$ is attainable. Note: no converse (like discrete case). Not very useful.

Proposition: For the Black-Scholes model any contingent claim $X$ with $E {[ {| \frac{X}{S_0 (T)} |} ]} \lt \infinity$ is attainable.

Risk-neutral pricing in the Black-Scholes model

Under $Q$ :

European options ($X = f(S_1 (T))$ )

Then $\pi_X (0) = e^{-rt} E_Q [ f(S_1 (T) ] = ... = e^{-rt} \int_{-\infinity}^\infinity f(s_0 exp((r-\sigma^2/2)T + \sigma \sqrt{T}x) \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$ .

European call ($X = (S_1 (T) - K)_{+}$ )

$\pi_X(0) = S_1(0) \Phi(z_0 + \sigma \sqrt{T}) - K e^{-rT} \Phi(z_0)$ , where $z_0 = \frac{(r-\sigma^2/2)T + log(S_1(0) / K)}{\sigma \sqrt{T}}$ – the Black-Scholes model.

European put ($Y = (K - S_1 (T) )_{+}$ )

Can value using integral as above, or using put-call parity: $X - Y = S_1(T) - K$ , so $Y = X - S_1(T) + K$ .

\[ \pi_Y (0) = E_Q [Y e^{-rT} ]

Replicating strategies

Theorem: If the arbitrage value of claim $X$ satisfies $\tilde{\pi}_X (t) = G(t, S_1(t))$ then a replicating strategy for $X$ is $\phi_1 (t) = e^{rt} G_s(t, S_1(T))$ and $\phi_0(t) = G(t, S_1(t)) - S_1(t)G_s(t, S_1(t))$ .

Proof: $\tilde{\pi} = G(t , S_1(t)) = G(t, s_1(0) exp{( (r - \sigma^2 /2)t + \sigma \tilde{W}_t)}$ . By Ito’s lemma, $d\tilde{\pi}_X (t) = (F_t + \frac{1}{2} F_ww)dt + F_w d\tilde{W}_t$ Drift = 0, so $ = G_s(t, S_1(t)) S_1(t) \sigma d\tilde{W}_t = h(t) d\tilde{W}_t$ , say.

Recall that in a Black-Scholes model, if a strategy has $d\tilde{\pi}_X (t) = h(t) d\tilde{W}_t$ it is replicated by $\phi$ with $\phi_1 (t) = h(t) / \sigma \tilde{S}_t (t)$ , hence $\phi_1(t) = e^{rt} G_s(t, S_1(t))$ .

Let’s apply it to the Euro call:

\[ \tilde{\pi}_X (t) = e^{rt} \int_{-\infinity}^{\infinity} (S_1(t) exp{( (r-\sigma^2/2)(T -t) + \sigma\sqrt{T - t}x)} - K)_{+} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx \] \[ \phi_1(t) = \Phi {( \frac{ log(S/K) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}} )} \]

The Black-Scholes PDE

Suppose the arbitrage value $\pi_X (T) = F(t, S(t))$ then $F(t,s)$ satisfies

\[ F_t + rsF_s + \frac{1}{2} \sigma^2 s^2 F_ss = rF$

By Ito’s lemma

\[ d\tilde{\pi}_X (t) = (h_t + \frac{1}{2} h_ss)dt + h_s d\tilde{W}_t \]

Setting $(h_t + \frac{1}{2} h_ss)$ to 0 ($\tilde{\pi}_X$ is a martingale) gives the Black-Scholes PDE.

The Greeks

These are the partial derivatives of arbitrage values with respect to various parameters. If $\pi_X (0) = p = g(S_1(0), T, r, \sigma)$ then:

Process of holding a replicating portfolio and attempting to keep up to date with current $\Delta$ called delta holding. Delta-gamma hedging uses both $\Delta$ and $\gamma$ .

Volatility ($\sigma$ )

$\sigma$ is the only parameter that is hard to estimate. Can estimate using historical or implied volatility.

Historical volatility

Fit a GBM to historical data on stock prices. $S_1(t) = S_1(0) exp {( (b-\frac{1}{2}\sigma^2)t + \sigma W_t )}$

Suppose we have observations $s_0, s_1, ..., s_n$ of stock prices at times $t_i = i \Delta t$ , $i = 0, ..., n$ . By the model $log(S_i / S_{i-1})$ are independent $N((b-\frac{1}{2}\sigma^2)t, \sigma^2 \Delta t)$ , so a sample standard deviation of $log(S_i / S_{i-1})$ is an estimate of $\sigma$ .

\[ \hat{\sigma} = \sqrt{ \frac{ \sum (x_i - \bar{x})^2}{(n-1)\Delta t} } \]

Implied volatility

We know $p, s, T, r$ so solve $p = g(s,T,\sigma,r)$ for $\sigma$ . Note $\nu = \frac{\partial p}{\partial \sigma}$ so $\frac{1}{\nu} = \frac{\partial \sigma}{\partial p}$ . If $\nu$ is large, $\sigma$ not very sensitive to changes in $p$ , so easy to estimate $\sigma$ well and vice versa. When computing $\sigma$ from different options often get a “volatility smile” (because GBM isn’t adequate for modelling tail behaviour).

Path-dependent options

Payoff dependents on the full path, eg. the knockout barrier function. For most exotic options numerical methods (or discretisation + CRR model) are required.

Black-Scholes models with multiple risky assets

Consider a model with $d+1$ assets

Equivalent martingale meausre

Look for equivalent measure $Q$ with:

\[ \frac{dQ}{dP} = exp{(\sum^n {( \int^T_0 \gamma_i (t) dW_j (t) - \frac{1}{2} \int^T_0 \gamma(t)^2 dt )} )} \]

Under Q $\tilde{W}_k (t) = W_j(t) - \int^T_0 (s) ds$ , which are Brownian motion for $j= 1, ..., n$ , so:

\[ dS_i(t) = b_i(t)S_i(t)dt + S_i(t) \sum \sigma_ij(t) )d\tilde{W}_j(t)

The first part must be 0 so there is no drift. Hence we need $\gamma_1, ..., \gamma_n$ s.t. $\sum \sigma_ij(t) \gamma_j(t) = r - b_i(t)$ for $i = 1,...,d$ . A special case is when $\sigma_ij$ and $b_i$ are constant. Then $\gamma_j$ will be constants too, satisfying a system of $d$ linear equations, eg. $\Delta \gamma = r 1_d - b$ , where $\Delta = (\sigma_ij)$ , $b = (b_i)$ .

Pricing contingent claims

Can have claims with payoff based on more than one risky asset, eg. exchange option $X = (S_2(T) - S_1(T))_{+}$ . As usual $\pi_X (t) = S_o (t) E_Q {[ \frac{X}{S_0(T)} | \mathcal{F}_t ]}$ , but now we will have a joint density to integrate out over.

Simulation

Instead of tricky integrals, price contingent claims by simulating asset price paths.

If we generate random variates $X_1, ..., X_n$ indepedent with the distribution $X / S_0(T)$ then $\bar{X} = \frac{1}{n} \sum X_i$ estimates $\pi_X (0)$ with usual CLT confidence intervals.

To generate the $X_i$ , need to simulate asset price paths $((S_1(t)), (S_2(t)), ...)$ . With a Black-Scholes model $S_i (t) = exp((r- \frac{1}{2} \sigma^2)t + \sigma \tilde{W}_t$ , so we need to simulate Brownian motion.

What to do next depends on the type of claim. For example, with Euro options only need final value so only need $\tilde{W}_T ~ N(0, T)$ . But in general will need the whole price path. Will discretise into $m$ steps of size $\Delta t = T / m$ , then fill in other values via linear interpolation. This works well for complicated options. It doesn’t work well for American options because it doesn’t provide optimal strategy.

Variance reduction techniques

High accuracy required for financial application, usually want error to be less than 1 part per thousand. Standard errors $\in o(\sqrt{n})$ , so therefore useful to use variance reduction techniques.

Antithetic variates

The simplest of these is antithetic variates – if $(W_t)$ is a Brownian motion, then so is $(-W_t)$ . Then let $X = \frac{1}{2} (f(S_i^{(i')} (T)) - f(S_i^{(i)} (T)))$ . Variance will be reduced because prices from positive and negative paths are likely to be negatively correlated.

Control variates

Idea: Use known “model payoffs” $Y$ , similar to $X$ , with $E[Y] = \mu_Y$ . $\frac{1}{n} \sum (X_i + c(Yi - \mu_Y))$ , estimates $E[X]$ as $E[X + c(Y - \mu_Y)] = E[Y]$ .

This will be better than simple sampling if $Var(X + c(Y - \mu_Y)) \lt Var(X)$ , ie. $Var(X) + 2cCov(X,Y) + Var(Y) \lt Var(X)$ . A pilot sample may help to estimate the best value $c^{*}$ of $c$ . Alternatively could just take $c$ to be $-1$ , ie use $\frac{1}{n} \sum (X_i - Y_i + \mu_Y)$ . Relies on $Var(X-Y)$ being smaller than $Var(X)$ .

Another approach uses several model payoffs $Y^{(1)}, Y^{(2)}, ..., Y^{(k)}$ with $E[Y^{(j)}] = \mu_j$ . Then estimate $E[X]$ using $\frac{1}{n} \sum (X_i + \sum c_j(Y^{(j)}_i - \mu_j))$ . This works as above. You can find the $c_j$ ’s by regressing $X_i$ on $Y^{(1)}_i, ...$ .

Importance sampling

Idea: Use $E_K [X \frac{dQ}{dR} ]$ , ie. simulate from distribution of $X \frac{dQ}{dR}$ under R rather than $X$ under $Q$ . Hopefully $Var_R (X \frac{dQ}{dR})$ will be smaller than $Var_Q (X)$ .

\[ Var_R(X \frac{dQ}{dR}) = E_R [ X^2 {(\frac{dQ}{dR})}^2] - \mu^2 Var_R(X \frac{dQ}{dR}) = E_Q [ X^2 {(\frac{dQ}{dR})}^2 \frac{dR}{dQ}] - \mu^2 Var_R(X \frac{dQ}{dR}) = E_R [ X^2 \frac{dQ}{dR}] - \mu^2 \]

We want $\frac{dQ}{dR}$ to be small. (Note: $\frac{dQ}{dR} \gt 0$ and $E_Q [ {(\frac{dQ}{dR})}^{-1}] = 1$ ). Can do this with GBM. If $S_1(t) = S_1(0) exp( (r-\frac{1}{2}\sigma^2)t \sigma \tilde{W}_t)$ we can find an equivalent measure $R$ s.t. $\frac{dR}{dQ} = exp(c\tilde{W}_T - c^2 T /2)$ , under which $\bar{W}_T = \tilde{W}_T - ct$ is a Brownian motion. So simulating the assert price under $R$ is just a matter of simulating a GBM.

Also:

\[ \frac{dQ}{dR} = exp (-\tilde{W}_T + \frac{1}{2} c^2 T) \frac{dQ}{dR} = exp (-\bar{W}_T - \frac{1}{2} c^2 T) \]