MTH 4500: Introductory Financial Mathematics

Barrier Options

Random walks on integers

Let $$\Omega=\{-1,1\}^n$$. Each element of $$\omega\in\Omega$$ can be seen as $$\omega=(\omega_1, \dots, \omega_n)$$, where $$\omega_i\in\{-1,1\}$$. Let $$W_0=a$$ and $$W_k(\omega)= a+\sum_{i=1}^k\omega_i$$, for $$k\in\{1,2,\dots, n\}$$.

Every fixed $$\omega$$ uniquely determines the sequence $$W_0(\omega)$$, $$W_1(\omega)$$, $$W_2(\omega)$$, $$\dots$$, $$W_n(\omega)$$ that can be graphically represented as a path in the coordinate plane.

A simple symmetric random walk is the one in which all outcomes have the same probability. The cardinality of the set $$\Omega$$ is $$2^n$$, i.e. $$|\Omega|=2^n$$ hence for each $$\omega\in \Omega$$ we have $$\mathbb P(\{\omega\})=\frac1{2^n}$$.

Let us for a moment study the distribution of the endpoint $$W_n(\omega)$$. Obviously, $$W_n\in\{a,a\pm 1, a\pm 2, \dots, a\pm n\}$$. However, we can reduce the set of possibilities for $$W_n$$ if we know whether $$n$$ is even or odd. If $$2\mid n$$ then there exists a positive integer $$m$$ such that $$n=2m$$. We then have $W_n=W_{2m}\in\{a, a\pm 2, a\pm 4, \dots, a\pm 2m\}.$ If $$2\nmid n$$ then there exists a non-negative integer $$m$$ such that $$n=2m+1$$. In this case we can conclude that $W_{n}=W_{2m+1}\in\{a\pm 1, a\pm 3, a\pm 5, \dots, a\pm (2m+1)\}.$

Recall that the definition of binomial coefficient $$\binom mr=\frac{m(m-1)\cdots (m-r+1)}{r!}$$ can be extended to include $$m\in \mathbb R$$ and $$n\in\mathbb N_0$$. Specifically, if $$m < r$$ and $$m\in\mathbb N$$ then $$\binom mr=0$$.

Distribution of endpoints of $$W_n(\omega)$$

Theorem For every $$b\in\mathbb Z$$ that satisfies $$-n+a\leq b\leq n+a$$ the following identity holds: $\mathbb P(W_n=b)=\binom{n}{\frac12(n+b-a)}\cdot 2^{-n}.$

Reflection principle

Let $$N_n(a,b)$$ be the number of paths from $$(0,a)$$ to $$(n,b)$$. Let $$N_n^0(a,b)$$ be the number of paths from $$(0,a)$$ to $$(n,b)$$ that contain at least one point on the $$x$$-axis (i.e. $$W_k=0$$ for some $$k\in\{0,1,2,\dots, n\}$$).

Theorem (Reflection principle) For any two positive integers $$a$$ and $$b$$ the following holds: $N_n^0(a,b)=N_n(-a,b).$

Corollary Let $$a$$ and $$b$$ be positive integers and $$W_m$$ for $$m\in\{0,1,\dots, n\}$$ a symmetric random walk. Then $\mathbb P_a\left(\min_{0\leq m\leq n}W_m\leq 0, W_n=b\right)=\frac1{2^n}N(-a,b)=\mathbb P_{-a}\left(W_n=b\right),$ where $$\mathbb P_x$$ is the probability when $$W_0=x$$.

Ballot theorem Let $$b > 0$$. The number of paths from $$(0,0)$$ to $$(n,b)$$ which do not visit the $$x$$-axis (apart from the starting point) is equal to $$\frac{b}nN_n(0,b)$$.

Interpretation: In a city with population $$n$$ everyone votes independently for one of the two candidates $$A$$ or $$C$$. Suppose that $$A$$ scores $$a$$ votes and $$C$$ scores $$c$$ votes ($$a+c=n$$), where $$a > c$$. What is the probability that $$A$$ was always in the lead.?

In this case we take $$W_0=$$, vote for $$A$$ is an _quot_up_quot_ movement in the random walk and vote for $$C$$ is a down movement of the random walk. The final position is $$W_n=a-c=b$$. We apply the ballot theorem:

$\mathbb P\left(\left. A \mbox{ always in the lead}\right| A\mbox{ gets }a \mbox{ votes}\right)=\frac{\frac{a-c}nN_n(0,a-c)}{N_n(0,a-c)}=\frac{a-c}n=\frac{a-c}{a+c}.$

Binomial asset pricing model

Let $$X_i$$ be the random variable that is $$1$$ if the stock went up in the $$i$$-th period, and $$0$$ otherwise. Then the price of the stock can be expressed in terms of $$X_i$$s in the following way: $S_n=S\cdot u^{\sum_{i=1}^nX_i}\cdot d^{n-\sum_{i=1}^nX_i}.$ In order to reduce the problem to the study of the random walk, let us recall that $$\Omega=\{-1,1\}^n$$ and let us assume that $\mathbb P(\omega)=p^{k_{\omega}}(1-p)^{n-k_{\omega}},$ where $$k_{\omega}$$ is the number of $$+1$$‘s in $$\omega$$. We can now set $$X_i=\frac{\omega_i+1}2$$, and obtain \begin{eqnarray*}S_n&=&S\cdot u^{\frac12W_n+\frac n2}\cdot d^{\frac n2-\frac{W_n}2}=S\left(\frac ud\right)^{\frac12W_n}\cdot \left(ud\right)^{\frac n2}\newline&=& Se^{W_n\cdot\frac{\ln u-\ln d}2+\frac{n(\ln u+\ln d)}2}.\end{eqnarray*} For simplicity, let us assume that $$ud=1$$. Then $$S_n=Se^{\ln uW_n}$$.

We will consider up-and-out call option with maturity time $$n$$ with the payoff: $\mbox{Payoff} = \left(S_n-K\right)^+\cdot 1_{S_n^* < B},$ where $$K$$ is the strike, $$B$$ is the barrier, and $$S_n^*=\max_{0\leq m\leq n}S_m$$.

We will use two methods to price such an option.

• $$1^{\circ}$$ First method. This is a forward method based on _quot_path counting._quot_ This method uses reflection principle and allows for closed form formulas. It can be extended to continuous setting into Black-Scholes model.

• $$2^{\circ}$$ Second method. This is a backward recursive method. It does not result in closed formulas but in an algorithm. This method can be used for a variety of options, including the American options. Continuous time generalizations lead to the principle of dynamic programming in optimization.

We will consider the following numeric example. Let $$u=1.25$$, $$d=0.8$$, $$S=60$$, $$r=0.05$$, $$n=4$$, $$K=90$$, $$B=115$$. Then the stock price tree and the payoffs of the option are given in the following table: $\begin{array}{llllll} &&&&146.484375&0\newline &&&117.1875&&\newline &&93.75&&93.75&3.75\newline &75&&75&&\newline 60&&60&&60&0\newline &48&&48&&\newline &&38.4&&38.4&0\newline &&&30.72&&\newline &&&&24.576&0\end{array}$ Then we obtain $$p_*=\frac{1+r-d}{u-d}=0.555555556$$.

• $$1^{\circ}$$ First method. The price is given by: \begin{eqnarray*} \mbox{Price}&=&\frac1{(1+r)^4}\mathbb E_*\left[\mbox{Payoff}\right]\newline&=&\frac1{1.05^4}\cdot 3.75\cdot \left(p_*\right)^3\cdot \left(1-p_*\right)^1\cdot \left(\mbox{Number of paths that end up at }93.75\mbox{ and stay below }115\right)\newline &=&\frac1{1.05^4}\cdot 3.75\cdot \left(0.555555556\right)^3\cdot 0.444444444\cdot 3\newline&=&0.705334769. \end{eqnarray*}

• $$2^{\circ}$$ Second method. Let $$\Pi_{i,j}$$ denote the value of the option after $$i$$ periods if the stock went up $$j$$ times. We have that $$\Pi_{4,0}=\Pi_{4,1}=\Pi_{4,2}=0$$, $$\Pi_{4,3}=3.75$$, $$\Pi_{4,4}=0$$. We now calculate backwards the price of the option. The option price tree is now given by the following table $\begin{array}{lllll} &&&&0\newline &&&0&\newline &&0.83984211&&3.75\newline &0.888721809&&1.984126984&\newline 0.705334769&&1.049802637&&0\newline &0.555451131&&0&\newline &&0&&0\newline &&&0&\newline &&&&0 \end{array}$ Let us point out that $$\Pi_{3,3}=0$$ because the stock hits the barrier at time $$3$$ if it went up in each of the periods. For every other cell in the table, the price is calculated in the standard way.

We will now explain how in general the method 1 can be used to calculate the price of the option. The price is given by $\mbox{Price}= \frac1{(1+r)^n}\mathbb E_*\left[\left(S_n-K\right)^+\cdot 1_{S_n^* < B}\right].$ The expectation can be now calculated in the following way: \begin{eqnarray*} \mathbb E_*\left[\left(S_n-K\right)^+\cdot 1_{S_n^* < B}\right]&=&\sum_{b:K < b < B}(b-K)\cdot\mathbb P_*\left(S_n^* < B,S_n=b\right). \end{eqnarray*} Let us now use that $$S_n=Se^{\ln uW_n}$$. Therefore $$S_n=b$$ can be re-written as $$W_n=\frac{\ln b-\ln S}{\ln u}$$. Similarly, $$S_n < B$$ can be written in an equivalent form as $$W_n^* < \frac{\ln B-\ln S}{\ln u}$$. Thus $\mbox{Price}= \frac1{(1+r)^n}\cdot \sum_{K < b < B}(b-K)\mathbb P_*\left(W_n^* < \frac{\ln B-\ln S}{\ln u}, W_n=\frac{\ln b-\ln S}{\ln u}\right).$

Problem Assume that the stock price follows binomial model with $$n=50$$ steps. The initial price of the stock at time $$0$$ is $$S_0=1024$$ and in each step the price of the stock goes up by the factor $$u=1.25$$ or down by the factor $$d=0.8$$. The simple interest rate in each period is $$r=10\%$$. Determine the price of the derived security that pays USD $$1$$ if the price of the stock at time $$n$$ is exactly $$2500$$ and during the entire time interval $$[0,50]$$ the stock has never reached the level $$3125$$ or gone over the level $$3125$$.