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.

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 \(p_*=\frac{1+r-d}{u-d}=0.555555556\).

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 \(3125\).