Chapter 2 Exercises

Exercise: Using Definition 2.1.1 show the following: (i) If \(A\) is an event and $A^c$ denotes its complement, then $\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$

Solution Using the fact that $A^c$ is the event complement of $A$, we can say that:

Equivalently, we have that:

Thus, we can express the probabilities under the probability measure $\mathbb{P}$:

as required.

(ii) If $A_1, A_2, \dots, A_N$ is a finite set of events, then:

If the events $A_1, A_2, \dots, A_N$ are disjoint, then equality holds in (2.8.1).

Solution

For our disjoint case, where $A_1, \dots, A_N$ for $A_i \hat A_j = \emptyset$ for $i \neq j$

Generally, for $A_1, \dots, A_N$ is not disjoint, we apply the inclusion-exclusion principle, to prevent overlapping events:

Thus, all the probabilities are non-negative, and the sum of the higher order intersections is subtracted in the inclusion-exclusion principle.

Exercise Consider the stock price $S_3$ in Figure 2.3.1. (i) What is the distribution of $S_3$ under the risk-neutral probabilities $\tilde{p} = \frac{1}{2}, \tilde{q} = \frac{1}{2}$?

Solution

(ii) Compute $\tilde{\mathbb{E}}S_1, \tilde{\mathbb{E}}S_2, \tilde{\mathbb{E}}S_3$. What is the average rate of growth of the stock price under $\tilde{P}$?

Thus, we get that the average rate of growth is:

Exercise Show that a convex function of a martingale is a submartingale. In other words, let $M_1, \dots, M_N$ be a martingale and let $\phi$ be a convex function. Show that $\phi(M_0), \phi(M_1), \dots, \phi(M_N)$ is a submartingale.

Solution To show that if $M_1, M_2, \dots, M_N$ is a martingale and $\phi$ is a convex function, then the sequence $\phi(M_0), \dots, \phi(M_N)$ is a submartingale, we proceed as follows:

Again, we define a martingale, $M_1, M_2, \dots$, as one that satisfies the property:

We want to show that:

Using convexity of $\phi$, we can apply Jensen’s inequality, which states that for any convex function $\phi$ and a random variable $X$, we have the following:

which mean that $\phi(M_0), \phi(M_1), \dots, \phi(M_N)$ is a submartingale. Thus, a convex function of a martingale is indeed a submartingale.

Exercise: Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2}$, as is the probability of tail. Let $X_j = 1$ if the $j$th toss results in a head and $X_j = -1$ if the $j$th toss results in a tail. Consider the stochastic process $M_0, M_1, M_2, \dots$ defined by $M_0 = 0$ and:

This is called a symmetric random walk; with each head, it steps up one, and with each tail, it steps down one. (i) Using the properties of Theorem 2.3.2, show that $M_0, M_1, M_2, \dots$ is a martingale.

Solution Each $X_j$ is a random variable taking values $1$ and $-1$ with equal probability, so it had finite expectation:

Since we have that:

Thus, we see that $M_n$ is a martingale, and satisfies all conditions such that $M_0, M_1, M_2, \dots$ is a martingale by definition of conditional expectation. Simply put, we pulled out what was know, in this case it was the $X_{n + 1}$ in the final two steps of the proof above, and computed directly to get $M_n$, giving, by definition, the required equation for a martingale.

(ii) Let $\sigma$ be a positive constant and for $n \ge 0$ define:

Show that $S_0, S_1, S_2, \dots$ is a martingale. Note that even though the symmetric random walk $M_n$ has no tendency to grow, the geometric symmetric random walk $\exp{\sigma M_n}$ does have a tendency to grow. This is the result of putting a martingale into the convex exponential function. In order to again have a martingale, we must discount the geometric symmetric random walk, using the term $\frac{2}{e^\sigma + e^{-\sigma}}$ as the discount rate. This term is strictly less than one unless $\sigma = 0$.

Solution

Exercise Let $M_0, M_1, M_2, \dots$ be the symmetric random walk of the previous exercise and we define a new variable $I_0 = 0$ and:

(i) Show that:

Solution We start with the definition of $I_n$:

We can use the relationship $M_{j+1} - M_j = X_{j+1}$, since $M_n = \sum_{i=1}^n X_i$, to rewrite:

Now consider $M_n^2$:

Since each $M_j = \sum_{i=1}^j X_i$ represents the accumulated sum of $X_i$ terms up to step $j$ this is exactly the definition of $2 I_n$, which, after substituting back, gives:

(ii) Let $n$ be an arbitrary non-negative integer, and let $f(i)$ be an arbitrary function of a variable $i$. In terms of $n$ and $f$, define another function $g(i)$ satisfying:

Note that although the function $g(I_n)$ on the right-hand side of this equation may depend on $n$, the only random variable that may appear in its argument is $I_n$; the random variable $M_n$ may not appear. You will need to use the formula in part $(i)$. The conclusion of part $(ii)$ is that the process $I_0, I_1, I_2, \dots$ is a Markov process.

Solution

Exercise: Discrete-time Stochastic Integral Suppose $M_0, M_1, \dots, M_N$ is a martingale, and let $\Delta_0, \Delta_1, \dots, \Delta_{N - 1}$ be an adapted process. Define the discrete-time stochastic integral (sometimes called a martingale transform) $I_0, I_1, \dots, I_N$ by setting $I_0 = 0$ and:

Show that $I_0, I_1, \dots, I_N$ is a martingale.

Solution First we assume the following, then the computation is pretty straight forward:

• $I_n$ is adapted to $F_n$, the neutral filtration of $M_n$
• $\mathbb{E}[I_n] < \infty$ for all $n$

• $\mathbb{E}n [I{n + 1}

  F_n] = I_n$

Thus, we have that $I_0, I_1, \dots, I_n$ is a martingale.

Exercise In a binomial model: Give an example of a stochastic process that is a martingale but is not Markov.

Solution Let $S_n$ denote the price of an asset at time $n$, following a binomial model: \(S_{n+1} = \begin{cases} u S_n & \text{with probability } p, \\ d S_n & \text{with probability } 1-p, \end{cases}\) where $u > 1$ (up factor), $d < 1$ (down factor), and $0 < p < 1$ (probability of an up move).

Define the filtration $\mathcal{F}_n$, representing the information available up to time $n$. We define the process $Y_n$ as: \(Y_n = S_n + Z_n,\) With $Z_n$ defined as:

\(Z_n = \sum_{j=0}^{n-1} S_j,\) the cumulative sum of past prices up to time $n-1$. To check if $Y_n$ is a martingale, we compute the conditional expectation $\mathbb{E}[Y_{n+1} | \mathcal{F}n]$: \(\mathbb{E}[Y_{n+1} | \mathcal{F}_n] = \mathbb{E}[S_{n+1} | \mathcal{F}_n] + \mathbb{E}[Z_{n+1} | \mathcal{F}_n].\) Since $S_n$ is a martingale under the appropriate risk-neutral measure: \(\mathbb{E}[S_{n+1} | \mathcal{F}_n] = S_n.\) For $Z{n+1}$, observe that: \(Z_{n+1} = Z_n + S_n \quad \implies \quad \mathbb{E}[Z_{n+1} | \mathcal{F}_n] = Z_n + S_n.\) Thus, $Y_n$ is a martingale. \(\mathbb{E}[Y_{n+1} | \mathcal{F}_n] = S_n + (Z_n + S_n) = Y_n.\) The Markov property requires: \(\mathbb{P}(Y_{n+1} | \mathcal{F}_n) = \mathbb{P}(Y_{n+1} | Y_n).\) However, $Y_n$ depends on the entire history of $S_n$ through \(Z_n = \sum_{j=0}^{n-1} S_j.\) The future evolution of $Y_n$ is influenced by $Z_n$, which cannot be determined solely from $Y_n$ at time $n$. Thus, $Y_n$ does not satisfy the Markov property.

Exercise Consider an $N$-period binomial model.

(i) Let $M_0, M_1, \dots, M_N$ and $M_0’, M_1’, \dots, M_N’$ be martingales under the risk-neutral measure $\tilde{P}$. Show that if $M_N = M_N’$ (for every possible outcome of the sequence of coin tosses), then, for each $n$ between $0$ and $N$, we have $M_n = M_n’$ (for every possible outcome of the sequence of coin tosses).

Solution Let $\tilde{\mathbb{P}}$ be the risk-neutral measure under which both $M_n$ and $M_n’$ are martingales. The martingale property means: \(\mathbb{E}_{\tilde{\mathbb{P}}}[M_{n+1} | \mathcal{F}_n] = M_n, \quad \mathbb{E}_{\tilde{\mathbb{P}}}[M_{n+1}' | \mathcal{F}_n] = M_n'.\) We are given that: \(M_N = M_N' \quad \text{for all possible outcomes of the sequence of coin tosses}.\) We aim to show: \(M_n = M_n' \quad \text{for all } n \text{ (from } 0 \text{ to } N\text{)}.\) We use backward induction, starting from $n = N$ and moving backward to $n = 0$. Base Case ($n = N$): From the problem statement, we know: \(M_N = M_N' \quad \text{for all outcomes}.\) Inductive Hypothesis: Assume that $M_{k+1} = M_{k+1}’$ for all outcomes for some $k \leq N-1$. We will show that $M_k = M_k’$. Inductive Step: Since $M_n$ and $M_n’$ are martingales under $\tilde{\mathbb{P}}$, we have: \(M_k = \mathbb{E}_{\tilde{\mathbb{P}}}[M_{k+1} | \mathcal{F}_k], \quad M_k' = \mathbb{E}_{\tilde{\mathbb{P}}}[M_{k+1}' | \mathcal{F}_k].\) By the inductive hypothesis, $M_{k+1} = M_{k+1}’$ for all outcomes. Therefore: \(\mathbb{E}_{\tilde{\mathbb{P}}}[M_{k+1} | \mathcal{F}_k] = \mathbb{E}_{\tilde{\mathbb{P}}}[M_{k+1}' | \mathcal{F}_k].\) Thus: \(M_k = M_k'.\)

(ii) Let $V_N$ be the payoff at time $N$ of some derivative security. This is a random variable that can depend on all $N$ coin tosses. Define recursively $V_{N - 1}, V_{N - 2}, \dots, V_0$ by the algorithm (1.2.16) of Chapter 1. Show that:

is a martingale under $\tilde{\mathbb{P}}$.

Solution The algorithm defines the value of the derivative security at earlier times as: \(V_n = \frac{1}{1 + r} \tilde{\mathbb{E}}[V_{n+1} | \mathcal{F}_n],\) where:

• $V_n$ is the value of the derivative security at time $n$,
• $\tilde{\mathbb{E}}$ is the expectation under the risk-neutral measure,
• $\mathcal{F}_n$ is the filtration (information up to time $n$),
• $r$ is the risk-free rate.

To show that the given sequence is a martingale under $\tilde{\mathbb{P}}$, we need to verify that:

Start with $V_n$, which is recursively defined as:

\(\frac{V_n}{(1 + r)^n} = \frac{1}{(1 + r)^n} \cdot \frac{1}{1 + r} \tilde{\mathbb{E}}[V_{n+1} | \mathcal{F}_n].\) Simplify the factor $\frac{1}{(1 + r)^n} \cdot \frac{1}{1 + r}$: \(\frac{V_n}{(1 + r)^n} = \frac{1}{(1 + r)^{n+1}} \tilde{\mathbb{E}}[V_{n+1} | \mathcal{F}_n].\) Since conditional expectation is linear and $\frac{V_{n+1}}{(1 + r)^{n+1}}$ is $\mathcal{F}_n$-measurable:

This proves that: \(V_0, \frac{V_1}{1 + r}, \dots, \frac{V_{N-1}}{(1 + r)^{N-1}}, \frac{V_N}{(1 + r)^N}\) is a martingale under the risk-neutral measure $\tilde{\mathbb{P}}$, as the conditional expectation at each step satisfies the martingale property:

(iii) Using the risk-neutral pricing formula:

we define:

Show that:

is a martingale.

Solution

(iv) Conclude that $V_n = V_n’$ for every $n$ (i.e., the algorithm (1.2.16) of Theorem 1.2.2 of Chapter 1 gives the same derivative security prices as the risk-neutral pricing formula (2.4.11)).

\begin{document}

\begin{tikzpicture}

% Tree Nodes
\node (S0) at (0,0) {$S_0$};
\node (S1H) at (2,2.5) {$S_1(H) = uS_0$};
\node (S1T) at (2,-2.5) {$S_1(T) = dS_0$};
\node (S2HH) at (4,4.5) {$S_2(HH) = u^2S_0$};
\node (S2HT) at (4,0) {$S_2(HT) = S_2(TH) = udS_0$};
\node (S2TT) at (4,-4.5) {$S_2(TT) = d^2S_0$};
\node (S3HHH) at (6,5.5) {$S_3(HHH) = u^3S_0$};
\node (S3HHT) at (6,3) {$S_3(HHT) = S_3(HTH) = S_3(THH) = u^2dS_0$};
\node (S3HTT) at (6,-3) {$S_3(HTT) = S_3(THT) = S_3(TTH) = ud^2S_0$};
\node (S3TTT) at (6,-5.5) {$S_3(TTT) = d^3S_0$};

% Connections
\draw[->, thick] (S0) -- (S1H);
\draw[->, thick] (S0) -- (S1T);
\draw[->, thick] (S1H) -- (S2HH);
\draw[->, thick] (S1H) -- (S2HT);
\draw[->, thick] (S1T) -- (S2HT);
\draw[->, thick] (S1T) -- (S2TT);
\draw[->, thick] (S2HH) -- (S3HHH);
\draw[->, thick] (S2HH) -- (S3HHT);
\draw[->, thick] (S2HT) -- (S3HHT);
\draw[->, thick] (S2HT) -- (S3HTT);
\draw[->, thick] (S2TT) -- (S3HTT);
\draw[->, thick] (S2TT) -- (S3TTT);

% Time Labels
\node at (0,-1) {$t = 0$};
\node at (2,-3) {$t = 1$};
\node at (4,-5) {$t = 2$};
\node at (6,-6) {$t = 3$};

\end{tikzpicture}

\end{document}

Solution This can be simply proven via induction: Let’s start with our definitions,

Base Case: $n = N \implies V_N = V_N’$ Inductive Step: We want to show $V_k = V_k’$ for all $k = N - 1, N - 2, \dots, 0$ Assume $V_{k + 1} = V_{k + 1}’$ for $k \le N - 1$

Thus, we have $V_k = V_k’$. Therefore, we’ve shown that $V_n = V_n’$ for all $n = 0, 1, \dots, N$. Therefore, 1.2.16 gives the same derivative security prices as the risk-neutral pricing formula.

Exercise: Stochastic Volatility, Random Interest Rate Consider a two-period stochastic volatility, random interest rate model of the type described in Exercise 1.9, shown in the image.

The stock prices and interest rates are shown in the following:

\begin{document}

\begin{tikzpicture}

% Tree Nodes
\node (S0) at (0,0) {$S_0$};
\node (S1H) at (2,2.5) {$S_1(H) = uS_0$};
\node (S1T) at (2,-2.5) {$S_1(T) = dS_0$};
\node (S2HH) at (4,4.5) {$S_2(HH) = u^2S_0$};
\node (S2HT) at (4,0) {$S_2(HT) = S_2(TH) = udS_0$};
\node (S2TT) at (4,-4.5) {$S_2(TT) = d^2S_0$};
\node (S3HHH) at (6,5.5) {$S_3(HHH) = u^3S_0$};
\node (S3HHT) at (6,3) {$S_3(HHT) = S_3(HTH) = S_3(THH) = u^2dS_0$};
\node (S3HTT) at (6,-3) {$S_3(HTT) = S_3(THT) = S_3(TTH) = ud^2S_0$};
\node (S3TTT) at (6,-5.5) {$S_3(TTT) = d^3S_0$};

% Connections
\draw[->, thick] (S0) -- (S1H);
\draw[->, thick] (S0) -- (S1T);
\draw[->, thick] (S1H) -- (S2HH);
\draw[->, thick] (S1H) -- (S2HT);
\draw[->, thick] (S1T) -- (S2HT);
\draw[->, thick] (S1T) -- (S2TT);
\draw[->, thick] (S2HH) -- (S3HHH);
\draw[->, thick] (S2HH) -- (S3HHT);
\draw[->, thick] (S2HT) -- (S3HHT);
\draw[->, thick] (S2HT) -- (S3HTT);
\draw[->, thick] (S2TT) -- (S3HTT);
\draw[->, thick] (S2TT) -- (S3TTT);

% Time Labels
\node at (0,-1) {$t = 0$};
\node at (2,-3) {$t = 1$};
\node at (4,-5) {$t = 2$};
\node at (6,-6) {$t = 3$};

\end{tikzpicture}

\end{document}

(i) Determine the risk-neutral probabilities:

such that the time-zero value of an option that pays off $V_2$ at time two is given by the risk-neutral pricing formula:

Solution We first want to find the expected value of $S_2$ given the following:

Then using the martingale condition and assuming $(1 + r_0)(1 + r_2) = 1$:

Given our parameters:

And thus, we can directly compute:

Thus our final probability is the following:

(ii) Let $V_2 = (S_2 - 7)^+$. Compute $V_0, V_1(H)$ and $V_(T)$.