Chapter 2 - Probability Theory on Coin Toss Space

Finite Probability Spaces

A finite probability space is a mathematical model for situations where a random experiment has a finite number of possible outcomes. For instance, in the binomial model discussed earlier, a coin was tossed a finite number of times.

Definition: A finite probability space consists of two components: a sample space $\Omega$ and a probability measure $\mathbb{P}$. These are defined as follows:

• Sample Space: $\Omega$ is a non-empty finite set representing all possible outcomes of the experiment.
• Probability Measure: $\mathbb{P}$ is a function that assigns a value in $[0, 1]$ to each element $\omega \in \Omega$ such that:

An event is any subset of $\Omega$. The probability of an event $A \subseteq \Omega$ is defined as:

This setup models the behavior of random experiments. Specifically:

• $\mathbb{P}(\omega)$ is the probability of the outcome $\omega$.
• $\mathbb{P}(A)$ is the probability that the outcome falls within the subset $A$.

If $\mathbb{P}(A) = 0$, the outcome is certain not to be in $A$. Conversely, if $\mathbb{P}(A) = 1$, the outcome is certain to be in $A$. Since every outcome must belong to $\Omega$, we have:

Even outcomes with zero probability can be included in $\Omega$, as they represent events that are theoretically possible but practically impossible.

Key Property: If $A$ and $B$ are disjoint subsets of $\Omega$, the probability of their union is given by:

Definition: Let $(\Omega, \mathbb{P})$ be a finite probability space. A random variable is a real-valued function defined on $\Omega$. (We sometimes also permit a random variable to take the values $+\infty$ and $-\infty$).

In the example of modelling stock prices in the binomial tree, we have written the arguments of $S_0, S_1, S_2$ and $S_3$ as $\omega_1 \omega_2 \omega_3$ even through some of these random variables do not depend on all the coin tosses. In particular, $S_0$ is actually not random because it takes the value $4$, regardless of how the coin tosses turn out; such a random variable is sometimes called a degenerate random variable. We write the argument of a random variable as $\omega$ even when $\omega = \omega_1 \omega_2 \omega_3$. We shall use these two notations interchangeably. It is even more common to write random variables without any arguments.

The distribution of a random variable is a specification of the probabilities that the random variable takes various values. A random variable is not a distribution, and a distribution is not a random variable.

e.g., Toss a coin three times, so the set of possible outcomes is:

Define the random variables:

Let’s define a probability measure on $\Omega$:

Or shorten the notation to $\tilde{\mathbb{P}}(X = j)$ which refers to the probability of a subset of $\Omega$, the set of elements $\omega$ for which $X(\omega) = j$. Under $\mathbb{P}$ the probability that $X$ takes values $0, 1, 2, 3$ are:

where $X$ counts the number of heads in a given 3-toss sequence. We could also list the table of probabilities where $Y$ counts the number of tails:

where $Y$ is the random variable that models the number of tails for a 3-toss sequence of coin tosses $\omega_1 \omega_2 \omega_3$.

Definition: Let $X$ be a random variable defined on a finite probability space $(\Omega, \mathbb{P})$. The expectation (or expected value) of $X$ is defined to be:

When we compute the expectation using the risk-neutral probability measure $\tilde{\mathbb{P}}$, we use the notation:

The variance of $X$ is:

Clearly, expectation is linear, since if $X$ and $Y$ are random variables and $c_1$ and $c_2$ are constants, we have that:

In particular, we have that $l(x) = ax + b$ is a linear function of a dummy variable $x$ ($a$ and $b$ are constants), then $\mathbb{E} [l(X)] = l(\mathbb{E} X)$. When dealing with convex functions, we have the following inequality.

Theorem: Jensen’s Inequality Let $X$ be a random variable on a finite probability space, and let $\phi(x)$ be a convex function of a dummy variable $x$, then:

Proof: We first argue that a convex function is the maximum of all linear functions that lie below it, i.e., for every $x \in \mathbb{R}$:

Since we only consider linear functions that lie below $\phi$, it is clear that:

On the other hand, let $x$ be an arbitrary point in $\mathbb{R}$. Because $\phi$ is convex, there is always a linear function $l$ that lies below $\phi$ and for which $\phi(x) = l(x)$ for this particular $x$. We call this a support line of $\phi$ at $x$: Therefore, we have that:

which gives us:

One consequence of Jensen’s inequality is that:

Conditional Expectations

We chose risk-neutral probabilities $\tilde{p}, \tilde{q}$ for our binomial model:

We use can use these to check that:

Thus, for every time $n$ and sequence $\omega_1 \dots \omega_n$ of coin tosses:

Thus, we say that the stock price at time $n$ is the discounted weighted average of the two possible stock prices at time $n + 1$ where $\tilde{p}, \tilde{q}$ are the weights used in averaging. Moreover, we simplify:

where $\tilde{\mathbb{E}}n [S{n + 1}]$ is the conditional expectation of $S_{n + 1}$ based on the information at time $n$, where the conditional expectation can be regarded as an approximation of the value of $S_{n + 1}$ given the first $n$ coin tosses.

Definition: Let $n$ satisfy $1 \le n \le N$ and let $\omega_1 \dots \omega_n$ be given and for the moment, fixed. There are $2^{N - n}$ possible continuations of $\omega_{n + 1} \dots \omega_N$ of the sequence fixed $\omega_1 \dots \omega_n$. Let $# H(\omega_{n + 1} \dots \omega_N)$ be the number of heads in $\omega_{n + 1} \dots \omega_N$ and $# T(\omega_{n + 1} \dots \omega_N)$ be the number of tails in $\omega_{n + 1} \dots \omega_N$. Now, we define:

where $\tilde{\mathbb{E}}_n [X]$ is the conditional expectation of $X$ based on information at time $n$.

Definition: The two extreme cases of conditioning are $\tilde{\mathbb{E}}_0 [X]$, the conditional expectation of $X$ based on no information, which we define by:

and $\tilde{\mathbb{E}}_N [X]$, the conditional expectation of $X$ based on knowledge of all $N$ coin tosses which we define by:

Theorem: Fundamental Properties of Conditional Expectations Let $N$ be a positive integers and $X, Y$ be random variables depending on the first $N$ coin tosses. Let $0 \le n \le N$ be given. The following properties hold:

e.g., Using $p = \frac{2}{3}, q = \frac{1}{3}$, we can express the Linearity of Conditional Expectations

Similarly, we can show that:

e.g., Iterated Conditioning. We first estimate $S_3$ based on the information at time 2:

And now we can estimate, based on the information at time 1:

The estimate of the estimate is an average of averages, and it is not surprising that we can get the same result by a more comprehensive averaging. This more comprehensive averaging occurs when we estimate $S_3$ directly based on the information at time 1:

e.g., Independence. The quotient $\frac{S_2}{S_1}$ takes either the value 2 or $\frac{1}{2}$ depending on whether the second coin toss results in head or tails, respectively. In particular, $\frac{S_2}{S_1}$ does not depend on the first coin toss. We complete:

Martingales

expresses the fact that under the risk-neutral measure, for a stock that pays no dividend, the best estimate based on the information at time $n$ of the value of the discounted stock price at time $n + 1$ is the discounted stock price at time $n$. The risk-neutral probabilities are chosen to enforce this. Moreover, processes that satisfy this condition are called martingales.

Definition: Martingale Consider a binomial asset pricing model. Let $M_0, M_1, \dots, M_N$ be a sequence of random variables, with each $M_n$ depending only on the first $n$ coin tosses ($M_0$ is constant). We call this stochastic process adapted:

Theorem: Expectation of a martingale is constant over time. i.e., If $M_0, M_1, \dots, M_N$ is a martingale, then we have that:

Indeed, taking expectations of both side gives that:

Theorem Consider the general binomial model with $0 < d < 1 + r < u$. Let the risk-neutral probabilities be given by:

Then, under the risk-neutral measure, the discounted stock price is a martingale, i.e., the following equation:

holds at every time $n$ and for every sequence of coin tosses.

Proof:

Key Concept: When an investor who, at each time $n$, takes a position of $\Delta_n$ shares of stock and holds this position until $n + 1$, when he takes a new position of $\Delta_{n + 1}$ shares, the portfolio variable $\Delta_n$ may depend on the first $n$ coin tosses and $\Delta_{n + 1}$ may depend on the first $n + 1$ coin tosses. In other words:

is an adapted process in the sense that, for each $\Delta_n$ for $n = 0, 1, \dots, N - 1$, depends only on the first $n$ coin tosses. Moreover, if the investor begins with initial wealth $X_0$ and $X_n$, which denotes their wealth at each time $n$, then the evolution of his wealth is governed by the wealth equation:

Note: Each $X_n$ depends only on the first $n$ coin tosses.

Theorem: Consider the binomial model with $N$ periods. Let $\Delta_0, \Delta_1, \dots, \Delta_{N - 1}$ be an adapted portfolio process, let $X_0$ be a real number (constant), and let the wealth process $X_1, \dots, X_N$ be generated recursively by:

Then, the discounted wealth process:

is a martingale under the risk-neutral measure, i.e.,

Proof:

Corollary: Under the conditions of the theorem above:

The expected value of a martingale cannot change with time and so must always be equal to the time-zero value of the martingale. Applying this to our $\tilde{\mathbb{P}}$-martingale $\frac{X_n}{(1 + r)^n}$ for $n = 0, 1, \dots, N$, we can get the equation above.

Another consequence of the Theorem above is another version of the risk-neutral pricing formula. Define $V_N$ to be a random variable that represents the value of a derivative security paying off at time $N$ depending on the first $N$ coin tosses. We know that there is initial wealth $X_0$ and a replicating portfolio process $\Delta_0, \dots, \Delta_{N - 1}$ that generates a wealth process $X_1, \dots, X_N$ satisfying $X_n = V_N$, no matter how the coin tossing turns out. Because $\frac{X_n}{(1 + r)^n}$ for $n = 0, 1, \dots, N$ is a martingale, the multi-step ahead property shown previously implies:

However, we may have derivative securities, such as bonds, that do NOT pay off on a single date, rather, make series of payments. For such a security, we have the following pricing and hedging formulas:

Theorem: Consider an $N$-period binomial asset pricing-model with $0 < d < 1 + r < u$ and with risk-neutral probability measure $\tilde{\mathbb{P}}$. Let $C_0, C_1, \dots, C_N$ be a sequence of random variables such that each $C_n$ depends only on $\omega_n, \dots, \omega_n$. The price at time $n$ of the derivative security that makes payments $C_n, \dots, C_N$ at time $n, \dots, N$, respectively, is:

The price process:

We define the Delta-Hedging Formula as:

where $n$ ranges between $0$ and $N - 1$. If we set $X_0 = V_0$ and define recursively forward in time the portfolio values $X_1, X_2, \dots X_N$ by:

then we have:

for all $n$ and all $\omega_1, \dots, \omega_n$.

We call $V_n$ the Net Present Value at time $n$ of the sequence of payments $C_n, \dots, C_N$. It’s just the sum of the value $\tilde{\mathbb{E}}_n \left[ \frac{C_k}{(1 + r)^{k - n}} \right]$ of each of the payments $C_k$ to be made at times $k = n, k = n + 1, \dots, k = N$. Note, that the payment at time $n$ is included. This payment $C_n$, depends on only the first $n$ tosses and so can be taken outside the conditional expectation $\tilde{\mathbb{E}}_n$ such that:

In the case that $n = N$, we have that:

An agent short cash flows $C_0, \dots, C_n$ hedges the payments by investing in a stock and a money market account. At each time $n$, before making the payment $C_n$, the portfolio value is $X_n$. After the payment, the agent takes a stock position $\Delta_n$ according to the Delta Hedging formula, ensuring the portfolio evolves to $X_{n+1}$ before $C_{n+1}$. Starting with $X_0 = V_0$, this strategy ensures $X_N = V_N = C_N$ at the final time $N$, allowing the agent to make the last payment $C_N$ and reduce the portfolio to zero, achieving a perfect hedge.

Proof: We can proceed by induction on $n$ and show that the by the hypothesis $X_n (\omega_1 \dots \omega_n) = V_n (\omega_1 \dots \omega_n)$ for some $n = 0, 1, \dots, N - 1$ and all $\omega_1 \dots \omega_n$.

Via iterated conditioning, we have that:

Markov Processes

Definition Consider the binomial asset pricing model. Let $X_0, X_1, \dots, X_N$ be an adapted process. If, for every $n$ between $0$ and $N - 1$ and for every function $f(x)$ there is another function $g(x)$ (depending on $n$ and $f$) each that:

we say that $X_0, X_1, \dots, X_N$ is a Markov process. In other words, the information about the coin tosses one needs in order to evaluate $\mathbb{E}n [f(X{n + 1})]$ is summarized by $X_n$. Thus, the expected value of some function of the next state $f(X_{n + 1})$ can be entirely determined by applying a function $g$ to the current state $X_n$. Thus to predict $X_{n + 1}$ you only need to know $X_n$.

e.g., Stock Price The stock price at time $n + 1$ is given in terms of the stock price at time $n$ by:

The conditional expectation is:

which depends only on $\omega_1 \dots \omega_n$ via $S_n (\omega_1 \dots \omega_n)$. Therefore:

where $g(x) = pf(ux) + qf(dx)$. This shows that the stock price process is Markov under both the risk-neutral and actual probability measures. To price a derivative security with payoff $V_N = v_N(S_N)$, the risk-neutral pricing formula is used:

but $V_N = v_N(S_N)$ and the stock price process is Markov:

for some function $v_{N - 2}$. Generally, $V_n = v_n(S_n)$ for some function $v_n$.

The martingale property is the special case of:

Moreover, we require that for every $f$ there is a corresponding $g$ such that the expectation above holds. The Markov property requires only that:

for some function $g$ and does not require that the function $g$ be given by $g(x) = x$. Thus, NOT every Markov process is a martingale.

Lemma: (Independence) In the $N$-period binomial asset pricing model, let $n$ be an integer between 0 and $N$. Suppose the random variables $X^1, \dots, X^K$ depend only on coin tosses 1 through $n$ and the random variables $Y^1, \dots, Y^L$ depend only on coin tosses $n + 1$ through $N$. Let $f(x^1, \dots, x^K, y^1, \dots, y^L)$ be a function of dummy variables $x^1, \dots, x^K$ and $y^1, \dots, y^L$ and define:

Assume that $K = L = 1$ then $g(x) = \mathbb{E} f(x, Y)$. Then,

and $\mathbb{E}_n [f(X, Y)] = g(X)$. We hold $f$ constant by replacing it with random variable $X$ by an arbitrary fixed dummy variable $x$ and compute the conditional expectation of $f(x, Y)$.

e.g., Consider the maximum-to-date process $M_n = \max{S_k}_{0 \le k \le n}$ with $p = \frac{2}{3}, q = \frac{1}{3}$.

\begin{document}

\begin{tikzpicture}

% Tree Nodes
\node (M0) at (0,0) {$M_0 = 4$};
\node (M1H) at (2,2.5) {$M_1(H) = 8$};
\node (M1T) at (2,-2.5) {$M_1(T) = 4$};
\node (M2HH) at (4,4.5) {$M_2(HH) = 16$};
\node (M2HT) at (4,0) {$M_2(HT) = 8$};
\node (M2TH) at (4,-1) {$M_2(TH) = 4$};
\node (M2TT) at (4,-4.5) {$M_2(TT) = 4$};
\node (M3HHH) at (6,5.5) {$M_3(HHH) = 32$};
\node (M3HHT) at (6,3) {$M_3(HHT) = 16$};
\node (M3HTH) at (6,1) {$M_3(HTH) = 8$};
\node (M3HTT) at (6,-1) {$M_3(HTT) = 8$};
\node (M3THH) at (6,-3) {$M_3(THH) = 8$};
\node (M3THT) at (6,-4.5) {$M_3(THT) = 4$};
\node (M3TTT) at (6,-6.5) {$M_3(TTT) = 4$};

% Connections
\draw[->, thick] (M0) -- (M1H);
\draw[->, thick] (M0) -- (M1T);
\draw[->, thick] (M1H) -- (M2HH);
\draw[->, thick] (M1H) -- (M2HT);
\draw[->, thick] (M1T) -- (M2TH);
\draw[->, thick] (M1T) -- (M2TT);
\draw[->, thick] (M2HH) -- (M3HHH);
\draw[->, thick] (M2HH) -- (M3HHT);
\draw[->, thick] (M2HT) -- (M3HTH);
\draw[->, thick] (M2HT) -- (M3HTT);
\draw[->, thick] (M2TH) -- (M3THH);
\draw[->, thick] (M2TH) -- (M3THT);
\draw[->, thick] (M2TT) -- (M3TTT);

% Time Labels
\node at (0,-1) {$t = 0$};
\node at (2,-3.5) {$t = 1$};
\node at (4,-5.5) {$t = 2$};
\node at (6,-7.5) {$t = 3$};

\end{tikzpicture}

\end{document}

Clearly, we have that:

Since $M_2(TH) = M_2(TT) = 4$ there cannot be a function $g$ such that we have that:

The maximum-to-date process is not Markov because recording only that the value of the maximum-to-date at time two is 4, without recording the value of the stock price at time two, neglects information relevant to the evolution of the maximum-to-date process after time two.

Recovering a Markov property: We add one or more state variables, recovering the Markov property, having determined a way to describe the state of the market in terms of these variables.

Definition: Consider the binomial asset pricing model. Let ${(X^1_n, \dots, X_n^K);\quad n = 0, 1, \dots, N}$ be a $K$-dimensional adapted process, i.e., $K$ one-dimensional adapted processes. If, for every $n$ between $0$ and $N - 1$, and for every function $f(x^1, \dots, x^K)$ there is another function $g(x^1, \dots, x^K)$ (depending on $n$ and $f$) such that:

we say that ${(X^1_n, \dots, X_n^K);\quad n = 0, 1, \dots, N}$ is a $K$-dimensional Markov process.

Remark: If $X_0, X_1, \dots, X_N$ is a Markov process and $n \le N - 2$, then the one-step ahead Markov property implies that for every function $h$, there is a function $f$ such that:

Then we can take conditional expectations on both sides based on the information at time $n$, and using iterated conditioning, we get:

For our binomial pricing model suppose we have a Markov process $X_0, X_1, \dots, X_N$ under the risk-neutral probability measure $\tilde{\mathbb{P}}$ and we have a derivative security whose payoff $V_N$ at time $N$ is a function $v_N$ of $X_N$ i.e., $V_N = v_N(X_N)$. The difference between $V_N$ and $v_N$ is that the argument of the former sequence of coin tosses, whereas the argument of the latter is a real number.

In place of $x$ we substitute the random variable $X_N$ to get:

Getting the risk-neutral pricing formula says that the price of this derivative security at earlier times $n$ is:

Thus we get that the price of the derivative security at time $n$ is a function of $X_n$:

An $N$-period binomial model with derivative security whose payoff at time $N$ is a function $v_N(S_N, M_N)$ of the stock price and the maximum stock price.

Take $V_N = (M_N - K)^{+}$. The stock price does not appear in $V_N$ we need to execute the pricing algorithm below because the maximum-to-date process is NOT Markov by itself. Thus, for any $n$ between $0$ and $N$, there is a non-random function $v_n(s, m)$ such that the price of the option at time $n$ is:

Theorem Let $X_0, X_1, \dots, X_N$ be a Markov process under the risk-neutral probability measure $\tilde{\mathbb{P}}$ in the binomial model. Let $V_N(x)$ be a function of the dummy variable $x$ and consider a derivative security whose payoff at time $N$ is $v_N(X_N)$. Then for each $n$ between $0$ and $N$, the price $V_n$ of this derivative security is some function $v_n$ of $X_n$,

There is a recursive algorithm for computing $v_n$ whose exact formula depends on the underlying Markov process $X_0, X_1, \dots, X_N$. Analogous results hold if the underlying Markov process is multi-dimensional.

Summary of Key Results, Theorems, and Algorithms

Finite Probability Spaces

1. Definition: A finite probability space consists of:
   • Sample Space ($\Omega$): A finite set of outcomes.
   • Probability Measure ($\mathbb{P}$): Assigns probabilities to outcomes such that $\sum_{\omega \in \Omega} \mathbb{P}(\omega) = 1$.
2. Probability of an Event:

1. Key Property: For disjoint subsets $A, B \subseteq \Omega$:

1. Expectation of a Random Variable:

1. Variance:

1. Jensen’s Inequality: For a convex function $\phi$ and random variable $X$:

Conditional Expectations

1. Conditional Expectation: \(\tilde{\mathbb{E}}_n[X] = \sum_{\omega_{n+1} \dots \omega_N} \tilde{p}^{\# H}\tilde{q}^{\# T} X(\omega_1 \dots \omega_N)\)

2. Key Properties:
   • Linearity: $\mathbb{E}_n[c_1X + c_2Y] = c_1\mathbb{E}_n[X] + c_2\mathbb{E}_n[Y]$
   • Taking Out What is Known: If $X$ depends only on the first $n$ coin tosses, $\mathbb{E}_n[XY] = X \cdot \mathbb{E}_n[Y]$.
   • Iterated Conditioning: $\mathbb{E}_n[\mathbb{E}_m[X]] = \mathbb{E}_n[X]$ for $n \leq m$.
   • Independence: If $X$ depends on tosses $n+1$ to $N$, $\mathbb{E}_n[X] = \mathbb{E}[X]$.
3. Risk-Neutral Stock Pricing: \(S_n = \frac{1}{1 + r} \tilde{\mathbb{E}}_n[S_{n+1}]\)

Martingales

1. Definition: A process $M_n$ is a martingale if: \(M_n = \mathbb{E}_n[M_{n+1}]\)

2. Properties:
   • Expectation of a martingale is constant over time: $\mathbb{E}[M_n] = M_0$.
   • Risk-neutral discounted stock price is a martingale: \(\frac{S_n}{(1 + r)^n} = \tilde{\mathbb{E}}_n\left[\frac{S_{n+1}}{(1 + r)^{n+1}}\right]\)

3. Wealth Process: \(X_{n+1} = \Delta_n S_{n+1} + (1 + r)(X_n - \Delta_n S_n)\) The discounted wealth process is also a martingale.

4. Risk-Neutral Pricing: \(V_n = \tilde{\mathbb{E}}_n\left[\frac{V_N}{(1 + r)^{N-n}}\right]\)

Markov Processes

1. Definition: A process $X_n$ is Markov if: \(\mathbb{E}_n[f(X_{n+1})] = g(X_n)\)

2. Stock Price: The stock price process is Markov: \(S_{n+1} = \begin{cases} uS_n, & \text{if } \omega_{n+1} = H, \\ dS_n, & \text{if } \omega_{n+1} = T. \end{cases}\)

3. Pricing with Markov Processes: \(V_n = v_n(S_n)\) where $v_n$ satisfies the recursive relation: \(v_n(s) = \frac{1}{1 + r}[\tilde{p} v_{n+1}(us) + \tilde{q} v_{n+1}(ds)]\)