Conditioning and Independence
Stochastic processes describe how uncertainty unfolds over time. At each date some information has been revealed and some has not, and the quantities we care about, such as the probability of a default or the expected value of a future payoff, must be computed given what is known at that date. Conditioning is the operation that updates probabilities and expectations in light of new information. Independence describes the opposite situation, in which learning one thing tells us nothing about another.
These notes continue the notes on probability basics and use the same setting: a discrete probability space (\Omega, \operatorname{P}) with a finite or countably infinite set of outcomes, where every subset of \Omega is an event. We first define conditional probability and use it to describe independence. We then define the conditional expectation of a random variable given another, describe the information carried by a random variable, and finally define the conditional expectation given an information set. The last notion, together with its properties, is the tool we will use to define martingales.
For more complete treatments, see Blitzstein and Hwang (2019) and Grimmett and Stirzaker (2001).
Conditional Probability
For two events A and B with \operatorname{P}(B) > 0, the conditional probability of A given B is \operatorname{P}(A \mid B) = \frac{\operatorname{P}(A \cap B)}{\operatorname{P}(B)}. Once we learn that B occurred, the outcomes outside B are no longer possible. Conditioning keeps the outcomes in B and rescales their probabilities by 1 / \operatorname{P}(B) so that they add up to one.
Example 1 Suppose we roll a fair six-sided die and define the events A = \{2, 4, 6\} \quad \text{and} \quad B = \{4, 5, 6\}. Then A \cap B = \{4, 6\}, \qquad \operatorname{P}(B) = \frac{3}{6}, \qquad \operatorname{P}(A \cap B) = \frac{2}{6}. Therefore, the conditional probability of A given B is \operatorname{P}(A \mid B) = \frac{\operatorname{P}(A \cap B)}{\operatorname{P}(B)} = \frac{2/6}{3/6} = \frac{2}{3}. Learning that the roll is at least four raises the probability that it is even from 1/2 to 2/3.
Rearranging the definition gives the multiplication rule \operatorname{P}(A \cap B) = \operatorname{P}(A \mid B) \operatorname{P}(B). If the events B_{1}, B_{2}, \ldots form a partition of \Omega, meaning that they are pairwise disjoint and their union is \Omega, then countable additivity and the multiplication rule give the law of total probability, \operatorname{P}(A) = \sum_{k} \operatorname{P}(A \cap B_{k}) = \sum_{k} \operatorname{P}(A \mid B_{k}) \operatorname{P}(B_{k}), where the sum runs over the events with \operatorname{P}(B_{k}) > 0. The probability of A is a weighted average of its conditional probabilities, with weights given by the probabilities of the events in the partition. We will meet this idea again when we average conditional expectations.
Independence
Independent Events
Two events A and B are independent if \operatorname{P}(A \cap B) = \operatorname{P}(A) \operatorname{P}(B). When \operatorname{P}(B) > 0, dividing by \operatorname{P}(B) shows that independence is equivalent to \operatorname{P}(A \mid B) = \operatorname{P}(A): learning that B occurred does not change the probability of A. The definition in terms of the product has the advantage of being symmetric in A and B and of making sense even when one of the events has probability zero.
Example 2 Suppose that the weather tomorrow can be either sunny, fair or rainy. In addition, a certain stock tomorrow can either go up or down in price.
We can define W = \{\text{sunny}, \text{fair}, \text{rainy}\} and S = \{\text{up}, \text{down}\}. The set of outcomes can be described as all possible pairwise combinations of weather tomorrow and the stock price movement. The sample space \Omega is then the Cartesian product of W and S, i.e., \Omega = W \times S.
We can then define the weather events \begin{aligned} \text{Sunny} & = \{(\text{sunny}, \text{up}), (\text{sunny}, \text{down})\}, \\ \text{Fair} & = \{(\text{fair}, \text{up}), (\text{fair}, \text{down})\}, \\ \text{Rainy} & = \{(\text{rainy}, \text{up}), (\text{rainy}, \text{down})\}. \end{aligned}
The table below describes the probabilities for tomorrow’s weather.
| Weather | Sunny | Fair | Rainy |
|---|---|---|---|
| Probability | 0.3 | 0.5 | 0.2 |
Similarly, the stock events can be defined as \begin{aligned} \text{Up} & = \{(\text{sunny}, \text{up}), (\text{fair}, \text{up}), (\text{rainy}, \text{up})\}, \\ \text{Down} & = \{(\text{sunny}, \text{down}), (\text{fair}, \text{down}), (\text{rainy}, \text{down})\}. \\ \end{aligned}
The probabilities of the stock price going up or down are described in the table below.
| Stock | Up | Down |
|---|---|---|
| Probability | 0.6 | 0.4 |
If the weather does not affect the likelihood of the stock going up or down, the conditional probability that the stock goes up should be the same whatever the weather, so that \operatorname{P}(\text{Up} \mid \text{Sunny}) = \operatorname{P}(\text{Up}) = 0.6, and similarly for the other combinations. By the multiplication rule, this means that \operatorname{P}(\text{Stock} \cap \text{Weather}) = \operatorname{P}(\text{Stock}) \operatorname{P}(\text{Weather}), where \text{Stock} is either \text{Up} or \text{Down}, and \text{Weather} is either \text{Sunny}, \text{Fair}, or \text{Rainy}.
The table below describes the combined probabilities of the stock price movement and the weather tomorrow that are consistent with the independence of those events.
| Stock\Weather | Sunny | Fair | Rainy |
|---|---|---|---|
| Up | 0.18 | 0.30 | 0.12 |
| Down | 0.12 | 0.20 | 0.08 |
In each column, the stock goes up 60% of the time and down 40% of the time. The weather does not change the relative proportions of the probabilities for the stock price.
The previous example shows how to generate independent events out of two discrete probability spaces (\Omega_{1}, \operatorname{P}_{1}) and (\Omega_{2}, \operatorname{P}_{2}). If we define \Omega = \Omega_{1} \times \Omega_{2} and let \operatorname{P}(\omega_{1}, \omega_{2}) = \operatorname{P}_{1}(\omega_{1}) \operatorname{P}_{2}(\omega_{2}) for each \omega_{1} \in \Omega_{1} and \omega_{2} \in \Omega_{2}, the pair (\Omega, \operatorname{P}) is a well-defined discrete probability space. In this new probability space, the events A = \{\omega_{1}\} \times \Omega_{2} and B = \Omega_{1} \times \{\omega_{2}\} are independent for any \omega_{1} \in \Omega_{1} and \omega_{2} \in \Omega_{2}.
Proof
We have that \begin{aligned} \operatorname{P}(A) & = \sum_{\omega_{2}' \in \Omega_{2}} \operatorname{P}(\omega_{1}, \omega_{2}') \\ & = \sum_{\omega_{2}' \in \Omega_{2}} \operatorname{P}_{1}(\omega_{1}) \operatorname{P}_{2}(\omega_{2}') \\ & = \operatorname{P}_{1}(\omega_{1}) \sum_{\omega_{2}' \in \Omega_{2}} \operatorname{P}_{2}(\omega_{2}') \\ & = \operatorname{P}_{1}(\omega_{1}). \end{aligned} Similarly, \operatorname{P}(B) = \operatorname{P}_{2}(\omega_{2}). Since A \cap B = \{(\omega_{1}, \omega_{2})\}, we have that \operatorname{P}(A \cap B) = \operatorname{P}_{1}(\omega_{1}) \operatorname{P}_{2}(\omega_{2}) = \operatorname{P}(A) \operatorname{P}(B), proving that A and B are independent.This product construction is how we build models of repeated experiments, such as a sequence of independent coin flips, where the outcome of each flip is unaffected by the others.
Example 3 The sample space \Omega is independent of any event A \subset \Omega, since \operatorname{P}(A \cap \Omega) = \operatorname{P}(A) = \operatorname{P}(A) \operatorname{P}(\Omega). Learning that \Omega occurred, which we knew all along, does not change the probability of A. For the same reason, the empty set is independent of every event.
Independent Random Variables
Two random variables X and Y are independent if the events \{X = x\} and \{Y = y\} are independent for all values x and y. Thus, if X and Y are independent we have that \operatorname{P}(X = x, Y = y) = \operatorname{P}(X = x) \operatorname{P}(Y = y) \quad \text{for all } x, y, or equivalently, the joint probability mass function factors into the product of the marginals, p_{X, Y}(x, y) = p_{X}(x) \, p_{Y}(y). More generally, the random variables X_{1}, X_{2}, \ldots, X_{n} are independent if their joint probability mass function factors as \operatorname{P}(X_{1} = x_{1}, \ldots, X_{n} = x_{n}) = \operatorname{P}(X_{1} = x_{1}) \cdots \operatorname{P}(X_{n} = x_{n}) \quad \text{for all } x_{1}, \ldots, x_{n}. An infinite sequence X_{1}, X_{2}, \ldots is independent if every finite subcollection is independent. The coin flips that drive a random walk are the leading example.
If X and Y are independent, anything we compute from X alone is also independent of anything we compute from Y alone. This fact, and its consequence for expectations, is used constantly in the rest of the course.
Proof
For any values a and b, the event \{f(X) = a\} is the disjoint union of the events \{X = x\} over the values x with f(x) = a, and similarly for \{g(Y) = b\}. Therefore, \begin{aligned} \operatorname{P}(f(X) = a, g(Y) = b) & = \sum_{x : f(x) = a} \; \sum_{y : g(y) = b} p_{X}(x) \, p_{Y}(y) \\ & = \sum_{x : f(x) = a} p_{X}(x) \sum_{y : g(y) = b} p_{Y}(y) \\ & = \operatorname{P}(f(X) = a) \operatorname{P}(g(Y) = b), \end{aligned} so f(X) and g(Y) are independent. For the expectation, \begin{aligned} \operatorname{E}(f(X) g(Y)) & = \sum_{x \in R_{X}} \sum_{y \in R_{Y}} f(x) g(y) \, p_{X}(x) \, p_{Y}(y) \\ & = \sum_{x \in R_{X}} f(x) \, p_{X}(x) \sum_{y \in R_{Y}} g(y) \, p_{Y}(y) \\ & = \operatorname{E}(f(X)) \operatorname{E}(g(Y)). \end{aligned} Taking f(x) = x and g(y) = y gives \operatorname{E}(X Y) = \operatorname{E}(X) \operatorname{E}(Y).The notes on generating functions rely on this result with f(x) = z^{x} and g(y) = z^{y}, which shows that the generating function of a sum of independent random variables is the product of their generating functions.
Equation (1) also implies that independent random variables are uncorrelated. Indeed, \operatorname{Cov}(X, Y) = \operatorname{E}(XY) - \operatorname{E}(X) \operatorname{E}(Y) = \operatorname{E}(X) \operatorname{E}(Y) - \operatorname{E}(X) \operatorname{E}(Y) = 0. As a consequence, the variance of a sum of independent random variables is the sum of their variances, as shown for uncorrelated random variables in the notes on probability basics. The converse is not true: random variables can be uncorrelated without being independent.
Example 4 Consider a random variable X that takes the values \{1, 0, -1\}, each with probability 1 /3. We compute: \operatorname{E}(X) = \frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}(-1) = 0, and \operatorname{E}(X^{3}) = \frac{1}{3}(1^{3}) + \frac{1}{3}(0^{3}) + \frac{1}{3}((-1)^{3}) = 0. Now, define Y = X^{2}. The covariance between X and Y is \operatorname{Cov}(X, Y) = \operatorname{E}(XY) - \operatorname{E}(X)\operatorname{E}(Y) = \operatorname{E}(X^{3}) - \operatorname{E}(X)\operatorname{E}(X^{2}) = 0. However, X and Y are not independent. Since Y = 0 exactly when X = 0, \operatorname{P}(X = 0, Y = 0) = \frac{1}{3} \neq \frac{1}{3} \times \frac{1}{3} = \operatorname{P}(X = 0) \operatorname{P}(Y = 0). The random variables are uncorrelated, yet Y is completely determined by X.
Covariance measures only the linear association between two random variables. In the example, Y depends on X through the symmetric function x^{2}, and positive and negative values of X cancel in the covariance.
Conditional Expectation Given a Random Variable
For x such that p_{X}(x) > 0, the conditional probability mass function of Y given X = x is p_{Y \mid X}(y \mid x) = \frac{p_{X, Y}(x, y)}{p_{X}(x)} = \operatorname{P}(Y = y \mid X = x). The conditional expectation of Y given X = x is the expectation of Y computed with this conditional pmf, \operatorname{E}(Y \mid X = x) = \sum_{y \in R_{Y}} y\, p_{Y \mid X}(y \mid x). This is a number that depends on x. Evaluating it at the observed value of X gives a random variable, denoted \operatorname{E}(Y \mid X), which assigns to each outcome \omega the value \operatorname{E}(Y \mid X = X(\omega)). In words, \operatorname{E}(Y \mid X) is our best forecast of Y once we have observed X, and it is random because it depends on what we observe.
When X and Y are independent the conditional pmf equals the marginal pmf, p_{Y \mid X}(y \mid x) = p_{Y}(y), and therefore \operatorname{E}(Y \mid X) = \operatorname{E}(Y). Observing X does not change our forecast of Y.
A key result in probability theory is the law of iterated expectations: \begin{aligned} \operatorname{E}(\operatorname{E}(Y \mid X)) & = \sum_{x \in R_{X}} \operatorname{E}(Y \mid X = x)\, p_{X}(x) \\ & = \sum_{x \in R_{X}} \sum_{y \in R_{Y}} y\, p_{Y \mid X}(y \mid x)\, p_{X}(x) \\ & = \sum_{x \in R_{X}} \sum_{y \in R_{Y}} y\, p_{X, Y}(x, y) \\ & = \sum_{y \in R_{Y}} y\, p_{Y}(y) \\ & = \operatorname{E}(Y). \end{aligned} The expected value of the conditional expectation equals the expected value of Y itself. Averaging our forecasts over everything we might observe gives back the unconditional forecast, in the same way that the law of total probability averages conditional probabilities.
Example 5 The notes on probability basics used the following joint pmf of a random variable X taking values in \{-1, 0, 1\} and a random variable Y taking values in \{0, 1, 2, 3\}: \begin{array}{c|cccc} X \setminus Y & 0 & 1 & 2 & 3 \\ \hline -1 & 0.12500 & 0.09375 & 0.06250 & 0.03125 \\ 0 & 0.06250 & 0.12500 & 0.12500 & 0.06250 \\ 1 & 0.03125 & 0.06250 & 0.09375 & 0.12500 \end{array} We compute the conditional expectation of Y given X = 1. Adding up the row X = 1, p_{X}(1) = 0.03125 + 0.06250 + 0.09375 + 0.12500 = 0.3125. Dividing each entry of the row by p_{X}(1), the conditional pmf is p_{Y \mid X}(y \mid 1) = 0.1,\; 0.2,\; 0.3,\; 0.4 \quad \text{for } y = 0, 1, 2, 3. Therefore, \operatorname{E}(Y \mid X = 1) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 2. The same calculation for the other rows gives \operatorname{E}(Y \mid X = -1) = 1 and \operatorname{E}(Y \mid X = 0) = 1.5, with p_{X}(-1) = 0.3125 and p_{X}(0) = 0.375. The law of iterated expectations then yields \operatorname{E}(Y) = 1 (0.3125) + 1.5 (0.375) + 2 (0.3125) = 1.5.
Information
Information Generated by a Random Variable
To describe how information arrives over time, we need a way to say what we know after observing a random variable. Consider a sample space with four possible outcomes \Omega = \{ \omega_{1}, \omega_{2}, \omega_{3}, \omega_{4} \} and three random variables X, Y, and Z with the following values:
| Outcome | Probability | X | Y | Z |
|---|---|---|---|---|
| \omega_{1} | 0.10 | -10 | 20 | 15 |
| \omega_{2} | 0.30 | -5 | 10 | -10 |
| \omega_{3} | 0.40 | 5 | 0 | 15 |
| \omega_{4} | 0.20 | 10 | 0 | -10 |
Observing X reveals which outcome occurred, since X takes a different value at each outcome. Observing Y does not: if Y = 0 we only know that either \omega_{3} or \omega_{4} occurred. Observing Z leaves two possible outcomes whatever its value.
Each random variable splits the sample space into the sets on which it takes each of its values. For Y, these sets are \{Y = 20\} = \{\omega_{1}\}, \qquad \{Y = 10\} = \{\omega_{2}\}, \qquad \{Y = 0\} = \{\omega_{3}, \omega_{4}\}, and for Z, \{Z = 15\} = \{\omega_{1}, \omega_{3}\}, \qquad \{Z = -10\} = \{\omega_{2}, \omega_{4}\}. These sets form a partition of \Omega, called the partition generated by the random variable. Observing the random variable tells us which cell of the partition contains the true outcome, and nothing more. We can then decide whether any event made of whole cells has occurred, but not events that split a cell. The collection of all events that are unions of cells is the information set generated by the random variable. We denote it by \mathcal{F}_{Y} for Y, and similarly for other random variables: \mathcal{F}_{Y} = \{ \emptyset, \{\omega_{1}\}, \{\omega_{2}\}, \{\omega_{1}, \omega_{2}\}, \{\omega_{3}, \omega_{4}\}, \{\omega_{1}, \omega_{3}, \omega_{4}\}, \{\omega_{2}, \omega_{3}, \omega_{4}\}, \Omega\}, and \mathcal{F}_{Z} = \{ \emptyset, \{\omega_{1}, \omega_{3}\}, \{\omega_{2}, \omega_{4}\}, \Omega\}. Since X separates all four outcomes, \mathcal{F}_{X} contains every subset of \Omega. A finer partition corresponds to more information, and in this example \mathcal{F}_{Z} \subset \mathcal{F}_{X} and \mathcal{F}_{Y} \subset \mathcal{F}_{X}. Neither \mathcal{F}_{Y} nor \mathcal{F}_{Z} contains the other: knowing Y does not tell us whether \{\omega_{1}, \omega_{3}\} occurred, and knowing Z does not tell us whether \{\omega_{1}\} occurred. In the general theory, the information set generated by a random variable is called the \sigma-algebra generated by it.
Observing several random variables X_{1}, \ldots, X_{n} together generates the partition whose cells are the sets \{X_{1} = x_{1}, \ldots, X_{n} = x_{n}\}, and the corresponding information set is denoted by \mathcal{F}_{X_{1}, \ldots, X_{n}}. For simplicity, in what follows we write \mathcal{F} for an information set generated by a partition of \Omega into cells C_{1}, C_{2}, \ldots with positive probabilities.
A random variable W is known given \mathcal{F}, or \mathcal{F}-measurable, if it is constant on each cell of the partition, so that observing the information in \mathcal{F} reveals the value of W. In the example, Y^{2} is known given \mathcal{F}_{Y} but Z is not, since Z takes the values 15 and -10 on the cell \{Y = 0\} = \{\omega_{3}, \omega_{4}\}. A random variable is known given \mathcal{F}_{Y} exactly when it can be written as a function of Y.
Conditional Expectation Given Information
The conditional expectation \operatorname{E}(Y \mid X) averages Y over the outcomes that are consistent with the observed value of X. The same idea works for any information set. Given \mathcal{F}, we know which cell of the partition contains the outcome, so our best forecast of a random variable Y is its average over that cell.
On each cell, \operatorname{E}(Y \mid \mathcal{F}) equals the expectation of Y computed with the probabilities conditional on that cell. By construction, \operatorname{E}(Y \mid \mathcal{F}) is constant on each cell, so it is known given \mathcal{F}. When \mathcal{F} = \mathcal{F}_{X}, the cells are the sets \{X = x\} and the definition reduces to \operatorname{E}(Y \mid X). When \mathcal{F} = \{\emptyset, \Omega\} contains no information, the only cell is \Omega and \operatorname{E}(Y \mid \mathcal{F}) = \operatorname{E}(Y).
Example 6 Using the table above, we compute \operatorname{E}(X \mid \mathcal{F}_{Z}). The partition generated by Z has two cells, C_{1} = \{\omega_{1}, \omega_{3}\} and C_{2} = \{\omega_{2}, \omega_{4}\}, each with probability 0.5. On each cell we average X: \begin{aligned} \operatorname{E}(X \mid \mathcal{F}_{Z}) & = \frac{(-10)(0.10) + 5 (0.40)}{0.5} = 2 \qquad \text{on } C_{1}, \\ \operatorname{E}(X \mid \mathcal{F}_{Z}) & = \frac{(-5)(0.30) + 10 (0.20)}{0.5} = 1 \qquad \text{on } C_{2}. \end{aligned} If we observe Z = 15 our forecast of X is 2, and if we observe Z = -10 it is 1. Averaging the forecasts over the two cells gives 2 (0.5) + 1 (0.5) = 1.5, which is \operatorname{E}(X) = -1 - 1.5 + 2 + 2 = 1.5, as the law of iterated expectations requires.
The following properties are all we need to work with conditional expectations in the rest of the course. They hold for general information sets, but we prove them only for information sets generated by partitions.
Proof
Fix a cell C of the partition that generates \mathcal{F}. Linearity follows from the linearity of \operatorname{E}(\cdot \, \mathbf{1}_{C}).
If W is known given \mathcal{F}, it takes a constant value w on C, so W Y \mathbf{1}_{C} = w Y \mathbf{1}_{C} and \frac{\operatorname{E}(W Y \mathbf{1}_{C})}{\operatorname{P}(C)} = w \frac{\operatorname{E}(Y \mathbf{1}_{C})}{\operatorname{P}(C)}, which is W \operatorname{E}(Y \mid \mathcal{F}) on C. Taking Y = 1 gives \operatorname{E}(W \mid \mathcal{F}) = W.
If Y is independent of the event C, then \operatorname{P}(Y = y, C) = \operatorname{P}(Y = y) \operatorname{P}(C) for every y, so that \operatorname{E}(Y \mathbf{1}_{C}) = \sum_{y} y \, \operatorname{P}(Y = y, C) = \operatorname{P}(C) \sum_{y} y \, \operatorname{P}(Y = y) = \operatorname{P}(C) \operatorname{E}(Y), and dividing by \operatorname{P}(C) gives \operatorname{E}(Y \mid \mathcal{F}) = \operatorname{E}(Y) on C.
For the tower property, since \mathcal{G} contains less information than \mathcal{F}, each cell D of the partition generating \mathcal{G} is a union of cells C of the partition generating \mathcal{F}. On each such C, \operatorname{E}(Y \mid \mathcal{F}) equals \operatorname{E}(Y \mathbf{1}_{C}) / \operatorname{P}(C), so that \operatorname{E}\left(\operatorname{E}(Y \mid \mathcal{F}) \mathbf{1}_{D}\right) = \sum_{C \subset D} \frac{\operatorname{E}(Y \mathbf{1}_{C})}{\operatorname{P}(C)} \operatorname{P}(C) = \sum_{C \subset D} \operatorname{E}(Y \mathbf{1}_{C}) = \operatorname{E}(Y \mathbf{1}_{D}). Dividing by \operatorname{P}(D) shows that \operatorname{E}(\operatorname{E}(Y \mid \mathcal{F}) \mid \mathcal{G}) = \operatorname{E}(Y \mid \mathcal{G}) on D. Taking \mathcal{G} = \{\emptyset, \Omega\} gives \operatorname{E}(\operatorname{E}(Y \mid \mathcal{F})) = \operatorname{E}(Y).Each property has a simple interpretation. Taking out what is known says that quantities we have already observed behave like constants when we forecast. The independence property says that information unrelated to Y does not change our forecast. The tower property says that if we first forecast with more information and then average that forecast using less information, we obtain the forecast based on less information: today’s forecast of tomorrow’s forecast is today’s forecast.
Filtrations
Information accumulates over time. A sequence of information sets \mathcal{F}_{0} \subset \mathcal{F}_{1} \subset \mathcal{F}_{2} \subset \cdots, where \mathcal{F}_{n} describes what is known at time n, is called a filtration. The inclusions express that nothing is forgotten: every event whose occurrence is known at time m is still known at any later time n > m.
Example 7 Let X_{1}, X_{2}, \ldots be independent coin flips, each equal to +1 or -1 with probability 1/2, and let \mathcal{F}_{n} = \mathcal{F}_{X_{1}, \ldots, X_{n}} be the information generated by the first n flips, with \mathcal{F}_{0} = \{\emptyset, \Omega\}. An infinite sequence of flips has uncountably many outcomes, but any statement about the first n flips only needs the finite sample space \{-1, +1\}^{n}. The cells of the partition generating \mathcal{F}_{n} are the 2^{n} possible sequences of the first n flips. Each flip splits every cell in two, so the partitions become finer and \mathcal{F}_{n} \subset \mathcal{F}_{n + 1}.
The position of the random walk S_{n} = X_{1} + \cdots + X_{n} is known given \mathcal{F}_{n}, whereas the next flip X_{n + 1} is independent of every event in \mathcal{F}_{n}. The properties of conditional expectation then give \operatorname{E}(S_{n + 1} \mid \mathcal{F}_{n}) = \operatorname{E}(S_{n} \mid \mathcal{F}_{n}) + \operatorname{E}(X_{n + 1} \mid \mathcal{F}_{n}) = S_{n} + \operatorname{E}(X_{n + 1}) = S_{n}. Given everything observed up to time n, the best forecast of the position at time n + 1 is the current position.
A process with this property is called a martingale, and it is the subject of the notes on discrete-time martingales. The calculation in the example uses exactly three of the properties above: linearity, taking out what is known, and independence. The tower property enters when we forecast more than one period ahead.