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  1. Random
  2. 16. Brownian Motion
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5. The Brownian Bridge

Basic Theory

Definition and Constructions

In the most common formulation, the Brownian bridge process is obtained by taking a standard Brownian motion process \( \bs{X} \), restricted to the interval \( [0, 1] \), and conditioning on the event that \( X_1 = 0 \). Since \( X_0 = 0 \) also, the process is tied down at both ends, and so the process in between forms a bridge (albeit a very jagged one). The Brownian bridge turns out to be an interesting stochastic process with surprising applications, including a very important application to statistics. In terms of a definition, however, we will give a list of characterizing properties as we did for standard Brownian motion and for Brownian motion with drift and scaling.

A Brownian bridge is a stochastic process \( \bs{X} = \{X_t: t \in [0, 1]\} \) with state space \( \R \) that satisfies the following properties:

  1. \( X_0 = 0 \) and \( X_1 = 0 \) (each with probability 1).
  2. \( \bs{X} \) is a Gaussian process.
  3. \( \E(X_t) = 0 \) for \( t \in [0, 1] \).
  4. \( \cov(X_s, X_t) = \min\{s, t\} - s t \) for \( s, \, t \in [0, 1] \).
  5. With probability 1, \( t \mapsto X_t \) is continuous on \( [0, 1] \).

So, in short, a Brownian bridge \( \bs{X} \) is a continuous Gaussian process with \( X_0 = X_1 = 0 \), and with mean and covariance functions given in (c) and (d), respectively. Naturally, the first question is whether there exists such a process. The answer is yes, of course, otherwise why would we be here? But in fact, we will see several ways of constructing a Brownian bridge from a standard Brownian motion. To help with the proofs, recall that a standard Brownian motion process \( \bs{Z} = \{Z_t: t \in [0, \infty)\} \) is a continuous Gaussian process with \( Z_0 = 0 \), \( \E(Z_t) = 0 \) for \( t \in [0, \infty) \) and \( \cov(Z_s, Z_t) = \min\{s, t\} \) for \( s, \, t \in [0, \infty) \). Here is our first construction:

Suppose that \(\bs{Z} = \{Z_t: t \in [0, \infty)\} \) is a standard Brownian motion, and let \( X_t = Z_t - t Z_1 \) for \( t \in [0, 1] \). Then \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge.

Proof:
  1. Note that \( X_0 = Z_0 = 0 \) and \( X_1 = Z_1 - Z_1 = 0 \).
  2. Linear combinations of the variables in \( \bs{X} \) reduce to linear combinations of the variables in \( \bs{Z} \) and hence have normal distributions. Thus \( \bs{X} \) is a Gaussian process.
  3. \( E(X_t) = \E(Z_t) - t \E(Z_1) = 0 \) for \( t \in [0, 1] \)
  4. \( \cov(X_s, X_t) = \cov(Z_s - s Z_1, Z_t - t Z_1) = \cov(Z_s, Z_t) - t \, \cov(Z_s, Z_1) - s \, \cov(Z_1, Z_t) + s t \, \cov(Z_1, Z_1) = \min\{s, t\} - s t - s t + s t \) for \( s, \, t \in [0, 1] \).
  5. \( t \mapsto X_t \) is continuous on \( [0, 1] \) since \( t \mapsto Z_t \) is continuous on \( [0, 1] \).

Let's see the Brownian bridge in action.

Run the simulation of the Brownian bridge process in single step mode a few times.

For the Brownian bridge \( \bs{X} \), note in particular that \( X_t \) is normally distributed with mean 0 and variance \( t (1 - t) \) for \( t \in [0, 1] \). Thus, the variance increases and then decreases on \( [0, 1] \) reaching a maximum of \( 1/4 \) at \( t = 1/2 \). Of course, the variance is 0 at \( t = 0 \) and \( t = 1 \), since \( X_0 = X_1 = 0 \) deterministically.

Open the simulation of the Brownian bridge process. Vary \( t \) and note the change in the probability density function and moments. For various values of \( t \), run the simulation 1000 times and compare the empirical density function and moments to the true density function and moments.

Conversely to the construction above, we can build a standard Brownian motion on the time interval \( [0, 1] \) from a Brownian bridge.

Suppose that \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge, and suppose that \( Z \) is a random variable with a standard normal distribution, independent of \( \bs{X} \). Let \( Z_t = X_t + t Z \) for \( t \in [0, 1] \). Then \( \bs{Z} = \{Z_t: t \in [0, 1]\} \) is a standard Brownian motion on \( [0, 1] \).

Proof:
  1. Note that \( Z_0 = X_0 = 0 \).
  2. Linear combinations of the variables in \( \bs{Z} \) reduce to linear combinations of the variables in \( \bs{X} \) and hence have normal distributions. Thus \( \bs{Z} \) is a Gaussian process.
  3. \( \E(Z_t) = \E(X_t) + t \E(Z) = 0 \) for \( t \in [0, 1] \).
  4. \( \cov(Z_s, Z_t) = \cov(X_s + s Z, X_t + t Z) = \cov(X_s, X_t) + t \, \cov(X_s, Z) + s \, \cov(X_t, Z) + s t \, \var(Z) = \min\{s, t\} - s t + 0 + 0 + s t = \min\{s, t\} \) for \( s, \, t \in [0, 1] \).
  5. \( t \mapsto Z_t \) is continuous on \( [0, 1] \) since \( t \mapsto X_t \) is continuous on \( [0, 1] \).

Here's another way to construct a Brownian bridge from a standard Brownian motion.

Suppose that \( \bs{Z} = \{Z_t: t \in [0, \infty)\} \) is a standard Brownian motion. Define \( X_1 = 0 \) and \[ X_t = (1 - t) Z\left(\frac{t}{1 - t}\right), \quad t \in [0, 1) \] Then \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge.

Proof:
  1. Note that \( X_0 = Z_0 = 0 \) and by definition, \( X_1 = 0 \).
  2. Linear combinations of variables in \( \bs{X} \) reduce to linear combinations of variables in \( \bs{Z} \) and hence have normal distributions. Thus \( \bs{X} \) is a Gaussian process.
  3. For \( t \in [0, 1] \), \[ \E(X_t) = (1 - t) \E\left[Z\left(\frac{t}{1 - t}\right)\right] = 0 \]
  4. If \( s, \, t \in [0, 1) \) with \( s \lt t \) then \( s \big/ (1 - s) \lt t \big/ (1 - t) \) so \[ \cov(X_s, X_t) = \cov\left[(1 - s) Z\left(\frac{s}{1 - s}\right), (1 - t) Z\left(\frac{t}{1 - t}\right)\right] = (1 - s)(1 - t) \frac{s}{1 - s} = s (1 - t) \]
  5. Finally, \( t \mapsto X_t \) is continuous with probability 1 on \( [0, 1) \), and with probability 1, \( X_t = (1 - t) Z\left[t \big/ (1 - t)\right] \to 0 \) as \( t \uparrow 1 \).

Conversely, we can construct a standard Brownian motion from a Brownian bridge.

Suppose that \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge. Define \[ Z_t = (1 + t) X\left(\frac{t}{1 + t}\right), \quad t \in [0, \infty) \] Then \( \bs{Z} = \{Z_t: t \in [0, \infty)\} \) is a standard Brownian motion process.

Proof:
  1. Note that \( Z_0 = X_0 = 0 \)
  2. Linear combinations of the variables in \( \bs{Z} \) reduce to linear combinations of the variables in \( X \), and hence have normal distributions. Thus \( \bs{Z} \) is a Gaussian process.
  3. For \( t \in [0, \infty) \), \[ \E(Z_t) = (1 + t) \E\left[X\left(\frac{t}{1 + t}\right)\right] = 0 \]
  4. If \( s, \, t \in [0, 1] \) with \( s \lt t \) Then \( s \big/ (1 + s) \lt t \big/ (1 + t) \) so \[ \cov(Z_s, Z_t) = \cov\left[(1 + s) X\left(\frac{s}{1 + s}\right), (1 + t) X\left(\frac{t}{1 + t}\right)\right] = (1 + s)(1 + t) \left[\frac{s}{1 + s} - \frac{s}{1 + s}\frac{t}{1 + t}\right] = s \]
  5. Since \( t \mapsto X_t \) is continuous, \( t \mapsto Z_t \) is continuous

We return to the comments at the beginning of this section, on conditioning a standard Brownian motion to be 0 at time 1. Unlike the previous two constructions, note that we are not transforming the random variables, rather we are changing the underlying probability measure.

Suppose that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a standard Brownian motion. Then conditioned on \( X_1 = 0 \), the process \( \{X_t: t \in [0, 1]\} \) is a Brownian bridge process.

Proof:

Part of the argument is based on properties of the multivariate normal distribution. The conditioned process is still continuous and is still a Gaussian process. In particular, suppose that \( s, \, t \in [0, 1] \) with \( s \lt t \). Then \( (X_t, X_1) \) has a joint normal distribution with parameters specified by the mean and covariance functions of \( \bs{X} \). By standard computations, the conditional distribution of \( X_t \) given \( X_1 = 0 \) is normal with mean 0 and variance \( t (1 - t) \). Similarly, the joint distribution of \( (X_s, X_t, X_1) \) is normal with parameters specified by the mean and covariance functions of \( \bs{X} \). Again, by standard computations, the conditional distribution of \( (X_s, X_t) \) given \( X_1 = 0 \) is bivariate normal with 0 means and with \( \cov(X_s, X_t \mid X_1 = 0) = s (1 - t) \).

Finally, the Brownian bridge can be defined in terms a stochastic integral

Suppose that \( \bs{Z} = \{Z_t: t \in [0, \infty)\} \) is standard Brownian motions. Define \( X_1 = 1 \) and \[ X_t = (1 - t) \int_0^t \frac{1}{1 - s} \, dZ_s, \quad t \in [0, 1) \] Then \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge process.

Proof:
  1. Note that \( X_0 = 0 \) and by definition, \( X_1 = 0 \).
  2. Since the integrand in the stochastic integral is deterministic, \( \bs{X} \) is a Gaussian process.
  3. \( \bs{X} \) is continuous on \( [0, 1) \) with probability 1, as a basic property of stochastic integrals. Moreover, \( X_t \to 0 \) as \( t \uparrow 1 \) as a consequence of the martingale inequality.
  4. \( \E(X_t) = 0 \) since the stochastic integral has mean 0.
  5. Suppose that \( s, \, t \in [0, 1] \) with \( s \le t \). Then \[ \cov(X_s, X_t) = \cov\left[(1 - s) \int_0^s \frac{1}{1 - u} \, dZ_u, (1 - t)\left(\int_0^s \frac{1}{1 - u} \, dZ_u + \int_s^t \frac{1}{1 - u} \, dZ_u\right)\right] \] But \( \int_0^s \frac{1}{1 - u} \, dZ_u \) and \( \int_s^t \frac{1}{1 - u} \, dZ_u \) are independent, \[ \cov(X_s, X_t) = (1 - s)(1 - t) \var\left(\int_0^s \frac{1}{1 - u} \, dZ_u\right) \] But then by the Ito isometry, \[ \cov(X_s, X_t) = (1 - s)(1 - t) \int_0^s \frac{1}{(1 - u)^2} \, du = (1 - s)(1 - t) \left(\frac{1}{1 - s} - 1\right) = (1 - t)s \]

In differential form, the process above can be written as \[ d X_t = \frac{X_t}{1 - t} \, dt + dZ_t, \; X_0 = 0 \]

The General Brownian Bridge

The processes constructed above (in several ways!) is the standard Brownian bridge. it's a simple matter to generalize the process so that it starts at \( a \) and ends at \( b \), for arbitrary \( a, \, b \in \R \).

Suppose that \( \bs{Z} = \{Z_t: t \in [0, 1]\} \) is a standard Brownian bridge process. Let \( a, \, b \in \R \) and define \( X_t = (1 - t) a + t b + Z_t\) for \( t \in [0, 1] \). Then \( \bs{X} = \{X_t: t \in [0, 1]\} \) is a Brownian bridge process from \( a \) to \( b \).

Of course, any of the constructions above for standard Brownian bridge can be modified to produce a general Brownian bridge. Here are the characterizing properties.

The Brownian bridge process \( \bs{X} = \{X_t: t \in [0, 1]\} \) from \( a \) to \( b \) is characterized by the following properties:

  1. \( X_0 = a \) and \( X_1 = b \) (each with probability 1).
  2. \( \bs{X} \) is a Gaussian process.
  3. \( \E(X_t) = (1 - t) a + t b \) for \( t \in [0, 1] \).
  4. \( \cov(X_s, X_t) = \min\{s, t\} - s t \) for \( s, \, t \in [0, 1] \).
  5. With probability 1, \( t \mapsto X_t \) is continuous on \( [0, 1] \).

Applications

The Empirical Distribution Function

We start with a problem that is one of the most basic in statistics. Suppose that \( T \) is a real-valued random variable with an unknown distribution. Let \( F \) denote the distribution function of \( T \), so that \( F(t) = \P(T \le t) \) for \( t \in \R \). Our goal is to construct an estimator of \( F \), so naturally our first step is to sample from the distribution of \( T \). This generates a sequence \( \bs{T} = (T_1, T_2, \ldots) \) of independent variables, each with the distribution of \( T \) (and so with distribution function \( F \)). Think of \( \bs{T} \) as a sequence of independent copies of \( T \). For \( n \in \N_+ \) and \( t \in \R \), the natural estimator of \( F(t) \) based on the first \( n \) sample values is \[ F_n(t) = \frac{1}{n}\sum_{i=1}^n \bs{1}(T_i \le t) \] which is simply the proportion of the first \( n \) sample values that fall in the interval \( (-\infty, t] \). Appropriately enough, \( F_n \) is known as the empirical distribution function corresponding to the sample of size \( n \). Note that \( \left(\bs{1}(T_1 \le t), \bs{1}(T_2 \le t), \ldots\right) \) is a sequence of independent, identically distributed indicator variables (and hence is a sequence of Bernoulli trials), and corresponds to sampling from the distribution of \( \bs{1}(T \le t) \). The estimator \( F_n(t) \) is simply the sample mean of the first \( n \) of these variables. The numerator, the number of the original sample variables with values in \( (-\infty, t] \), has the binomial distribution with parameters \( n \) and \( F(t) \). Like all sample means from independent, identically distributed samples, \( F_n(t) \) satisfies some basic and important properties. A summary is given below, but to make sense of some of these facts, you need to recall the mean and variance of the indicator variable that we are sampling from: \( \E\left[\bs{1}(T \le t)\right] = F(t) \), \( \var\left[\bs{1}(T \le t)\right] = F(t)\left[1 - F(t)\right] \)

For fixed \( t \in \R \),

  1. \( \E\left[F_n(t)\right] = F(t) \) so \( F_n(t) \) is an unbiased estimator of \( F(t) \)
  2. \( \var\left[F_n(t)\right] = F(t)\left[1 - F(t)\right] \big/ n \) so \( F_n(t) \) is a consistent estimator of \( F(t) \)
  3. \( F_n(t) \to F(t) \) as \( n \to \infty \) with probability 1, the strong law of large numbers.
  4. \( \sqrt{n}\left[F_n(t) - F(t)\right] \) has mean 0 and variance \( F(t)\left[1 - F(t)\right] \) and converges to the normal distribution with these parameters as \( n \to \infty \), the central limit theorem.

The theorem above gives us a great deal of information about \( F_n(t) \) for fixed \( t \), but now we want to let \( t \) vary and consider the expression in (d), namely \( t \mapsto \sqrt{n}\left[F_n(t) - F(t)\right] \), as a random process for each \( n \in \N_+ \). The key is to consider a very special distribution first.

Suppose that \( T \) has the standard uniform distribution, that is, the continuous uniform distribution on the interval \( [0, 1] \). In this case the distribution function is simply \( F(t) = t \) for \( t \in [0, 1] \), so we have the sequence of stochastic processes \( \bs{X}_n = \left\{X_n(t): t \in [0, 1]\right\} \) for \( n \in \N_+ \), where \[ X_n(t) = \sqrt{n}\left[F_n(t) - t\right] \] Of course, the previous results apply, so the process \( \bs{X}_n \) has mean function 0, variance function \( t \mapsto t(1 - t) \), and for fixed \( t \in [0, 1] \), the distribution \( X_n(t) \) converges to the corresponding normal distribution as \( n \to \infty \). Here is the new bit of information, the covariance function of \( \bs{X}_n \) is the same as that of the Brownian bridge!

\( \cov\left[X_n(s), X_n(t)\right] = \min\{s, t\} - s t \) for \( s, \, t \in [0, 1] \).

Proof:

Suppose that \( s \le t \). From basic properties of covariance, \[ \cov\left[X_n(s), X_n(t)\right] = n \, \cov\left[F_n(s), F_n(t)\right] = \frac{1}{n} \cov\left(\sum_{i=1}^n \bs{1}(T_i \le s), \sum_{j=1}^n \bs{1}(T_j \le t)\right) = \frac{1}{n} \sum_{i=1}^n \sum_{j=1}^n \cov\left[\bs{1}(T_i \le s) \bs{1}(T_j \le t)\right] \] But if \( i \ne j \), the variables \( \bs{1}(T_i \le s) \) and \( \bs{1}(T_j \le t) \) are independent, and hence have covariance 0. On the other hand, \[ \cov\left[\bs{1}(T_i \le s), \bs{1}(T_i \le t)\right] = \P(T_i \le s, T_i \le t) - \P(T_i \le s) \P(T_i \le t) = \P(T_i \le s) - \P(T_i \le s) \P(T_i \le t) = s - st \] hence \[ \cov\left[X_n(s), X_n(t)\right] = \frac{1}{n} \sum_{i=1}^n \cov\left[\bs{1}(T_i \le s), \bs{1}(T_i \le t)\right] = s - s t\]