# 18.3: 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 $\cov\left[X_n(s), X_n(t)\right] = \frac{1}{n} \sum_{i=1}^n \cov\left[\bs{1}(T_i \le s), \bs{1}(T_i \le t)\right] = s - s t$