# Brownian Bridges

A Brownian bridge can be defined as standard Brownian motion conditioned on hitting zero at a fixed future time T, or as any continuous process with the same distribution as this. Rather than conditioning, a slightly easier approach is to subtract a linear term from the Brownian motion, chosen such that the resulting process hits zero at the time T. This is equivalent, but has the added benefit of being independent of the original Brownian motion at all later times.

Lemma 1 Let X be a standard Brownian motion and ${T > 0}$ be a fixed time. Then, the process

 $\displaystyle B_t = X_t - \frac tTX_T$ (1)

over ${0\le t\le T}$ is independent from ${\{X_t\}_{t\ge T}}$.

Proof: As the processes are joint normal, it is sufficient that there is zero covariance between them. So, for times ${s\le T\le t}$, we just need to show that ${{\mathbb E}[B_sX_t]}$ is zero. Using the covariance structure ${{\mathbb E}[X_sX_t]=s\wedge t}$ we obtain,

 $\displaystyle {\mathbb E}[B_sX_t]={\mathbb E}[X_sX_t]-\frac sT{\mathbb E}[X_TX_t]=s-\frac sTT=0$

as required. ⬜

This leads us to the definition of a Brownian bridge.

Definition 2 A continuous process ${\{B_t\}_{t\in[0,T]}}$ is a Brownian bridge on the interval ${[0,T]}$ if and only it has the same distribution as ${X_t-\frac tTX_T}$ for a standard Brownian motion X.

In case that ${T=1}$, then B is called a standard Brownian bridge.

There are actually many different ways in which Brownian bridges can be defined, which all lead to the same result.

• As a Brownian motion minus a linear term so that it hits zero at T. This is definition 2.
• As a Brownian motion X scaled as ${tT^{-1/2}X_{T/t-1}}$. See lemma 9 below.
• As a joint normal process with prescribed covariances. See lemma 7 below.
• As a Brownian motion conditioned on hitting zero at T. See lemma 14 below.
• As a Brownian motion restricted to the times before it last hits zero before a fixed positive time T, and rescaled to fit a fixed time interval. See lemma 15 below.
• As a Markov process. See lemma 13 below.
• As a solution to a stochastic differential equation with drift term forcing it to hit zero at T. See lemma 18 below.

There are other constructions beyond these, such as in terms of limits of random walks, although I will not cover those in this post. Continue reading “Brownian Bridges”

# Independence of Normals

A well known fact about joint normally distributed random variables, is that they are independent if and only if their covariance is zero. In one direction, this statement is trivial. Any independent pair of random variables has zero covariance (assuming that they are integrable, so that the covariance has a well-defined value). The strength of the statement is in the other direction. Knowing the value of the covariance does not tell us a lot about the joint distribution so, in the case that they are joint normal, the fact that we can determine independence from this is a rather strong statement.

Theorem 1 A joint normal pair of random variables are independent if and only if their covariance is zero.

Proof: Suppose that X,Y are joint normal, such that ${X\overset d= N(\mu_X,\sigma^2_X)}$ and ${Y\overset d=N(\mu_Y,\sigma_Y^2)}$, and that their covariance is c. Then, the characteristic function of ${(X,Y)}$ can be computed as

 \displaystyle \begin{aligned} {\mathbb E}\left[e^{iaX+ibY}\right] &=e^{ia\mu_X+ib\mu_Y-\frac12(a^2\sigma_X^2+2abc+b^2\sigma_Y^2)}\\ &=e^{-abc}{\mathbb E}\left[e^{iaX}\right]{\mathbb E}\left[e^{ibY}\right] \end{aligned}

for all ${(a,b)\in{\mathbb R}^2}$. It is standard that the joint characteristic function of a pair of random variables is equal to the product of their characteristic functions if and only if they are independent which, in this case, corresponds to the covariance c being zero. ⬜

To demonstrate necessity of the joint normality condition, consider the example from the previous post.

Example 1 A pair of standard normal random variables X,Y which have zero covariance, but ${X+Y}$ is not normal.

As their sum is not normal, X and Y cannot be independent. This example was constructed by setting ${Y={\rm sgn}(\lvert X\rvert -K)X}$ for some fixed ${K > 0}$, which is standard normal whenever X is. As explained in the previous post, the intermediate value theorem ensures that there is a unique value for K making the covariance ${{\mathbb E}[XY]}$ equal to zero. Continue reading “Independence of Normals”

# Multivariate Normal Distributions

I looked at normal random variables in an earlier post but, what does it mean for a sequence of real-valued random variables ${X_1,X_2,\ldots,X_n}$ to be jointly normal? We could simply require each of them to be normal, but this says very little about their joint distribution and is not much help in handling expressions involving more than one of the ${X_i}$ at once. In case that the random variables are independent, the following result is a very useful property of the normal distribution. All random variables in this post will be real-valued, except where stated otherwise, and we assume that they are defined with respect to some underlying probability space ${(\Omega,\mathcal F,{\mathbb P})}$.

Lemma 1 Linear combinations of independent normal random variables are again normal.

Proof: More precisely, if ${X_1,\ldots,X_n}$ is a sequence of independent normal random variables and ${a_1,\ldots,a_n}$ are real numbers, then ${Y=a_1X_1+\cdots+a_nX_n}$ is normal. Let us suppose that ${X_k}$ has mean ${\mu_k}$ and variance ${\sigma_k^2}$. Then, the characteristic function of Y can be computed using the independence property and the characteristic functions of the individual normals,

 \displaystyle \begin{aligned} {\mathbb E}\left[e^{i\lambda Y}\right] &={\mathbb E}\left[\prod_ke^{i\lambda a_k X_k}\right] =\prod_k{\mathbb E}\left[e^{i\lambda a_k X_k}\right]\\ &=\prod_ke^{-\frac12\lambda^2a_k^2\sigma_k^2+i\lambda a_k\mu_k} =e^{-\frac12\lambda^2\sigma^2+i\lambda\mu} \end{aligned}

where we have set ${\mu_k=\sum_ka_k\mu_k}$ and ${\sigma^2=\sum_ka_k^2\sigma_k^2}$. This is the characteristic function of a normal random variable with mean ${\mu}$ and variance ${\sigma^2}$. ⬜

The definition of joint normal random variables will include the case of independent normals, so that any linear combination is also normal. We use use this result as the defining property for the general multivariate normal case.

Definition 2 A collection ${\{X_i\}_{i\in I}}$ of real-valued random variables is multivariate normal (or joint normal) if and only if all of its finite linear combinations are normal.