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 be a fixed time. Then, the process
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 and , and that their covariance is c. Then, the characteristic function of can be computed as
for all . 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. ⬜
Example 1 A pair of standard normal random variables X,Y which have zero covariance, but is not normal.
As their sum is not normal, X and Y cannot be independent. This example was constructed by setting for some fixed , 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 equal to zero. Continue reading “Independence of Normals”→
I looked at normal random variables in an earlier post but, what does it mean for a sequence of real-valued random variables 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 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 .
Lemma 1 Linear combinations of independent normal random variables are again normal.
Proof: More precisely, if is a sequence of independent normal random variables and are real numbers, then is normal. Let us suppose that has mean and variance . Then, the characteristic function of Y can be computed using the independence property and the characteristic functions of the individual normals,
where we have set and . This is the characteristic function of a normal random variable with mean and variance . ⬜
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 of real-valued random variables is multivariate normal (or joint normal) if and only if all of its finite linear combinations are normal.