Question

6. Let tı < t2< ts and let B(t) be standard Brownian motion. What is the distribution of B(ti) B(t2)+ B(t3)?

Answer 1

We know that,

E[B(t)] = 0

Var[B(t)] = t

Cov(B(s),B(t))=min(s,t), for all s,t.

Let X = B(t1) + B(t2) + B(t3)

Since B(t) is a Gaussian process (Standard Brownian Motion), X is a normal random variable.

E[X] = E[B(t1) + B(t2) + B(t3)] = E[B(t1)] + E[B(t2)] + E[B(t3)]

= 0 + 0 + 0

= 0

Var[X] = Var[B(t1) + B(t2) + B(t3)]

= Var[B(t1)] + Var[B(t2)] + Var[B(t3)] + 2 Cov[B(t1) * B(t2)] + 2 Cov[B(t1) * B(t3)] + 2 Cov[B(t2) * B(t3)]

= t1 + t2 + t3 + 2 min(t1, t2) + 2 min(t1, t3) + 2 min(t2, t3)

= t1 + t2 + t3 + 2t1 + 2t1 + 2t2

= 5t1 + 3t2 + t3

Thus,

the distribution of B(t1) + B(t2) + B(t3) is a Normal distribution with mean = 0 and variance = 5t1 + 3t2 + t3