Question

If X(t), t>=0 is a Brownian motion process with drift mu and variance sigma squared for which X(0)=0, show that -X(t), t>=0 is a Brownian Motion process with drift negative mu and variance sigma squared.

Answer 1

Solution:

X(t), t>=0 is a Brownian motion process with drift parameter $\mu$ and variance parameter $\sigma^2.$Then, the process exhibits the following properties

1.

Prob X(0) 0 1P-X(0) = 0

2. For s,t∈[0,∞) with s<t, the distribution of X(t)−X(s) is the same as the distribution of X(t−s)

thedistributionof-X (t)-X(sissameas -)X-

3. X has independent increments. That is, for t1,t2,…,tn∈[0,∞) with t1<t2<⋯<tn, the random variables

X(t1), X(t2)−X(t1) ,…, X(tn)−X(tn−1) are independent.

are independent

X(t1),-X (t2)

Hence, -X(t) has independent increments.

4.

X (t)N(ut, ot)e 0, o] X(t)~N(mut, ot)

5. With probability 1,
t(t)
is continuous on
0, ox)
With probability
1,t X(t)
is continuous on [0,∞)

(as X is continuous random variable implies -X is also a continuous random variable)

Hence, -X(t) is a Brownian motion process with drift parameter$-\mu$ and variance parameter $\sigma^2$