Brownian Motion Homework: Computing Probability & Expectation

[theCoker]

Homework Statement

Let B_t be a standard Brownian motion. Let s<t:
a) Compute P(\sigma B_{t}+\mu t|B_{s}=c)
b) Compute E(B_{t}-t|B_{s}=c)

Homework Equations

Defition of brownian motion: B(t) is a (one-dim) brownian motion with variance \sigma^{2}if it satisfies the following conditions:
(a) B(0)=0
(b) independent increments
(c) stationary increments
(d) B(t)~normal(0,\sigma^{2}t)
(e) t\rightarrow B_{t} is continous

The Attempt at a Solution

I know the policy is the attempt to do the problem, but I don't even know where to start. Maybe the definition of conditional probability?

 

[Billy Bob]

Hmmm, is part (a) stated correctly?

Just thoughts here:

given B_s = c, can't you think of the start time as s and starting position as c.

Then isn't B_t - c ~ N(0,t-s), in other words B_t ~ N(c,t-s) ?

 

[theCoker]

Hmmm, is part (a) stated correctly?

assuming you are referring to (a) and not a):

(a) B(0)=0, *a convenient normalization.

sorry about that. thanks for the thoughts. assignment is due now =[

however, i am puzzled by this problem and would appreciate more thoughts.

 

[Billy Bob]

No I meant a)

 

[theCoker]

a) is stated exactly as my professor posed the question.

 

[Billy Bob]

It's just weird to me. It's like asking "what is P(Z+2)" instead of a sensible question like "what is P(Z>2)."

For b), can you find E(B_{t}|B_{s}=c)-E(t|B_{s}=c)

 

[theCoker]

From what the Prof. said today (last day of class)... X_{t}=\sigma B_{t}+\mu t where X_{t} is "Brownian motion with drift" and \mu t is the "drift term". It was also said that X_{t}~Normal(\mu t, \sigma^{2}t).

 

[theCoker]

Solution
a) P(\sigma B_{t}+\mu t=a|B_{s}=c)=P(B_{t}=\frac{a-\mu t}{\sigma}|B_{s}=c)=\frac{P(B_{t}=\frac{a-\mu t}{\sigma},B_{s}=c)}{P(B_{s}=c)}=P(B_{t}-B_{s}=\frac{a-\mu t}{\sigma}-c=f_{t-s}(\frac{a-\mu t}{\sigma}-c)

b) E(B_{t}-t|B_{s}=c)=E(B_{t}-B_{s}+B_{s}|B_{s}=c]-t=E(B_{t}-B_{s}|B_{s}=c)+c-t=c-t

For those that were interested.