# Brownian motion

1. May 7, 2009

### theCoker

1. The problem statement, all variables and given/known data
Let Bt be a standard Brownian motion. Let s<t:
a) Compute $$P(\sigma B_{t}+\mu t|B_{s}=c)$$
b) Compute $$E(B_{t}-t|B_{s}=c)$$

2. Relevant 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

3. 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?

Last edited: May 7, 2009
2. May 7, 2009

### 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) ?

3. May 7, 2009

### theCoker

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.

4. May 7, 2009

### Billy Bob

No I meant a)

5. May 7, 2009

### theCoker

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

6. May 7, 2009

### 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 $$E(B_{t}|B_{s}=c)-E(t|B_{s}=c)$$

7. May 7, 2009

### 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 $$X_{t}~Normal(\mu t, \sigma^{2}t)$$.

8. May 11, 2009

### 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.