Brownian motion

  • Thread starter theCoker
  • Start date
  • #1
9
0

Homework Statement


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

Homework Equations


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

Answers and Replies

  • #2
392
0
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
9
0
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.
 
  • #4
392
0
No I meant a)
 
  • #5
9
0
a) is stated exactly as my professor posed the question.
 
  • #6
392
0
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 [tex]E(B_{t}|B_{s}=c)-E(t|B_{s}=c)[/tex]
 
  • #7
9
0
From what the Prof. said today (last day of class)... [tex]X_{t}=\sigma B_{t}+\mu t[/tex] where [tex]X_{t}[/tex] is "Brownian motion with drift" and [tex]\mu t[/tex] is the "drift term". It was also said that [tex]X_{t}~Normal(\mu t, \sigma^{2}t)[/tex].
 
  • #8
9
0
Solution
a) [tex]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)[/tex]

b) [tex]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[/tex]

For those that were interested.
 

Related Threads on Brownian motion

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
4
Views
986
Replies
1
Views
844
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
0
Views
2K
Replies
0
Views
2K
Top