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Brownian motion

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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

    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. jcsd
  3. May 7, 2009 #2
    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) ?
     
  4. May 7, 2009 #3
    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.
     
  5. May 7, 2009 #4
    No I meant a)
     
  6. May 7, 2009 #5
    a) is stated exactly as my professor posed the question.
     
  7. May 7, 2009 #6
    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]
     
  8. May 7, 2009 #7
    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].
     
  9. May 11, 2009 #8
    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.
     
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