Solution 1

ORF 474 Problem 1.

Problem Set 1

Solution

a. We show this by checking the conditions of the definition in lecture notes 3. ˜ (1) From definition, W0 = Ws − Ws = 0. ˜ ˜ (2) For any 0 ≤ s1 < s2 , the increment Ws2 − Ws1 = (Ws+s2 − Ws ) − (Ws+s1 − Ws ) = Ws+s2 − Ws+s1 . Since W is a Brownian Motion, it has the Gaussian distribution N (0, s2 − s1 ). ˜ (3) For any k > 0 and 0 ≤ s1 < s2 < · · · < sk , the increments Ws2 − ˜ ˜ ˜ Ws1 , . . . , Wsk − Wsk−1 equals Ws+s2 − Ws+s1 , . . . , Ws+sk − Ws+sk−1 (same reason as above). From the independent increment property of the original Brownian Motion W , we know they are independent as well. ˜ (4) The path t → Wt , i.e. the path t → (Wt+s − Ws ) is continuous. This is because as function of t, the term Ws can be treated as a constant, and t + s is a continuous function of t. ˜ All in all, W is a Brownian Motion. b. For simplicity, we only show the second part of the definition holds true. For any ˜ ˜ 0 ≤ s1 < s2 , the increment Ws2 − Ws1 = cWs2 /c2 − cWs1 /c2 . Ws2 /c2 − Ws1 /c2 has Gaussian distribution N (0, (s2 − s1 )/c2 ), so cWs2 /c2 − cWs1 /c2 has Gaussian distribution N (0, s2 − s1 ). Problem 2. E(St ) = S0 exp(αt)E(exp σWt )

∞

= S0 exp(αt)

1 x2 exp(σx) √ exp(− )dx 2t 2π −∞ ∞ (x − σt)2 σ2 t 1 ) √ exp(− )dx = S0 exp(αt) exp( 2 2t 2π −∞ σ2 = S0 exp((α + )t) 2

So when 2α + σ 2 = 0 we would have E(St ) = S0 for all times t. Problem 3. a. Assume we’re given St1 , . . . , StN with t1 < t2 < . . . < tN . Then log(Stk /S0 ) = αtk + σWtk , and Wtk = σ −1 (log(Stk /S0 ) − αtk ). Now the independent increment property of Brownian Motion gives us Wtk − Wtk−1 = σ −1 (log 1 Stk − α(tk − tk−1 )) Stk−1

are independent. In order to apply the law of large numbers, we need to find independent identical distributed random variables. The above increments are independent already. To make them identically distributed, we normalize them to N (0, 1):

ξk = √

1 1 (Wtk − Wtk−1 ) = √ tk − tk−1 σ tk − tk−1

log

Stk −...