Question

PROBLEM 10.1.1 Suppose {B(t):120} represents a standard Brownian motion process. a. Determine P(1.2<B(9)<3.9). b. Determine P(0.8 < B (9)- B(5)<2). PROBLEM 10.1.2 (pg 639, #1, #3) Suppose {B(l):120} represents a standard Brownian motion process. a. What is the distribution of B(1) + 2B(11) ? Note: B(11) - B(1) + [B(11)- B(1)] b. Determine E[B(1)B(5)]. Note: B(5) - B(l) + [B(5)– B(1)]

Answer 1

IBCZ): z ob nepresents a standard Brownian motion process So B(t) satisfies 1) BO)=0 2) B(2+4) BCE) is independent with B(s) for all sst 3) Beztu)-BCZ) N(o, u). 4) B(t) is continuary in t. From property we can get BCV) = B(+)-BO) ON N6,+) 10.1.1 a BCG). No,g) :- P(1-2 <349) <3.9) = P(1.270 < 2693-0 <3.969) =P(0.4<2<1-3) = P(2<103) - P(2<0.4) = 0.9032 - 0.6554 = 0.2478 b) From property ② Beg)-B(5) WN(0,4) P(0.8<B(9)-B(5) <2) =P(0.8 <2<22) =P(0.44221) = P(2<1) -P(Z <0.4) = 0.8413-0.6554 = 0.1859 10.1.2 (2) a) since B )= BC)+(BCH)-BCD] & B) & (BCL)- BC) are independent by property also B(l) ~ N(0,1) & B(11)-B(1) UN (0,10) B - Bli) ~ N(o, 11) 28 (11) WN(0,44) O th) N property Home

B41)+2B(1) = B(1) +25BC9+[BC11-Bcu] = 3 BC1)+ 2[BCW-BG] since BC) & B(1)-B(1) are inclependent and BO) N60,1) 2011)-BCI) ~NCO,10) B01)+2B(u) = 3861)+2 [0003– Bcv] ~ N60,9+40) ~N(0,49) b) E (BCU) BCS)] = RE E [B(1). (B6v+[065)=B60] = ECB1) +B(1). [BC5)-BC]] = E [CB.C173] + E [oco).Coco+B6] .: E(x+4)=E(X)+E(4) – E[C8C112] + E[B-03-E [863)-BCV] .8 X&Y independent E (X4) = ECX)- E(4) = 1 to Ver (30)) = ECBC1) - ECBCU)] = 1 1 = E{(BC1)] -0 .:[[BC*)]=o vt