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STAT4207/5207 Homework assignment 4 Due by April 15, 2021. Write down the complete derivations for full credits. Question 3 and 4 are bonus questions (worth 12 and 13 points respectively). Exercise 1 (55 points) For a standard Brownian motion (Bt : t ≥ 0) a) Find P(B2 − B1 > 1|B0.5 = 2). b) Consider the process Xt = Bt − tB1 0 ≤ t ≤ 1. i) Derive E[Xt] and Cov(Xt, Xs) for some fixed t, s ∈ [0, 1] and specify the joint distribution of Xt and Xs. ii) Calculate P(X0.15 > 0.2). iii) Find E[Xs|Xt = 1] for some fixed s, t ∈ [0, 1] such that 0 < s=””>< t.=”” c)=”” consider=”” the=”” process=”” xt=”µt” +=”” σbt=”” t=”” ≥=”” 0.=”” for=”” constants=”” µ=””>∈ R, σ > 0. Derive E[Xt] and Cov(Xt, Xs) for some fixed t, s ∈ [0, 1] and specify the joint distribution of Xt and Xs. d) Consider the process Xt = e µt+σBt t ≥ 0. for constants µ ∈ R, σ > 0. i) Prove that, for Z ∼ N (0, 1), E[e λZ] = e λ 2/2 ∀λ ∈ R. ii) Use your previous answer and the self-similarity property of Brownian motion to show that EXt = e (µ+0.5σ 2 )t . iii) Derive Cov(Xt, Xs) for some fixed t, s ∈ [0, 1]. iv) Give the distribution of Xt for some fixed t > 0. e) Find the distribution of the random variable (Bt − Bs) 2 (Bv − Bt) 2 for 0 < s=””>< t=””>< v.=”” exercise=”” 2=”” (15=”” points)=”” suppose=”” a=”” factory=”” has=”” two=”” machines=”” that=”” are=”” maintained=”” by=”” a=”” single=”” technician.=”” furthermore,=”” suppose=”” that=”” machine=”” i=”1,” 2=”” functions=”” for=”” an=”” exponential=”” time=”” with=”” rate=”” µi=”” before=”” it=”” breaks=”” down.=”” the=”” repair=”” time=”” for=”” each=”” of=”” the=”” machines=”” is=”” exponentially=”” distributed=”” with=”” rate=”” α.=”” the=”” repair=”” and=”” lifetimes=”” of=”” the=”” two=”” machines=”” are=”” completely=”” independent.=”” model=”” this=”” system=”” as=”” a=”” continuous-time=”” markov=”” chain.=”” can=”” it=”” be=”” modeled=”” as=”” a=”” birth=”” and=”” death=”” process?=”” explain.=”” 1exercise=”” 3=”” (bonus=”” question)=”” consider=”” a=”” continuous=”” time=”” markov=”” chain=”” with=”” states=”” 0,=”” 1,=”” 2,=”” 3=”” and=”” generator=”” q=”” that=”” is=”” given=”” by=”” q=”” =”” −6=”” 3=”” 3=”” 0=”” 2=”” −4=”” 0=”” 2=”” 2=”” 0=”” −5=”” 3=”” 0=”” 2=”” 1=”” −3=”” =”” =”” a)=”” let=”” a=”1,” 3.=”” find=”” e[τ0,a].=”” b)=”” find=”” the=”” probability=”” that,=”” starting=”” in=”” state=”” 0,=”” state=”” 1=”” is=”” visited=”” before=”” state=”” 3.=”” exercise=”” 4=”” (bonus=”” question)=”” suppose=”” each=”” time=”” a=”” machine=”” is=”” repaired=”” it=”” remains=”” up=”” for=”” an=”” exponentially=”” distributed=”” amount=”” of=”” time=”” with=”” parameter=”” λ.=”” it=”” then=”” fails,=”” and=”” its=”” failure=”” is=”” either=”” of=”” two=”” types.=”” if=”” it=”” is=”” a=”” type=”” 1=”” failure,=”” the=”” time=”” to=”” repair=”” is=”” exponentially=”” distributed=”” with=”” parameter=”” µ1.=”” if=”” it=”” is=”” type=”” 2=”” failure,=”” the=”” repair=”” time=”” is=”” exponentially=”” distributed=”” with=”” parameter=”” µ2.=”” assume=”” that=”” a=”” failure=”” is,=”” independent=”” of=”” the=”” time=”” it=”” took=”” for=”” the=”” machine=”” to=”” fail,=”” a=”” type=”” 1=”” failure=”” with=”” probability=”” p=”” and=”” a=”” type=”” 2=”” failure=”” with=”” probability=”” 1=”” −=”” p.=”” a)=”” what=”” is=”” the=”” proportion=”” of=”” time=”” that=”” the=”” machine=”” is=”” down=”” due=”” to=”” a=”” type=”” 1=”” failure?=”” b)=”” what=”” is=”” the=”” proportion=”” of=”” time=”” that=”” the=”” machine=”” is=”” down=”” due=”” to=”” a=”” type=”” 2=”” failure?=”” c)=”” what=”” proportion=”” of=”” time=”” is=”” the=”” machine=””>

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