Solution Manual

alternative: Let (Xt)t∈I be a real-valued stochastic process which has a second moment (such that the covariance is defined!), set μt = EXt. For any finite set S ⊂ I we pick λs ∈ C, s ∈ S. Then ∑ s,t∈S Cov(Xs,Xt)λsλ̄t = ∑ s,t∈S E ((Xs − μs)(Xt − μt))λsλ̄t = E ⎛ ⎝ ∑ s,t∈S (Xs − μs)λs(Xt − μt)λt ⎞ ⎠ = E(∑ s∈S (Xs − μs)λs∑ t∈S (Xt − μt)λt) = E ⎛ ⎝ ∣∑ s∈S (Xs − μs)λs∣ ⎞ ⎠ ⩾ 0. Remark: Note that this alternative does not prove that the covariance is strictly positive definite. A standard counterexample is to take Xs ≡X. Problem 2.3 (Solution) These are direct & straightforward calculations. Problem 2.4 (Solution) Let ei = (0, . . . ,0,1 ́11111111111111111 ̧111111111111111111¶ i ,0 . . .) ∈ R be the ith standard unit vector. Then aii = ⟨Aei, ei⟩ = ⟨Bei, ei⟩ = bii. Moreover, for i ≠ j, we get by the symmetry of A and B ⟨A(ei + ej), ei + ej⟩ = aii + ajj + 2bij and ⟨B(ei + ej), ei + ej⟩ = bii + bjj + 2bij which shows that aij = bij . Thus, A = B. We have Let A,B ∈ Rn×n be symmetric matrices. If ⟨Ax,x⟩ = ⟨Bx,x⟩ for all x ∈ R, then A = B. Problem 2.5 (Solution) a) Xt = 2Bt/4 is a BM: scaling property with c = 1/4, cf. 2.12. b) Yt = B2t −Bt is not a BM, the independent increments is clearly violated: E(Y2t − Yt)Yt = E(Y2tYt) −EY 2 t = E(B4t −B2t)(B2t −Bt) −E(B2t −Bt) (B1) = E(B4t −B2t)E(B2t −Bt) −E(B2t −Bt) (B1) = −E(B t ) = −t ≠ 0. c) Zt = √ tB1 is not a BM , the independent increments property is violated: E(Zt −Zs)Zs = ( √ t − √ s) √ sEB 1 = ( √ t − √ s) √ s ≠ 0.