Show that the covariance matrix (C=left(t_{j} wedge t_{k} ight)_{j, k=1, ldots, n}) appearing in Theorem 2.6 is

Show that the covariance matrix \(C=\left(t_{j} \wedge t_{k}\right)_{j, k=1, \ldots, n}\) appearing in Theorem 2.6 is positive definite.

Data From Theorem 2.6

Transcribed Image Text:

2.6 Theorem. A one-dimensional Brownian motion (B+)to is a Gaussian process. For to 0 1, the vector I := (Bt,,..., Bt.) is a Gaussian random variable with mean vector m = 0 = R" and a strictly positive definite, symmetric Ex. 2.5 covariance matrix C = (tj ^ tk)j,k=1...,n: Ee(D) = e.c) Moreover, the probability distribution of I is given by (2.9) 1 1 P(I dx) = (2)/2 det C exp(-(x, Cx)) dx (2.10a) 1 1(x-x1-1) exp dx (2.10b) (2n)/2 (t-t-1) tj-tj-1 1=1 with x = (x1,...,xn) = R" and xo := 0. == Proof. Set A (Bt - Bto, Bt - Bt,..., Bt. -B-) and observe that we can write B(tk) B(to)=(Bt, - Bt,-1). Thus, r = = MA