Thus the mean value satisfies the linear differential Equation (4.3). 
The covariance matrix is more difficult to compute. In order to 
simplify the calculations, assume My = 0; hence, E[x(t)] = 0. This can 
always be achieved by subtracting mi, OMNES LOGS mame, 
Ris, eaicov Ix). x(t) = EleG) x (1 
Sy 
=i lee, t) x(t) + I d(s, 0) d “| x! (t) 
iE 
fs) 
= 6(s, t) Elx(t) x (t)] + | ®(s, 0) Eld w() x (t)] 
a 
=Ol (shat) eR (tema) (4.9) 
The integral is zero since w(0) and x(t) are independent for s > t. 
Set PG) — Rene) = Eee) x (t)]. Then P(t) is the variance and is 
therefore the function of interest. 
t 
E Ce 0) c+ | Oey Od a) 
0 
t alg 
oF (se O) cect { OGEs Oo) ad “) 
0 
ace, O) Ble &] © '@, 0) 
P(t) 
iC 
+ o(t, 0) E|c { d wi(a) o(t, 9) 
0 
t 
+ { ®(t, tT) Eld w(t) c | @(t, 0) 
(s) 
i 1E 
i | I ite, 2) Bld @@) da @)l Ot, o) 
OnaO 
50 
