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94) 0 This is an exact expression for I , since we can easily obtain an exact expression for the probability density of Z 2 . 95) 0 (where once again both integrals contain increments of the same Wiener process), we can calculate the expectation value of J . Note that in the deﬁnition of J we have ﬁxed the upper limits of both the inner and outer integrals to be T . We will consider replacing the upper limit of the inner integral with t shortly. Discretizing J , and taking the average, we have N −1 N J = lim fm gn N →∞ Wm Wn n=0 m=1 N −1 = lim N →∞ T fn gn ( Wn )2 = f (t)g(t)dt.

34) and thus At = U † DtU . Because of this, the power series ∞ n=0 (At)2 (At)3 (At)n = 1 + At + + + ··· n! 2 3! = 1 + U † DtU + U † DtU U † DtU (U † DtU )3 + + ··· 2 3! U † (Dt)2 U U † (Dt)3 U + + ··· 2 3! (Dt)2 (Dt)3 1 + Dt + + + ··· U 2 3! = 1 + U † DtU + = U† = U † eDt U. , which is the natural generalization of the exponential function for a matrix At. 36) f (A) ≡ U † f (D)U ≡ U † ⎜ . ⎟ U. . . ⎝ . . 38) is where ∞ eAt ≡ n (At)n = U † eDt U. n! 39) We can now also solve any linear vector differential equation with driving, just as we did for the single-variable linear equation above.

There are systematic numerical methods to ﬁnd the U and corresponding D for a given A, and numerical software such as Matlab and Mathematica include routines to do this. If A is two-by-two or three-by-three, then one can calculate U and D analytically, and we show how this is done in the next section. 29) D=⎜ . ⎟, . . . ⎝ . . 30) 22 Differential equations and this has the solution yn (t) = yn (0)eλn t . The solution for y is thus ⎛ λt ⎞ e 1 0 ··· 0 ⎜ 0 eλ2 t · · · 0 ⎟ ⎜ ⎟ y(t) = ⎜ . ⎟ y(0) ≡ eDt y(0), ..