Homework Assignment #5
1: A family {X(t) : t ≥ 0} of square integrable, R-valued random variables on a probability space
(Ω, F, P) is called a centered Gaussian process if its span in L2 (P; R)
Gaussian family.
is a centered,
Given such a process, define its covariance function C(s, t) ≡ EP X(s)X(t) .
(i) Given 0 ≤ t0 < · · · tn , let µt0 ,...,tn be the distribution of X(t0 ), . . . , X(tn ) and express the
Fourier transform of µt0 ,...,tn in terms of C. In particular, conclude that C uniquely determines
the distribution of {X(t) : t ≥ 0} as a stochastic process.
(ii) Assuming that for each T > 0 there exist a K(T ) < ∞ and αT > 0 such that
C(t, t) − 2C(s, t) + C(s, s) ≤ K(T )(t − s)αT for s, t ∈ [0, T ],
show that there exists a version that is continuous in the sense that there exists a family {X̃(t) :
t ≥ 0} such that t
X̃(t) is continuous and X̃(t) = X(t) (a.s.,P) for each t ≥ 0.
C(s,t)
X(s) is independent of X(s) and
(iii) Assume that C(s, s) > 0, and show that X(t) − C(s,s)
therefore that, for any Borel measurable ϕ : R −→ [0, ∞),
s
!
Z
C(s, t)
C(s, t)2
ϕ
X(s) + (C(t, t) −
y γ0,1 (dy)
C(s, s)
C(s, s)
R
is the conditional expectation of ϕ X(t) given σ({X(s)}).
(iv) Show that {X(t) : t ≥} is Markov in the sense that, for all 0 ≤ s < t and non-negative Borel
measurable ϕ’s
EP ϕ X(t) σ {X(τ ) : τ ∈ [0, s]} = EP ϕ X(t) σ {X(s)} ,
if and only if C(τ, t)C(s, s) = C(s, t)C(τ, s) for all 0 ≤ τ ≤ s < t.
(v): Take Ω = C [0, ∞); R , F = BΩ , and P = W. Let h : [0, ∞)2 −→ R be a Borel measurable
function for which
Z ∞
|h(t, τ )|2 dτ < ∞ for all t ≥ 0,
0
and set
Z
X(t) =
t
h(t, τ ) dw(τ ).
0
Show that {X(t) : t ≥} is a centered Gaussian process with covariance
Z ∞
C(s, t) =
h(s, τ )h(t, τ ) dτ.
0
Show that this process will have a continuous version if for each T > 0 there exist a K(T ) < ∞
and αT > 0 such that
Z ∞
|h(t, τ ) − h(s, τ )|2 dτ ≤ K(T )(t − s)αT for 0 ≤ s < t ≤ T.
0
2: Aside from Brownian motion, the most famous Gaussian process is the Ornstein-Uhlenbeck
process. Namely, continue in the setting of (v) above, and, for x ∈ R and w ∈ C [0, ∞); R , show
that there is precisely one solution X( · , x)(w) to the integral equations
Z
X(t, x) = x + w(t) −
(1)1
t
X(τ, x)(w) dτ.
0
In fact, use Riemann-Stieltjes integration to show that
−t
X(t, x) = e x + e
−t
Z
t
eτ dw(τ ).
0
In particular, conclude that {X(t, 0) : t ≥ 0} is a centered Gaussian process with covariance
C(s, t) =
e−|s−t| − e−(s+t
.
2
Although one can use (iv) above to check that {X(t, 0) : t ≥ 0} is Markov, there is a better way.
Namely, show that, for 0 ≤ s < t,
X(t, x)(w) = e
−(t−s)
−s−t
Z
X(s, x)(w) + e
= X t − s, X(s, x)(w) (δs w),
t
eτ dw(τ )
s
where δs w(t) = w(s + t) − w(s), and conclude that
W
E ϕ X(t, x) Bs =
Z
ϕ(y)p t − s, X(s, x), y) dy
where
p(τ, x, y) = π(1 − e
−2τ
)
− 21
y − e−τ x
exp −
1 − e−2τ
Finally, show that {X(t, 0) : t ≥ 0} has the same distribution as
−t
e w
e2t − 1
2
: t≥0 .
.