SOLUTIONS TO PROBLEM SET 7
1. It follows from Theorem 3.5.3 and Proposition 3.5.4 in the lecture note that we need to find
increasing and decreasing solutions to the equation
(1)
Af = λf,
where A = 2x
d2
d
+δ .
dx2
dx
Let ν = δ/2 − 1 and function g(x) via
f (x) = x−ν/2 g(x).
Calculation shows that g satisfies
g 0 (x)
ν2
λ
g = 0.
g (x) +
−
+
x
2x 4x2
√
Let us also define h(y) = g(x), where y = 2λx, then calculation shows that g solves
00
00
y 2 h (y) + yh0 (y) − (y 2 + ν 2 )h(y) = 0.
The independent solutions to this ODE are so called modified Bessel functions Iν (y) and Kν (y),
where Iν is a increasing function, Kν is a decreasing function, and both of them are positive.
Therefore,
√
ϕ+ (λ, x) = x−ν/2 Iν ( 2λx)
and ϕ− (λ, x) =
√
x−ν/2 Kν ( 2λx)
are two linearly independent solutions to (1). To check that ϕ+ (λ, x) is increasing, we use the
relation between Iν and Iν+1 : Iν0 = Iν+1 + (ν/x)Iν to obtain
√2λ ν−1
√
d −ν/2 √
0
ϕ (x) =
x
I( 2λx) =
x− 2 Iν+1 ( 2λx) > 0.
dx
2
Similarly, we can use the relation: Kν0 = Kν−1 − (ν/x)Kν to check that ϕ− (x) is decreasing.
It then follows that
Ex [e−λTy ] =



ϕ+ (λ,x)
ϕ+ (λ,y)
=


ϕ− (λ,x)
ϕ− (λ,y)
=
√
Iν ( 2λx)
√
I ( 2λy)
−ν/2 ν √
Kν ( 2λx)
x
√
y
Kν ( 2λy)
−ν/2
x
y
when x < y
when x > y.
Here when x > y, Iν should be replaced by Kν so that the Laplace transform on the left hand side
is less than 1.
2 and 3. Note that Problem 2 will be a special case of Problem 3 once we solve the 3rd question
with correlation.
Let X be a diffusion with generator L:
n
L=
n
X
1X
2
aij (x)∂ij
+
bi (x)∂i ,
2
i,j
i
1
2
SOLUTIONS TO PROBLEM SET 7
where a = σσ 0 . The observation process is of the form
Z t
Yt =
hs ds + Wt ,
0
RT
2 (Rn )
where W is a Brownian motion in Rn and E[ 0 |hs |2 ds] < ∞. We suppose that f ∈ CK
R
T
satisfying supt≤T E[f (Xt )2 ] < ∞ and E 0 |Lf (Xt )|2 dt < ∞. Then
Z t
Lf (Xs ) ds is a L2 martingale,
Mt = f (Xt ) − f (X0 ) −
0
and there exists an Ft -optional process α = (α1 , · · · , αn ) such that
Z t
[M, W i ]t =
αsi ds.
0
Define πt f =
(2)
E[f (Xt ) | FtY ],
then the Kushner-Stratonovic equation reads
Z t
Z t
πs Lf ds +
(πs hf − πs hπs f + πs αs ) dNs ,
πt f = π0 f +
0
0
where N is a FtY -Brownian motion in Rn .
The proof is almost the same with that in Theorem 4.1.2 in the lecture note, except the second
formula from the bottom of page 64 is replaced by
Z t
Z t
Z t
f (Xt )Yt =
f (Xs )dYs +
Ys df (Xs ) +
αs ds
0
Z
0
0
t
=
Z
f (Xs )(dWs + hs ds) +
0
Z
t
Z
t
Ys (dMs + Lf (Xs )ds) +
0
αs ds
0
t
(f (Xs )hs + Ys Lf (Xs ) + αs ) ds + Ft − martingale.
=
0
Specialize to the situation of Problem 2, where M has the following form:
Z t
Z t
Lf (Xs ) ds −
f 0 (Xs )σ(Xs )dBs ,
Mt = f (Xt ) − f (X0 ) −
0
0
where d[W, B]t = ρ(Xt , Yt ). Then α in the previous general setting is
αt = −f 0 (Xt )σ(Xt )ρ(Xt , Yt ).
Therefore Kushner-Stratonovic equation is in (2) where α is given as above.
4.
b denote the optional projection. Then
a) Let X
Z t
1 c2 [
Y
b
bs X\
Xt =
Xs − Xs Ys ) − X
s − Ys dBs
0 f (s)
Z t
1 c2 c 2
(Xs − Xs )dBsY ,
=
0 f (s)
where B Y is an F Y -Brownian motion given by
Z t
1 b
BtY = Yt −
(Xs − Ys )ds.
f
(s)
0
SOLUTIONS TO PROBLEM SET 7
3
b) Using Ito’s formula,
dXt2 = 2Xt dXt + σ 2 (t)dt.
Thus,
c2 = σ 2 (t)dt +
dX
t
c3 − X
c2 X
b
X
t
t t
dBtY .
f (t)
c3 − X
c2 X
c2
b
b
b2
The hint yields X
t
t t = 2v(t)Xt . Applying Ito’s formula to X − X gives
bt v(t)
bt
X
2v(t)X
v 2 (t)
dBtY − 2
dBtY − 2 dt
f (t
f (t)
f (t)
2
v (t)
dt.
=
σ 2 (t) − 2
f (t)
dv(t) = σ 2 (t)dt +
bt , v(t)) since X and Y are jointly Gaussian. When f (t) = s(t) − t, it is easily
c) Xt ∼ N (X
checked that s(t) − t satisfies the differential equation for v. Next note that
Z t
1 c2 c 2
b
Xt =
(Xs − Xs )dBsY
0 f (s)
Z t
1
v(s)dBsY = BtY .
=
0 f (s)