MSc Financial Engineering–Spring term 2012
Exercise 1: Consider the Vasicěk model, where we always assume that a > 0.
a.)Solve the Vasicěk SDE explicitly, and determine the distribution of r(t).
Hint: The distribution is Gaussian (why?), so it is enough to compute the expected value
and the variance.
b.)As t → ∞, the distribution of r(t)
√ tends to a limiting distribution. Show that this
is the Gaussian distribution N[b/a, σ 2a]. Thus we see that, in the limit ,r will indeed
oscillate around its mean reversion level b/a.
c.)Now assume that r(0) is a stochastic variable, independent of the Wiener process W ,
and by definition having the Gaussian the limit distribution in (b), for all values of t.
Thus we have found the stationary distribution for the the Vasicěk model.
Solution:
In the Vasicek model we have
dr(t) = (b − a(r))dt + σdWt
with a > 0
a.)Using the SDE above and multiplying both sides by eat , we have that,
eat dr(t) + r(t)aeat dt = beat dt + σeat dWt
d(eat r(t)) = beat dt + σeat dWt
Z t
Z t
as
at
eas dWs
e bds + σ
e r(t) = r(0) +
0
0
Z t
Z t
−a(t−s)
−at
e−a(t−s) dWs
e
ds + σ
r(t) = r(0)e + b
0
0
The first integral is deterministic and just a number. The second integral is a stoch. Ito
integral with a deterministicR integrand and therefore Gaussian, (See Math. Methods I),
t
with mean 0 and variance σ 2 0 e−2a(t−s) ds; hence r(t) is Gaussian with mean and variance:
b
E(r(t)) = r(0)e−at + (1 − e−at )
Z ta
e−2as ds
V(r(t)) = σ 2 e−2at
0
σ2
= (1 − e−2at )
2a
b.)As t → ∞, we have
b
b
lim E(r(t)) = r(0) lim e−at + (1 − lim e−at ) =
t→∞
t→∞
t→∞
a
a
and
lim V(r(t)) =
t→∞
σ2
σ2
(1 − lim e−2at ) =
t→∞
2a
2a
1
2
So, N[ ab , σ2a ] is the limiting distribution of r(t).
c.)The results in (a) and (b) are based on the assumption that the value r(0) is known.
2
If, instead we have that r(0) ∼ N[ ab , σ2a ], independent of the Brownian motion (Wt )t≥0 ,
then for any t we have that
b
b
b
E(r(t)) = E(r(0))e−at + (1 − e−at ) =
∵ E(r(0)) =
a
a
a
2
2
σ
σ
V(r(t)) = e−2at V (r(0)) + (1 − e−2at ) =
2a
2a
σ 2 e−2at σ 2
=
+ (1 − e−2at )
2a
2a
σ2
=
2a
Exercise 2: The object of this exercise is to indicate why the CIR model is connected
to squares of linear diffusions. Let Y be given as the solution to the following SDE:
√
dY = (2aY + σ 2 )dt + 2σ Y dWt ,
Y (0) = y0
p
Define the process Z by Z(t) = Y (t). It turns out that Z satisfies a stochastic differential equation. Which?
Solution:
dY (t) = (2aY (t) + σ 2 )dt + 2σ
Then by Ito, Z(t) =
p
Y (t)dWt
, Y (0) = y0 ,
p
Y (t) follows
1
3
1
1 1
dZ(t) = (Y (t)− 2 )dY (t) + (− (Y (t)− 2 ))(dY (t))2
2
2 4
1
1
1
1
1
= (2aY (t) 2 + σ 2 (Y (t)− 2 ))dt + σdWt − σ 2 Y (t)− 2 dt
2
2
1
= aY (t) 2 dt + σdWt
| {z }
Z(t)
so that dZ(t) = aZ(t)dt + σdWt which is a Vasicek equation.
2