A course in financial calculus chapter 4's Solutions

0
Since Itˆ o integrals are martingales with zero initial value, it follows that the differential of the expected value is
dE [Y T ] = 0 E [YT ] = E [exp{−u(0)rT }] = Y0 = exp{−u(T )r0 }. (1)
We set Yt = exp{−u(T − t)rt }, use that α = 0 and apply Iˆ o’s lemma to obtain
1 dYt = −rt Yt du(T − t) − u(T − t)Yt drt + u2 (T − t)Yt (drt )2 2 2 σ = −rt Yt βu(T − t)dt − rt Yt u2 (T − t)dt + u(T − t)Yt βrt dt 2 √ 1 2 −u(T − t)Yt σ rt dWt + u (T − t)Yt σ 2 rt dt 2 √ = −u(T − t)Yt σ rt dWt t √ u(T − s)Ys σ rs dWs . Yt = Y0 −
β¯ θ e−βT 2 σ2 − ¯ ¯ r0 1 − θ + e βT θ
E exp
= exp
1¯ θ 2
.
−1
˜= Next, we make the substitution θ obtain
¯ = 2θ ˜ 1 − e−βT 1 − e−βT , i.e. θ
and
E exp
σ2
4β ˜T θr (1 − e−βT )
We require r0 = 0 and see that the moment generating function of λrT e−βT r0
is
the same as the moment generating function of a non-central chi-square distribution with zero degrees of freedom and non-centrality parameter λ ≥ 0. The uniqueness of the moment generating function implies that λrT e−βT r0
3
Equation (1) can now be written as e−βT θ r0 σ2 σ2 1+ 2 θ − e−βT 2 θ β β
E [exp {−θrT }] = exp −
.
(4)
The first two derivatives of this expression w.r.t. θ are
(drt )2 = σ 2 rt dt. Applying Itˆ o’s lemma for α = 0, the differential of √ rt becomes
√ 1 √ 1 1 σ2 d rt = − β rt dt + σdWt − √ dt 2 2 8 rt √ 1 1 σ2 1 = − β rt + √ dt + σdWt . 2 4 rt 2 ii) Let τ = T − t. Applying a change of variable from u(t) to u(τ ) yields du(t) du(τ ) dτ du(τ ) = =− dt dτ dt dτ
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and
1
and we obtain du(τ ) σ2 = βu(τ ) + u2 (τ ). dτ 2 Now let f (t, x, y ) = exp{−x(t)y }. We first note that (du(τ ))2 = 0 and du(τ )drt = 0 since u(τ ) is a non-random function of time. The relevant partial derivatives are ∂f = −yf (t, x, y ), ∂x ∂f = −x(t)f (t, x, y ), ∂y ∂ 2f = x2 (t)f (t, x, y ). ∂y 2
σ2 2 1 − e−βT r0 + e−2βT r0 . β
The variance of rT is then given by
2 Var [rT ] = E rT − (E [rT ])2 σ2 = e−βT 1 − e−βT r0 . β
4
In order to compute the probability P [rT = 0], we first need to determine the type ¯ = − σ2 θ in of distribution that rT follows. We start by making the substitution θ 2β Equation (4) to get 2β ¯ θrT σ2
P [rT = 0] = exp −
1
.
Siegel, Andrew F.: The Noncentral Chi-Squared Distribution with Zero Degrees of Freedom and
Testing for Uniformity; Biometrika, Vol. 66, No. 2 (August 1979), pp. 381-386
∂ ˆT E exp θr ˆ ∂θ ∂2 ˆT E exp θr 2 ˆ ∂θ
=
∂ ∂θ ∂ E [exp {θrT }] = − E [exp {θrT }] ˆ ∂θ ∂θ ∂θ 2 ∂ ∂θ ∂2 = − 2 E [exp {θrT }] = 2 E [exp {θrT }] . ˆ ∂θ ∂θ ∂θ
2
2 r0
exp {. . . }
∂2 E [exp {−θrT }] = ∂θ2
2e−βT 1+ + 1+
σ − e−βT 2 β σ2 e−βT 2 θ β
σ2 θ 2β
3 r0
exp {. . . }
−
e−2βT
σ2 2β
θ−
e−βT
σ2 2β
2 4 r0
exp {. . . } .
θ
We note once again, that Equations (1) and (4) are not moment generating funcˆ = −θ such that tions due to the minus sign in the exponential. We thus define θ ˆT E exp θr is the moment generating function. We furthermore note that
σ2 dt 2
e−βt v (t) = −
σ 2 −βt e + C. 2β
We substitute back to obtain the solution for u(t) 1 . + eβt C
u(t) =
σ2 −2 β
(3)
Using u(0) = θ allows us to solve for the integration constant C . 1+
e ∂ E [exp {−θrT }] = − ∂θ = −
−βT
1+
σ2 θ 2β
σ − e−βT 2 θ − e−βT θ β
2
σ2 2β 2
σ − e−βT 2 β
2
1+ e−βT 1+
σ2 θ 2β σ2 2β
σ2 θ 2β
−
σ2 e−βT 2 θ β
r0 exp {. . . }
−
σ2 e−β2 θ β
P
.
Given that β > 0, we have λ > 0 and the scaling has no impact on the probability mass at zero. In this case, we can conclude that 2βe−βT r0 σ 2 (1 − e−βT )
ˆ = 0 yields the first two moments Evaluating these two derivatives at θ
E [rT ] =
2 E rT
∂ ˆT E exp θr ˆ ∂θ ∂2 ˆT E exp θr 2 ˆ ∂θ
= e−βT r0
ˆ=0 θ
=
= e−βT
ˆ=0 θ
e−βt
dv (t) σ2 − βe−βt v (t) = e−βt dt 2 −βt d e v (t) σ2 = e−βt . dt 2
We can now integrate both sides to obtain the solution for v (t)
d e−βt v (t)
=
e−βt
v (t) =
⇒
u(t) =
du(t) = −
Substituting into the ODE yields 2
−
β σ2 1 dv (t) = − − v 2 (t) dt v (t) 2v 2 (t) dv (t) σ2 = βv (t) + . dt 2
(2)
Equation (2) is now a linear ODE and we multiply both sides by the integration factor e−βt to obtain