hw6_solutions
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Solution for Homework 6. PDE for Finance, Spring 2003 Prepared by Fan-fu Feng. 1) (a) The HJB equation is 0 = ut + min Lu + running cost
α
= ut + min
α
1 −uw α + up θα + σ 2 upp + pα + θα2 2
3) For any fixed time t, yt = µt + σwt ∼ N (µt, σ 2 t). So its characteristic function is E [eiξyt ] = e(iµξ − 2 σ
1
2 ξ 2 )t
which is exactly the one given in equation (3) in the notes. (Notice that λ = 0) 4) Let {ξj } be a sequence of iid N (0, δ 2 ) random variables which is also independent of the Poisson process Nt and the Brownian motion Wt . Fix any time t > 0, then we know that
gN −1 θN −1
which will be a function of V and P . Let us first concentrate on the inner minimun - the one wrt θ - and take care of the outer one later. This will enable us to use the result from the self-financing case in the lecture notes. Since both are minimum, which order is taken does not matter. Now observe that the inner minimun is similar to the self-financing case with V N −1 = V + g , instead of VN −1 = V . min EVN −1 =V,PN −1 =P VN − F (PN ) + αgN −1 2 V + g + δP − F (PN )
2 2 2
θN −1
= min EVN −1 =V +g,PN −1 =P
θN −1
+ αg 2
2
= min JN −1 (V + g, P )
θN −1
+ αg = aN −1 (P ) V + g − bN −1 (P )
+ cN −1 (P ) + αg 2
(3)
where J is the performance function in the self-financing case, and the formulas of a N −1 , bN −1 and cN −1 are derived in the notes. Now one can find the minimum of (3), which is a a qua−V )a dratic polynomial in terms of g by differentiation, and the minimizer is g ∗ = (ba +α (the index N − 1 is dropped for convenience) After plugging in (3), we get the value function as αa |b − V |2 + c a+α which is again a quadratic polynomial for V , with coefficients α aN −1 a ˜N −1 = , ˜ bN −1 = bN −1 , c ˜N −1 = cN −1 aN −1 + α
1 1 g = = , so 2 (θ − 2g ) 2(θ − 2g ) 4θ and after some algebra, one sees that (c) Since g (s) = (d) from part (a), we see that α(s) = −
1 1 2 θ − 2g (r )
Nt
yt = µt + σwt +
j =1
ξj
4
Because a sum of independent Gaussian random variables is still Gaussian distributed, so 2 = σ 2 t + nδ 2 ) ( define νn
∞ ∞ n
P (yt = x) =
n=0 ∞
P (Nt = n)P (yt = x|Nt = n) =
n=0 2 P (Nt = n)P N (0, νn ) = x − µt n=0
P (Nt = n)P µt + σwt +
j =1 ∞
ξj = x
2 /(2ν 2 ) n
=
=
n=0(λt)n e−源自t · n!12 2πνn
e−(x−µt)
(*) In Merton’s paper, he uses the same idea to express the price of an European option with jumps in terms of the one without jumps. Let u λ (S, t) be the solution of the pricing equation (page 8 in section 8): ∂uλ + Lλ uλ − ruλ = 0, t < T ∂t with the final payoff uλ (S, t) = h(S ) and the boundary condition uλ (0, t) = 0 Thus uλ (S, t) is the price of the European option with final payoff h at expiration T . Define τ = T − t, k = E [eJ − 1] and also let u = u0 be the price of the option when there is no jump, then by the Feymamn-Kac formula: uλ (S, t) = ERN e−rτ h(Sτ ) = E e−rτ h Se(r−σ
1
2
(*) If one takes the running cost to be t p(s)α(s) ds instead of t p(s)α(s) + θ )α2 (s) ds. ˜ in the final cost, then for t < T : And also changes the parameter θ to θ u(w, p, t) = And the final time value is of course ˜ 2 u(w, p, T ) = pw + θw (1) is true because by an explicit computation, one see that for any fixed nonanticipating control α, the associated value function: θ ˜ − θ E w(T )2 (2) uα (w, p, t) = pw + w2 + θ 2 2 and since 0 ≤ w(T ) ≤ w with probability 1, we see clearly that ˜ > θ , with the choice of α to make w(T ) ≡ 0 (a) when θ 2 u(w, p, t) = θ uα (w, p, t) = pw + w2 α: nonanticipating 2 min min ˜ 2 uα (w, p, t) = pw + θw
θ 2 pw + 2 w ˜ 2 pw + θw
T
T
˜> if θ ˜< if θ
θ 2 θ 2
(1)
˜ < θ , with the choice of α to make w(T ) ≡ w (b) when θ 2 u(w, p, t) =
α: nonanticipating
Now let’s turn to (2). First observe that d(w2 ) = 2wdw = −2wαdt. Next, since dw has no stochastic term, d(pw) = pdw + wdp. Combining these gives (with a bit of manipulation) d(pw) = −pαdt − (θ/2)d(w 2 ) + wσdz,
T θ σw dz. pα ds + (w2 (t) − w2 (T )) + 2 t t The last integral has mean value zero, so for any (nonanticipating) strategy α the associated value is T
so
p(T )w(T ) − p(t)w(t) = −
T
uα (w, p, t) ≡ E
α
= ut + min
α
1 −uw α + up θα + σ 2 upp + pα + θα2 2
3) For any fixed time t, yt = µt + σwt ∼ N (µt, σ 2 t). So its characteristic function is E [eiξyt ] = e(iµξ − 2 σ
1
2 ξ 2 )t
which is exactly the one given in equation (3) in the notes. (Notice that λ = 0) 4) Let {ξj } be a sequence of iid N (0, δ 2 ) random variables which is also independent of the Poisson process Nt and the Brownian motion Wt . Fix any time t > 0, then we know that
gN −1 θN −1
which will be a function of V and P . Let us first concentrate on the inner minimun - the one wrt θ - and take care of the outer one later. This will enable us to use the result from the self-financing case in the lecture notes. Since both are minimum, which order is taken does not matter. Now observe that the inner minimun is similar to the self-financing case with V N −1 = V + g , instead of VN −1 = V . min EVN −1 =V,PN −1 =P VN − F (PN ) + αgN −1 2 V + g + δP − F (PN )
2 2 2
θN −1
= min EVN −1 =V +g,PN −1 =P
θN −1
+ αg 2
2
= min JN −1 (V + g, P )
θN −1
+ αg = aN −1 (P ) V + g − bN −1 (P )
+ cN −1 (P ) + αg 2
(3)
where J is the performance function in the self-financing case, and the formulas of a N −1 , bN −1 and cN −1 are derived in the notes. Now one can find the minimum of (3), which is a a qua−V )a dratic polynomial in terms of g by differentiation, and the minimizer is g ∗ = (ba +α (the index N − 1 is dropped for convenience) After plugging in (3), we get the value function as αa |b − V |2 + c a+α which is again a quadratic polynomial for V , with coefficients α aN −1 a ˜N −1 = , ˜ bN −1 = bN −1 , c ˜N −1 = cN −1 aN −1 + α
1 1 g = = , so 2 (θ − 2g ) 2(θ − 2g ) 4θ and after some algebra, one sees that (c) Since g (s) = (d) from part (a), we see that α(s) = −
1 1 2 θ − 2g (r )
Nt
yt = µt + σwt +
j =1
ξj
4
Because a sum of independent Gaussian random variables is still Gaussian distributed, so 2 = σ 2 t + nδ 2 ) ( define νn
∞ ∞ n
P (yt = x) =
n=0 ∞
P (Nt = n)P (yt = x|Nt = n) =
n=0 2 P (Nt = n)P N (0, νn ) = x − µt n=0
P (Nt = n)P µt + σwt +
j =1 ∞
ξj = x
2 /(2ν 2 ) n
=
=
n=0(λt)n e−源自t · n!12 2πνn
e−(x−µt)
(*) In Merton’s paper, he uses the same idea to express the price of an European option with jumps in terms of the one without jumps. Let u λ (S, t) be the solution of the pricing equation (page 8 in section 8): ∂uλ + Lλ uλ − ruλ = 0, t < T ∂t with the final payoff uλ (S, t) = h(S ) and the boundary condition uλ (0, t) = 0 Thus uλ (S, t) is the price of the European option with final payoff h at expiration T . Define τ = T − t, k = E [eJ − 1] and also let u = u0 be the price of the option when there is no jump, then by the Feymamn-Kac formula: uλ (S, t) = ERN e−rτ h(Sτ ) = E e−rτ h Se(r−σ
1
2
(*) If one takes the running cost to be t p(s)α(s) ds instead of t p(s)α(s) + θ )α2 (s) ds. ˜ in the final cost, then for t < T : And also changes the parameter θ to θ u(w, p, t) = And the final time value is of course ˜ 2 u(w, p, T ) = pw + θw (1) is true because by an explicit computation, one see that for any fixed nonanticipating control α, the associated value function: θ ˜ − θ E w(T )2 (2) uα (w, p, t) = pw + w2 + θ 2 2 and since 0 ≤ w(T ) ≤ w with probability 1, we see clearly that ˜ > θ , with the choice of α to make w(T ) ≡ 0 (a) when θ 2 u(w, p, t) = θ uα (w, p, t) = pw + w2 α: nonanticipating 2 min min ˜ 2 uα (w, p, t) = pw + θw
θ 2 pw + 2 w ˜ 2 pw + θw
T
T
˜> if θ ˜< if θ
θ 2 θ 2
(1)
˜ < θ , with the choice of α to make w(T ) ≡ w (b) when θ 2 u(w, p, t) =
α: nonanticipating
Now let’s turn to (2). First observe that d(w2 ) = 2wdw = −2wαdt. Next, since dw has no stochastic term, d(pw) = pdw + wdp. Combining these gives (with a bit of manipulation) d(pw) = −pαdt − (θ/2)d(w 2 ) + wσdz,
T θ σw dz. pα ds + (w2 (t) − w2 (T )) + 2 t t The last integral has mean value zero, so for any (nonanticipating) strategy α the associated value is T
so
p(T )w(T ) − p(t)w(t) = −
T
uα (w, p, t) ≡ E
