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Unformatted text preview: f S has a positive risk premium then Q will negative risk
premium.
Question 20.14.
As mentioned in the problem, σ dZ appears in both dS and dQ. One can think of dQ as an alternative
model for the stock (with dS being the standard geometric Brownian motion).
a)
If there were no jumps, dQ would also be geometric Brownian motion. Since it has the same
risk component, σ dZ , αQ must equal α . If we thought of Q as another traded asset, this naturally
follows from no arbitrage.
b)
If Y1 > 1 then there are only positive jumps. We would therefore expect αQ < α to compensate for this. Mathematically, dQ/Q − dS/S = αQ − α d t + dq1 . If a jump occurs, dq1 =
Y1 − 1 > 0; if αQ ≥ α we could buy Q and short S . the only risk we have is jump risk but this will
always be “good” news for our portfolio. In order to avoid this arbitrage αQ must be less than α .
130 Chapter 20 Brownian Motion and Itô’s Lemma If we use a weaker assumption k1 = E (Y1 − 1) > 0 and we assume the returns to S and Q should
be the same (this makes sense if we are looking at Q as an alternative model instead of another
stock) then we arrive at a similar result. The expected return to Q is αQ + λ1 k1 ; setting this equal
to α implies α − αQ = λ1 k1 > 0.
c)
Let α ∗ be the expected return of Q. Note that αQ is not the expected return, it is the expected
return conditional on no jumps occurring. We have the following relation ship,
α∗ = E dQ
/dt = αQ + k1 λ1 + k2 λ2
Q (16) where ki = E (Yi − 1). Hence αQ = α ∗ − k1 λ1 − k2 λ2 . If α ∗ = α (i.e. Q and S have the same
expected return) then α − αQ = k1 λ1 + k2 λ2 . The sign of which could be positive or negative if
there are no restriction on k1 and k2 . 131 Chapter 21
The BlackScholes Equation
Question 21.2.
If V (S, t) = AS a eγ t then Vt = γ V , VS = aS a −1 eγ t = aV /S , and VSS = a (a − 1) S a −2 eγ t =
a (a − 1) V /S 2 . Therefore the left hand side of the Black Scholes equation (21.11) is
Vt + (r − δ) VS S + VSS S 2 σ 2 /2 − rV = γ − r + (r − δ) a + σ2
a (a − 1) V .
2 (1) We can rewrite the coefﬁcient of V as
γ + (r − δ) a + σ2 2
σ2
σ2
a (a − 1) =
a + r −δ−
a + γ − r.
2
2
2 (2) From the quadratic formula, this has roots a= − r −δ− σ2
2 σ2 ± r −δ− σ2
2 2 2 − 4 σ2 (γ − r)
. σ2 (3) Simplifying,
a= 1 r −δ
−
2
σ2 ± r −δ 1
−
σ2
2 2 + 2 (r − γ )
.
σ2 (4) Note, for a given γ , these are the only values for a that will satisfy the PDE.
Question 21.4.
Deﬁning V (S, t) = Ke−r(T −t) + Se−δ(T −t) we have Vt = rKe−r(T −t) + δSe−δ(T −t) , VS = e−δ(T −t)
and VSS = 0. The Black Scholes equation is satisﬁed for Vt + (r − δ)VS S + VSS S 2 σ 2 /2 is
rKe−r(T −t) + δSe−δ(T −t) + (r − δ) e−δ(T −t) S (5) = r K e−r(T −t) + Se−δ(T −t) = rV . (6) This also follows from the result that linear combinations of solutions of the PDE are also solutions.
The boundary condition is V (S, T ) = K + ST , i.e. we receive one share and K dollars. Similarly,
a long forward contract with value Se−δ(T −t) − Ke−r(T −t) will solve the PDE.
132 Chapter 21 The BlackScholes Equation Question 21.6.
Let V (S, t) = e−r(T −t) N (d2 ); we must show V solves the PDE Vt + (r − δ) SVS + S 2 σ 2 VSS /2 =
rV . Note that
d2 = ln (S/K)
r − δ − σ 2 /2 √
T −t
+
√
σ
σ T −t (7) depends on both S and t . Beginning with the ﬁrst term in the PDE,
Vt = rV + e−r(T −t) N (d2 ) ln (S/K)
2σ (T − t)3/2 − r − δ − σ 2 /2 2σ (T − t)1/2
2 r − δ − σ 2 /2 √
e−r(T −t) N (d2 )
= rV +
d2 −
T −t .
2 (T − t)
σ (8) √
Since VS = e−r(T −t) N (d2 ) / S σ T − t the second term in the PDE is
(r − δ) SVS = r −δ
√
σ T −t e−r(T −t) N (d2 ) . (9) The second partial of V with respect to S is
VSS = √
e−r(T −t) N (d2 ) − N (d2 )
e−r(T −t) N (d2 )
=− 2 2
d2 + σ T − t
S 2 σ 2 (T − t)
S σ (T − t) (10) where we use the property N (x) = −xN (x). The third term in the PDE is therefore
√
e−r(T −t) N (d2 )
S 2 σ 2 VSS
=−
d2 + σ T − t .
2
2 (T − t) (11) Adding equations (8), (9), and (11), all terms cancel expect the rV term in equation (8); i.e. V
satisﬁes the PDE.
Question 21.8.
These bets are all or nothing options. The cash bets being worth, per dollar, e−rT N (d2 ) if we receive
$1 if ST > K and e−rT N (−d2 ) if we receive $1 if ST < K . The stock bets being worth, per share,
SN (d1 ) if we receive 1 share if ST > K and SN (−d1 ) if we receive 1 share if ST < K . (Note we
are assuming the current time is t = 0 and the bet is for the stock price T years from now).
√
,d
a)
By setting K = Se(r −δ)T √ 2 = −σ T /2 the value of the bet that the share price will exceed
the forward price is e−rT N(−σ T /2). This is always less than the opposite bet, which has value
√
e−rT N (σ T /2).
133 Part 5 Advanced Pricing Theor...
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 Spring '14
 NguyenThiMaiLan
 Derivatives

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