Risk Neutral Modelling Exercises
Geneviève Gauthier
Exercise 12.1. Assume that the price evolution of a given asset satis…es dX t = t X t dt + X t dW t
where t = (1 + sin (t)) and W = f W t : t 0 g is a ( ; F ; P ) Brownian motion.
a) Show that the SDE admits a unique strong solution.
b) The riskless asset as a time t instantaneous return at time of r t = r (1 + sin (t)). Find the SDE of the risky asset price in the risk neutral framework. What allows you to pretend that the risk neutral measure exists?
c) What is this risk neutral measure? Express it in function of P .
Exercise 12.2. The instantaneous interest rate f r t : t 0 g of the riskless asset satis…es the ordinary di¤erential equation
d r t = c ( r t ) dt;
which implies that
r t = + (r 0 ) e ct : The riskless asset price evolution f S t : t 0 g satis…es
S t = exp Z t
0
r u du :
a) Show that the price evolution satis…es
dS t = r t S t dt:
dS t = S t dt + S t dW t where f W t : t 0 g is a ( ; F ; P ) Brownian motion.
Show how to change the probability measure such that we get the risky asset price dynamics under the risk neutral measure.
Exercise 12.3. The prices of the assets are characterize with : dS 1 (t) = 1 S 1 (t) dt + 1 S 1 (t) dW 1 (t) ;
dS 2 (t) = 2 S 2 (t) dt + 2 S 2 (t) dW 1 (t) + 2 p
1 2 S 2 (t) dW 2 (t)
where W 1 and W 2 are independent ( ; F ; fF t : t 0 g ; P ) Brownian motions. The risk free interest rate r is assumed to be constant. fF t : t 0 g is the …ltration generated by the Brownian motions, with the usual regularity conditions.
Find the risk neutral market model.
Exercise 12.4. The value in American dollar of the yen satis…es dU (t) = U U (t) dt + U U (t) dW U (t) : The value in Canadian dollars of the yen evolves like
dC (t) = C C (t) dt + C C (t) dW C (t) : The value in Canadian dollars of an American dollar is
dV (t) = V V (t) dt + V V (t) dW V (t) :
W U ; W C and W V are ( ; F ; fF t : t 0 g ; P ) Brownian motions such that for all t > 0 Corr P [W U (t) ; W C (t)] = U C ,
Corr P [W U (t) ; W V (t)] = U V and Corr P [W V (t) ; W C (t)] = V C : Moreover,
dA (t) = r U A (t) dt; A (0) = 1 dB (t) = r C B (t) dt, B (0) = 1 dD (t) = r J D (t) dt; D (0) = 1
represent the evolutions of the riskless asset in the United-States, Canada and Japan, respec-
tively. More precisely, A (t) is in American dollars, B (t) is expressed in Canadian dollars
and D (t) is in yens.
Questions.
a) What are the conditions so that there is no arbitrage opportunity?
b) Is the market model complete?
c) What are the evolutions of the exchange rates in the risk neutral framework?
Interpret your results
Exercise 12.5. Assume that W , W c and W f are independent ( ; F ; fF t : 0 t T g ; P ) Brownian motions. Let
B e t W t + p
1 2 W f t , 0 t T
and B b t W t + p
1 2 W c t , 0 t T:
a) Show that
n B b t : 0 t T o
is a ( fF t g ; P ) Brownian motion.
b) Compute Corr P h b B t ; B e t i
for all 0 < t T:
The price evolution of two risky assets is dX t = X X t dt + X X t d B e t
= X X t dt + X X t dW t + X p
1 2 X t df W t and
dY t = Y Y t dt + Y Y t d B b t
= Y Y t dt + Y Y t dW t + Y p
1 2 Y t dc W t :
We can interpret W as the common chock, whereas W f and c W are the idiosyncratic chocks of X and Y respectively.
c) Determine the model in the risk neutral environment , assuming that the instan- taneous riskless interest rate is r:
d) Is this market model complete? Justify.
e) Can you price the contract that promises the di¤erence C = X T Y T between to
asset? Justify all steps: Interpret your results.
Solutions
1 Exercise 12.1
a) We need to verify that.
(i) j b (x; t) b (y; t) j + j (x; t) (y; t) j K j x y j ; 8 t 0 (ii) j b (x; t) j + j (x; t) j K (1 + j x j ) ; 8 t 0
(iii) E [X 0 2 ] < 1 : Let K = 2 j j + j j :
j b (x; t) b (y; t) j + j (x; t) (y; t) j
= j (1 + sin (t)) x (1 + sin (t)) y j + j x y j
= j j (1 + sin (t)) j x y j + j j j x y j
2 j j j x y j + j j j x y j = (2 j j + j j ) j x y j K j x y j j b (x; t) j + j (x; t) j
= j (1 + sin (t)) x j + j x j 2 j j j x j + j x j K j x j :
If E [X 0 2 ] < 1 ; then the three conditions are satis…ed.
b)
dX t = t X t dt + X t dW t
= r t X t dt + ( t r t ) X t dt + X t dW t
= r t X t dt + X t d W t + Z t
0
s r s ds :
Let
s = s r s
; s 0
and note that the function s ! s is continuous, which implies that the process f s : s 0 g
is predictable. Since Z T
0 2
s ds = Z T
0
s r s 2 ds
= Z T
0
(1 + sin (s)) r (1 + sin (s)) 2 ds
= 1 2
r 2
(4 4 cos T cos T sin T + 3T ) < 1 ;
then E P h
exp 1 2 R T 0
2 t dt
i
< 1 . We can apply the Cameron-Martin-Girsanov theorem :there is a martingale measure Q on ( ; F ) such that W f = n
f
W t : t 2 [0; T ] o
de…ne as
f
W t = W t + Z t
0
s ds; t 0:
is a ( ; F ; Q ) Brownian motion. Therefore,
dX t = r t X t dt + X t df W t :
c) According to the Cameron-Martin-Girsanov theorem, the Radon-Nikodym deriv- ative is
d Q
d P = exp
Z T
0
t dW t 1 2
Z T
0 2 t dt
= exp
" R T
0 (1 + sin t) r dW t
1 2 1 2
r 2
(4 4 cos T cos T sin T + 3T )
#
= exp
"
r W T r R T
0 sin t dW t
r 2
1 cos T 1 4 cos T sin T + 3 4 T
# :
Consequently,
Q (A) = E P
"
exp
"
r W T r R T
0 sin t dW t
r 2
1 cos T 1 4 cos T sin T + 3 4 T
#
A
#
:
2 Exercise 12.2
a) Let Y t = R t
0 r u du. Y is an Itô process (dY t = K t dt + H t dW t )for which K t = r t and H t = 0 since
Z T
0 j K s j ds = Z T
0 j r s j ds = Z T
0
+ (r 0 ) e cs ds Z T
0
+ (r 0 + ) e cs ds = T + (r 0 + ) 1 e cT c < 1 : Note that
dY t = r t dt and h Y i t = Z T
0
H s 2 ds = 0:
Let f (t; y) = e y . We have @f @t = 0, @f @y = @ @y
2f
2= f , f (t; Y t ) = S t and f (0; Y 0 ) = 1 = S 0 . From Itô’s lemma,
dS t = df (t; Y t )
= @f
@t (t; Y t ) dt + @f
@y (t; Y t ) dY t + 1 2
@ 2 f
@y 2 (t; Y t ) d h Y i t
= f (t; Y t ) dY t
= S t r t dt b)
dS t = S t dt + S t dW t
= r t S t dt + S t r t
dt + S t dW t
= r t S t dt + S t d W t + Z t
0
r u du
Let t = r
t. We need to verify that E P exp 1
2 Z T
0 2
t dt < 1
is verify to use Girsanov theorem.
E P exp 1 2
Z T
0 2
t dt = E P
"
exp 1 2
Z T
0
r t 2 dt
!#
= E P
"
exp 1 2
Z T
0
( + (r 0 ) e ct ) 2 dt
!#
= exp 1 2
Z T
0
(r 0 ) e ct 2 dt
!
exp Z T
0
C 1 + C 2 e ct + C 3 e 2ct dt
< 1
3 Exercise 12.4
3.1 The main ideas
To show that there is no arbitrage opportunity, we need to …nd a risk neutral measure under which the discounted price process of tradable assets are martingale.
What are these tradable assets? In the perspective of a local investor, these are the three riskless assets, all expressed in Canadian dollars. The value in Canadian dollars of the American riskless asset A (t) = V (t) A (t) satis…es
dA (t) = V (t) dA (t) + A (t) dV (t) + d h A; V i (t)
= (r U + V ) A (t) dt + V A (t) dW V (t) :
Similarly, the value in Canadian dollars of the Japanese riskless asset D (t) = C (t) D (t) is characterize with
dD (t) = C (t) dD (t) + D (t) dC (t) + d h D; C i (t)
= (r J + C ) D (t) dt + C D (t) dW C (t) :
But the value in Canadian dollars of the American riskless asset is also A (t) = C(t) U(t) A (t).
Indeed, U(t) 1 A (t) is the value in yen of the American bank account and C (t) U(t) 1 A (t) is the
value in Canadian dollars of U(t) 1 A (t) which is expressed in yens. Using Itô’s lemma, dA (t) = C (t)
U 2 (t) A (t) dU (t) + 1
U (t) A (t) dC (t) + C (t)
U (t) dA (t) + 1
2 2C (t)
U 3 (t) A (t) d h U i (t) 1
U 2 (t) A (t) d h C; U i (t)
= r U + C U + 2 U C U CU A (t) dt + C A (t) dW C (t) U A (t) dW U (t) :
Similarly, the value in Canadian dollars of the Japanese bank account satis…es D (t) = V (t) U (t) D (t). Itô’s lemma gives
dD (t) = (r J + U + V + U V U V ) D (t) dt + U D (t) dW U (t) + V D (t) dW V (t) :
Finally, C(t) 1 B (t) is the value in yen of the Canadian bank account; C(t) U(t) B (t) is expressed in American dollars and B (t) = V (t)U(t) C(t) B (t) is reverted back to Canadian dollars. Itô’s lemma gives
dB (t) = V (t) U (t)
C 2 (t) B (t) dC (t) + U (t)
C(t) B (t) dV (t) + V (t)
C(t) B (t) dU (t) + V (t) U (t)
C(t) dB (t) + 1
2
2V (t) U (t)
C 3 (t) B (t) d h C i (t) V (t)
C 2 (t) B (t) d h C; U i (t) U (t)
C 2 (t) B (t) d h C; V i (t) + 1
C(t) B (t) d h U; V i (t)
= C + V + U + r C + 2 C C U CU C V CV + U V U V B (t) dt
C B (t) dW C (t) + U B (t) dW U (t) + V B (t) dW V (t) :
Since the value in Canadian dollars of the American bank account has two di¤erent expressions, , A (t) and A (t), these two processes must be the same
dA (t) = (r U + V ) A (t) dt + V A (t) dW V (t) dA (t) = r U + C U + 2 U C U CU A (t) dt
+ C A (t) dW C (t) U A (t) dW U (t) : Therefore,
r U + V = r U + C U + 2 U C U CU (1)
V W V (t) = C W C (t) U W U (t) : (2)
Similarly, the value in Canadian dollars of the Japanese bank account have two expressions:
dD (t) = (r J + C ) D (t) dt + C D (t) dW C (t) dD (t) = (r J + U + V + U V U V ) D (t) dt
+ U D (t) dW U (t) + V D (t) dW V (t) Therefore
r J + C = r J + U + V + U V U V
C W C (t) = U W U (t) + V W V (t) : These four restrictions may be rewritten as
V = C U + 2 U C U CU ;
V = C U U V U V ;
V W V (t) = C W C (t) U W U (t) : From the two …rst ones, we conclude that
U = C CU V U V : From the last one,
U V = Corr [W U (t) ; W V (t)]
= Corr W U (t) ; C W C (t) U W U (t)
V
= C
V
Corr [W U (t) ; W C (t)] U
V
Corr [W U (t) ; W U (t)]
= C CU U
V
= C CU ( C CU V U V )
= U V V
which do not brings new information. Therefore
U = C CU V
The second step consist in …nding the risk neutral measures. The Choleski decomposition allows to express the dependent Brownian motions as a linear combination of independent Brownian motions:
W V (t) = a 11 B 1 (t) + a 12 B 2 (t) + a 13 B 3 (t) W U (t) = a 21 B 1 (t) + a 22 B 2 (t) + a 23 B 3 (t) W C (t) = a 31 B 1 (t) + a 32 B 2 (t) + a 33 B 3 (t) : Let
B e i (t) = B i (t) + Z t
0
i (s) ds; i = 1; 2; 3 and
W f V (t) = a 11 B e 1 (t) + a 12 B e 2 (t) + a 13 B e 3 (t) = W V (t) + Z t
0
V (s) ds W f U (t) = a 21 B e 1 (t) + a 22 B e 2 (t) + a 23 B e 3 (t) = W U (t) +
Z t
0
U (s) ds f
W C (t) = a 31 B e 1 (t) + a 32 B e 2 (t) + a 33 B e 3 (t) = W C (t) + Z t
0
C (s) ds where
V (t) = a 11 1 (t) + a 12 2 (t) + a 13 3 (t)
U (t) = a 21 1 (t) + a 22 2 (t) + a 23 3 (t)
C (t) = a 31 1 (t) + a 32 2 (t) + a 33 3 (t) : Therefore
dA (t) = (r U + V V V ) A (t) dt + V A (t) df W V (t)
dA (t) = r U + C U + 2 U C U CU C C + U U A (t) dt + C A (t) df W C (t) U A (t) df W U (t)
dD (t) = (r J + C C C ) D (t) dt + C D (t) df W C (t)
dD (t) = (r J + U + V + U V U V U U V V ) D (t) dt + U D (t) df W U (t) + V D (t) df W V (t)
dB (t) = C + V + U + r C + 2 C C U CU C V CV + U V U V
+ C C U U V V B (t) dt
C B (t) df W C (t) + U B (t) df W U (t) + V B (t) df W V (t) :
To risk neutralize the system, the drift parameter must be the Canadian risk free rate:
r U + V V V = r C
r U + C U + 2 U C U CU C C + U U = r C r J + C C C = r C (r J + U + V + U V U V U U V V ) = r C
C + V + U + r C + 2 C C U CU C V CV + U V U V
+ C C U U V V = r C :
On matrix form, the …rst three expressions become 2
4 V
0 0
0 U C
0 0 C
3 5
2 4 V U
C
3 5 +
2 4
r U + V r C
r U + C U + 2 U C U CU r C r J + C r C
3 5 =
2 4
0 0 0
3 5 :
The solution is 2 4 V U
C
3 5 =
2 4
1
V
(r U + V r C )
1
U
(r J r U + U 2 U + C U CU )
1
C
