MATH 623 FinalExam 11 solutions A++

(1) Suppose I wish to compute the value of a European Straddle option on a stock. The option pays (S) if
exercised when the stock price is S, where
(S) =
(
34 􀀀 S if S 34
S 􀀀 34 if S 34:
The current value of the stock is 33 and it expires 8 months from today. The stock volatility is = 0:31
and the risk free interest rate is r = 0:05. The value of the option is to be obtained by numerically solving
a terminal-boundary value problem for a PDE. The PDE is the Black-Scholes PDE transformed by the
change of variable S = ex, where S is the stock price. Hence the value u(x; t) of the option as a function
of the variables x and t with the units of t in years, is a solution of the di erential equation
@u
@t
+

r 􀀀
2
2

@u
@x
+
2
2
@2u
@x2
􀀀 ru = 0;
on the interval a < x < b; 0 t T, where t = 0 denotes today, with terminal condition u(x; T) =
    (x); a < x < b, and boundary conditions u(a; t) = f(t); u(b; t) = g(t); 0 t T. I use the explicit
Euler method to numerically solve the problem. This can be written in the form
u(x; t 􀀀 t) = Au(x; t) + Bu(x + x; t) + Cu(x 􀀀 x; t) ; a < x < b;
where A;B;C are constant parameters and a = 2:7; b = 4:3.
(a) Explain why the choices for a; b are suitable for this problem.
(b) Find the values of f(0) and g(T).
(c) In the numerical scheme I take x = 0:04; t = 0:015. Find the value of the parameter B.
(d) Explain why the choice x = 0:04; t = 0:015 in (c) is suitable for the numerical scheme but the
choice x = 0:04; t = 0:02 is not suitable.
(a) Soln: = :31; T = 2=3; K = 34. Hence we should choose b logK + 3
p
T = 4:28 and
a logK 􀀀 3
p
T = 2:76.
(b) Soln: We have f(t) = Ke􀀀r(T􀀀t) 􀀀 ea and g(t) = eb 􀀀 Ke􀀀r(T􀀀t). Hence f(0) = 18:0056 and
g(T) = 39:6998.
(c) Soln: The explicit Euler numerical scheme for the PDE is
u(x; t) 􀀀 u(x; t 􀀀 t)
t
+

r 􀀀
2
2

u(x + x; t) 􀀀 u(x; t)
2x
+
2
2

u(x + x; t) + u(x 􀀀 x; t) 􀀀 2u(x; t)
(x)2

􀀀 ru(x; t) = 0:
We conclude that
B =
2t
2(x)2 +

r 􀀀
2
2

t
2x
= :4508:
(d) Soln: For x = 0:04; t = 0:02 we have 2t=(x)2 = :901 < 1, so this choice for x; t yields
a stable numerical scheme. If x = 0:04; t = 0:015 we have 2t=(x)2 = 1:2 > 1, so this choice for
x; t yields an unstable numerical scheme.
2
(2) We wish to nd the value of an Asian option on a zero dividend stock which expires 9 months from today.
The payo on the option is the excess of the stock price at expiration over the continuous average of
the stock price during the 9 month life span of the option. The current price of the stock is 28 and its
volatility is 0:37 per annum. The risk free interest rate is 0:04 per annum. One can compute the value
of the Asian option by solving a terminal-boundary value problem for a PDE. The PDE for the function
w(; t) is given by
@w
@t
+ a()
@w
@
+ b()
@2w
@2 + c()w = 0;
where a(); b(); c() are known functions of . The equation is to be solved in the interval 0 < t < T; 0 <
< max; with terminal condition w(; T) = (), where t = 0 denotes today. We use the explicit Euler
method with = 0:025; t = 0:001 to numerically solve the problem. This can be written in the form
w(; t 􀀀 t) = A()w(; t) + B()w( + ; t) + C()w( 􀀀 ; t) ; 0 < max :
(a) Find the value of a(0:6) and (0:6).
(b) Find the values of A();B();C() when = 0.
(c) Suppose that in the numerical scheme we approximate the derivative @w(; t)=@ by a symmetric
di erence for > 0 > 0 and a forward di erence for 0 0. Explain why it makes sense to do
this and obtain a suitable value for 0.
(a) Soln: We have that a() = 1 􀀀 r and () = max[1 􀀀 =T; 0]. Since r = :04; T = 3=4; we
conclude that a(0:6) = :976 and (0:6) = :200.
(b) Soln: The boundary condition at = 0 is given by
@w(; t)
@t
+
@w(; t)
@
= 0;
which in the discrete approximation yields
w(; t) 􀀀 w(; t 􀀀 t)
t
+
w( + ; t) 􀀀 w(; t)

= 0:
Hence we have that
w(; t 􀀀 t) =
t

w( + ; t) +

1 􀀀
t


w(; t) :
We conclude that B(0) = t= = 0:04 and A(0) = 1 􀀀 B(0) = 0:96. Evidently C(0) = 0.
(c) Soln: We should try to approximate the derivative @w(; t)=@ by the second order accurate
approximation
@w(; t)
@
'
w( + ; t) 􀀀 w( 􀀀 ; t)
2
so that cumulative error is O[()2]. This is not possible for all > 0 as we can see that the boundary
condition forces us to use the forward di erence approximation
@w(; t)
@
'
w( + ; t) 􀀀 w(; t)

at = 0, which is rst order accurate. The numerical scheme with the second order di erence approx-
imation is only stable for >
p
=, but the scheme with the rst order di erence approximation is
stable for all 0. Hence we should take 0 =
p
= = :4273.