The purpose of this problem is to solve the Black-Scholes PDE with analytical techniques, which will lead us back to the Black-Scholes formula. The technique is very similar to the one used with the Feynman-Kac formula back in MATH 467. Let's consider the PDE given by with terminal condition f(T,x) af + Ət 1 02 ર .2მ2 f af მ2 +rx მე - rf = 0, = (x-K)+. The solution f(t, x) corresponds to the price of a call option (given the initial condition) at time t if the stock price is x. (a) The first two things that prevent us from solving this PDE directly are (i) the fact that we have a terminal condition, instead of an initial condition; (ii) the terms in front of the derivatives are not constant. To address these, we use the transformation g(t, x) = ƒ(T − t,e³), equivalent to f(t, x) = g(T-t, log(x)). Under this condition, determine the PDE and the initial condition satisfied by g. (b) Now, the PDE obtained in (a) should have an initial condition and constant coeffi- cients, but it still includes a first order derivative in x and the function g itself. In order to simplify this, use the transformation h(t, x) eax+bt g(t, x) and determine the appropriate values of a and b in order for h to solve a heat equation (see (c)). = (c) We know that the solution to the heat equation Əh a ²h Ət 2 მ2 = 0 with intial condition ho is given by 1 h(t, x) = L √ not 2πat (x-y)2 e 2at ho(y) dy. Use this result and the initial condition found in (b) to determine the function h. 2 (d) Work the transformations backwards to determine the function g, and then f. You should recover the Black-Scholes formula.
The purpose of this problem is to solve the Black-Scholes PDE with analytical techniques, which will lead us back to the Black-Scholes formula. The technique is very similar to the one used with the Feynman-Kac formula back in MATH 467. Let's consider the PDE given by with terminal condition f(T,x) af + Ət 1 02 ર .2მ2 f af მ2 +rx მე - rf = 0, = (x-K)+. The solution f(t, x) corresponds to the price of a call option (given the initial condition) at time t if the stock price is x. (a) The first two things that prevent us from solving this PDE directly are (i) the fact that we have a terminal condition, instead of an initial condition; (ii) the terms in front of the derivatives are not constant. To address these, we use the transformation g(t, x) = ƒ(T − t,e³), equivalent to f(t, x) = g(T-t, log(x)). Under this condition, determine the PDE and the initial condition satisfied by g. (b) Now, the PDE obtained in (a) should have an initial condition and constant coeffi- cients, but it still includes a first order derivative in x and the function g itself. In order to simplify this, use the transformation h(t, x) eax+bt g(t, x) and determine the appropriate values of a and b in order for h to solve a heat equation (see (c)). = (c) We know that the solution to the heat equation Əh a ²h Ət 2 მ2 = 0 with intial condition ho is given by 1 h(t, x) = L √ not 2πat (x-y)2 e 2at ho(y) dy. Use this result and the initial condition found in (b) to determine the function h. 2 (d) Work the transformations backwards to determine the function g, and then f. You should recover the Black-Scholes formula.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.3: The Natural Exponential Function
Problem 58E
Related questions
Question
Transcribed Image Text:The purpose of this problem is to solve the Black-Scholes PDE with analytical techniques,
which will lead us back to the Black-Scholes formula. The technique is very similar to the
one used with the Feynman-Kac formula back in MATH 467.
Let's consider the PDE given by
with terminal condition f(T,x)
af
+
Ət
1
02
ર
.2მ2 f
af
მ2
+rx
მე
- rf = 0,
=
(x-K)+. The solution f(t, x) corresponds to the price
of a call option (given the initial condition) at time t if the stock price is x.
(a) The first two things that prevent us from solving this PDE directly are
(i) the fact that we have a terminal condition, instead of an initial condition;
(ii) the terms in front of the derivatives are not constant.
To address these, we use the transformation g(t, x) = ƒ(T − t,e³), equivalent to
f(t, x) = g(T-t, log(x)). Under this condition, determine the PDE and the initial
condition satisfied by g.
(b) Now, the PDE obtained in (a) should have an initial condition and constant coeffi-
cients, but it still includes a first order derivative in x and the function g itself. In
order to simplify this, use the transformation h(t, x) eax+bt g(t, x) and determine
the appropriate values of a and b in order for h to solve a heat equation (see (c)).
=
(c) We know that the solution to the heat equation
Əh a ²h
Ət 2 მ2
= 0 with intial condition
ho is given by
1
h(t, x) = L √ not
2πat
(x-y)2
e 2at ho(y) dy.
Use this result and the initial condition found in (b) to determine the function h.
2
(d) Work the transformations backwards to determine the function g, and then f. You
should recover the Black-Scholes formula.
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