Initial value PDE: the Black-Scholes model (part a)

kreedman

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This turned my head around!

The celebrated Black-Scholes model [20] for the pricing of stock options is central in mathematical finance; see, for example, [156]. The PDE is given by:

. . . . .\(\displaystyle \L u_t\, +\, \frac{1}{2}\sigma^2 x^2 u_{xx}\, +\, rxu_x\, -\, ru\, =\, 0,\, \mbox{ }\, 0\, <\, x\, <\, \infty,\, \mbox{ }\, t\, \leq\, T\, \, \mbox{ }(1.27)\)

For the sake of completeness let us add that u is the sought value of the option under consideration, t is time, x is the current value of the underlying asset, r is the interest rate, \(\displaystyle \sigma\) is the volatility of the underlying asset, T is the expiry date, and E is the exercise price. In general, r and \(\displaystyle \sigma\) may vary, but they are assumed to be known constants, as are E and T.

For the European call option, we have the terminal condition:


. . . . .\(\displaystyle \L u(T,\,x)\, =\, max(x\,-\,E,\,0)\,\,\mbox{ }\,(1.28a)\)

...and the boundary conditions:

. . . . .\(\displaystyle \L u(t,\,0)\, =\, 0,\,\,u(t,\,x)\,\sim \, x\, -\, Ee^{-r(T\,-\,t)}\,\,\mbox{as}\,x\, \rightarrow \,\infty\,\,\mbox{ }\,(1.28b)\)

a) Show that the transformation:

. . . . .\(\displaystyle \L x\,=\,Ee^y ,\,\,t\,=\, T\,-\, \frac{2s}{\sigma^2} ,\, \, u\,=\, Ev(s,\,y)\)

...results in the initial value PDE:

. . . . .\(\displaystyle \L v_s\, =\, v_{yy}\, +\, (\kappa \,-\,1)v_y \, -\, \kappa v,\,\, \mbox{ }\, -\infty\, <\, y\, < \infty\)

. . . . .\(\displaystyle \L v(0,\, y)\, =\, max(e^y \, -\, 1,\, 0)\)

...where:

. . . . .\(\displaystyle \L \kappa \, =\, \frac{2s}{\sigma^2}\)
link to screen-shot
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Edited by stapel -- Reason for edit: Replacing graphic with text
 
Please show your work - so that we know exactly where you are having problem.
 
Well,

I just need to understand the question. WHat is it asking me?

And hints on how to solve such a problem.
 
Re: Initial value PDE

kreedman said:
This turned my head around!

The celebrated Black-Scholes model [20] for the pricing of stock options is central in mathematical finance; see, for example, [156]. The PDE is given by:

. . . . .\(\displaystyle \L u_t\, +\, \frac{1}{2}\sigma^2 x^2 u_{xx}\, +\, rxu_x\, -\, ru\, =\, 0,\, \mbox{ }\, 0\, <\, x\, <\, \infty,\, \mbox{ }\, t\, \leq\, T\, \, \mbox{ }(1.27)\)

For the sake of completeness let us add that u] is the sought value of the option under consideration, t is time, x is the current value of the underlying asset, r is the interest rate, \(\displaystyle \sigma\) is the volatility of the underlying asset, T is the expiry date, and E is the exercise price. In general, r and \(\displaystyle \sigma\) may vary, but they are assumed to be known constants, as are E and T.

For the European call option, we have the terminal condition:


. . . . .\(\displaystyle \L u(T,\,x)\, =\, max(x\,-\,E,\,0)\,\,\mbox{ }\,(1.28a)\)

...and the boundary conditions:

. . . . .\(\displaystyle \L u(t,\,0)\, =\, 0,\,\,u(t,\,x)\,\sim \, x\, -\, Ee^{-r(T\,-\,t)}\,\,\mbox{as}\,x\, \rightarrow \,\infty\,\,\mbox{ }\,(1.28b)\)

a) Show that the transformation:

. . . . .\(\displaystyle \L x\,=\,Ee^y ,\,\,t\,=\, T\,-\, \frac{2s}{\sigma^2} ,\, \, u\,=\, Ev(s,\,y)\)

...results in the initial value PDE:

. . . . .\(\displaystyle \L v_s\, =\, v_{yy}\, +\, (\kappa \,-\,1)v_y \, -\, \kappa v,\,\, \mbox{ }\, -\infty\, <\, y\, < \infty\)

. . . . .\(\displaystyle \L v(0,\, y)\, =\, max(e^y \, -\, 1,\, 0)\)

...where:

. . . . .\(\displaystyle \L \kappa \, =\, \frac{2s}{\sigma^2}\)
link to screen-shot
________________________________
Edited by stapel -- Reason for edit: Replacing graphic with text

Start with:

u = E* v(s,y)

\(\displaystyle u_t= E[v_s * (ds/dt) + v_y * (dy/dt)]\)

and

\(\displaystyle u_x= E[v_s * (ds/dx) + v_y * (dy/dx)]\)

and

\(\displaystyle u_{xx} = ...\)

and so on....
 
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