PDE change of variables Black-Scholes equation

In summary: Sure, no problem! Let's start by breaking down the equation:\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0We want to get this in terms of the new variables, so let's start by substituting in the definitions for \tau and x:\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}e^{2r\tau}(e^{2x}-1)\frac{\partial^2 V}{\partial x^2} + re^
  • #1
perishingtardi
21
1

Homework Statement


By changing variables from [itex](S,t,V)[/itex] to [itex](x,\tau,u)[/itex] where
[tex]\tau = T - t,[/tex]
[tex]x = \ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T-t),[/tex]
[tex]u=e^{r\tau}V,[/tex]
where [itex]r, \sigma, \tau, K[/itex] are constants, show that the Black-Scholes equation
[tex]\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0[/tex]
reduces to the diffusion equation
[tex]\frac{\partial u}{\partial \tau} - \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}=0.[/tex]

Homework Equations


Chain rule.

The Attempt at a Solution


I know that [tex]\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau}[/tex] and similarly for [itex]\partial / \partial S[/itex]. I don't know what to do with the second-order derivative though. Since [itex]\partial / \partial S[/itex] turns out to be [itex]\frac{1}{S}\frac{\partial}{\partial x}[/itex], I reckoned that [tex]\frac{\partial^2 V}{\partial S^2} = \frac{1}{S}\frac{\partial}{\partial x} \left( \frac 1 S \frac{\partial V}{\partial x} \right),[/tex] but that seems to just make things more complicated.
 
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  • #2
perishingtardi said:

Homework Statement


By changing variables from [itex](S,t,V)[/itex] to [itex](x,\tau,u)[/itex] where
[tex]\tau = T - t,[/tex]
[tex]x = \ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T-t),[/tex]
[tex]u=e^{r\tau}V,[/tex]
where [itex]r, \sigma, \tau, K[/itex] are constants, show that the Black-Scholes equation
[tex]\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0[/tex]
reduces to the diffusion equation
[tex]\frac{\partial u}{\partial \tau} - \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}=0.[/tex]

Homework Equations


Chain rule.


The Attempt at a Solution


I know that [tex]\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau}[/tex] and similarly for [itex]\partial / \partial S[/itex]. I don't know what to do with the second-order derivative though. Since [itex]\partial / \partial S[/itex] turns out to be [itex]\frac{1}{S}\frac{\partial}{\partial x}[/itex], I reckoned that [tex]\frac{\partial^2 V}{\partial S^2} = \frac{1}{S}\frac{\partial}{\partial x} \left( \frac 1 S \frac{\partial V}{\partial x} \right),[/tex] but that seems to just make things more complicated.

Use
[tex]
\frac{\partial^2 V}{\partial x^2} = S\frac{\partial }{\partial S}\left(S\frac{\partial V}{\partial S}\right)
= S\frac{\partial V}{\partial S} + S^2 \frac{\partial^2 V}{\partial S^2}[/tex]
to eliminate [itex]S^2 \frac{\partial^2 V}{\partial S^2}[/itex] to leave you with terms in [itex]\partial V/\partial t[/itex], [itex]S\partial V/\partial S[/itex], [itex]V[/itex] and [itex]{\partial^2 V}/{\partial x^2}[/itex]. Then you can tidy up the first-order derivatives.
 
  • #3
Sorry to resurrect a very old thread...but can anyone provide more guidance on this? I'm also stuck on this question. Any help would be greatly appreciated!

I see how the second derivative above helps...but I'm not sure whether to start with V or u. How do I translate between the two?

Also can you help with the partial derivatives of V wrt S and t?
 

FAQ: PDE change of variables Black-Scholes equation

1. What is a PDE change of variables in the Black-Scholes equation?

A PDE (partial differential equation) change of variables is a mathematical technique used to transform the original variables in a PDE, such as time and underlying asset price, into a new set of variables. This transformation can simplify the equation and make it easier to solve.

2. Why is a change of variables necessary in the Black-Scholes equation?

The Black-Scholes equation is a PDE that describes the behavior of financial derivatives, such as options, over time. However, the equation is complex and difficult to solve directly. A change of variables allows us to transform the equation into a simpler form, making it easier to solve and understand.

3. How is a change of variables performed in the Black-Scholes equation?

A change of variables in the Black-Scholes equation involves replacing the original variables, time and asset price, with new variables, typically called the "risk-neutral" variables. These new variables are defined based on the underlying assumptions of the Black-Scholes model and allow for a more simplified equation.

4. What are the benefits of using a change of variables in the Black-Scholes equation?

Using a change of variables in the Black-Scholes equation can lead to a better understanding of the behavior of financial derivatives and can make the equation more tractable for solving. It can also allow for the derivation of important formulas, such as the Black-Scholes option pricing formula, which would be difficult to obtain otherwise.

5. Are there any limitations to using a change of variables in the Black-Scholes equation?

While a change of variables can simplify the Black-Scholes equation, it is not a perfect solution. The new variables may introduce some distortion to the original equation, and the results may not be as accurate as solving the original equation directly. Additionally, the change of variables may not be applicable to all types of financial derivatives or market conditions.

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