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Consider the Black and Scholes equation $$V_t - \frac{1}{2}x^2 \sigma^2 V_{xx} - \rho xV_x + \rho V = 0 \quad \text{ in } (0,\infty)\times \mathbb{R},$$ with initial condition $$V(0,x) = V_0(x).$$

What changes of variables reduce this equation to $V_t - \frac{1}{2}V_{xx} = 0$? How does the initial condition change?

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  • $\begingroup$ Black-Scholes would be $V_t - \frac{1}{2}x^2 \sigma^2 V_{xx} - \rho xV_x + \rho V = 0$ $\endgroup$ – RRL Feb 28 '18 at 17:40
  • $\begingroup$ see the wikipedia page for black scholes $\endgroup$ – qbert Feb 28 '18 at 17:46
  • $\begingroup$ @RRL You're right. I edited the question. $\endgroup$ – user411609 Feb 28 '18 at 17:48
  • $\begingroup$ @qbert That was not very clear to me. $\endgroup$ – user411609 Feb 28 '18 at 17:48
  • $\begingroup$ @Rene that's sort of a different question. I spent some time trying to come up with the change of variables which is not trivial. However understanding it when someone gives it to you should be just the chain rule $\endgroup$ – qbert Feb 28 '18 at 17:58
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The Black-Scholes PDE is $V_t - \frac{1}{2}x^2 \sigma^2 V_{xx} - \rho xV_x + \rho V = 0 $.

Taking $U(t,y) = e^{\rho t}V(t,e^y)$ we have

$$U_t - \frac{\sigma^2}{2}U_{yy} - (\rho - \sigma^2/2)U_y = 0.$$

Next, taking $W(t,z) = U(t, \sigma z - (\rho - \sigma^2/2)t)$ we have

$$W_t - \frac{1}{2} W_{zz} = 0.$$

Putting it together, we have the transformation

$$W(t,z) = e^{\rho t}V\left(\, t,\, \exp[\, \sigma z - (\rho - \sigma^2/2)t)\, ]\, \right),$$

and the initial condition is

$$W(0,z) = V(0,e^{\sigma z})= V_0(e^{\sigma z}).$$

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