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Here is the original question, Suppose you have a call option on the square of a log-normal asset $V_t$. what equation does the price satisfy?

my question is how this corresponds to a change of variable stated in the solution to this question. in the original BS equation.enter image description here solution gives,

since $V_t$ is also a log-normal process, The BS equation governs this process will be similar to that governs $S_t$ where $S_t$ is a standard log-normal process.

here is the full solution,

enter image description here

junyaozheng98
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1 Answers1

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Notice that by Ito if we set $F(s,t)=s^2$ we have $$F_s(s,t)=2s,\,F_{ss}(s,t)=2$$ so $$dF=\bigg(2r S_t^2+2\frac{1}{2}\sigma^2S_t^2\bigg)dt+2\sigma S_t^2dW_t$$ $$\implies \frac{dF}{F}=(2r+\sigma^2)dt+2\sigma dW_t$$ So the PDE satisfied by a call option on such asset is $$C_t(x,t)+x(2r+\sigma^2)C_x(x,t)+2\sigma^2x^2C_{xx}(x,t)=rC(x,t)$$ with terminal condition $C(x,T)=(x-K)^+$. To see that this corresponds to a change of variable in the original PDE, set $x=s^2$ obtaining $$\begin{aligned}C_t(s,t)&=C_t(x,t)\\ C_s(s,t)&=2sC_x(x,t)\\ C_{ss}(s,t)&=2C_x(x,t)+4s^2C_{xx}(x,t) \end{aligned}$$ Substituting these back into the Black-Scholes PDE you obtain the PDE above.

Snoop
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