Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One step in the derivation of Black-Scholes

Assumptions:(1) ${\displaystyle \frac{\partial F}{\partial t}(t,x)+\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}F}{\partial x^{2}}(t,x)-rF(t,x)+rx\frac{\partial F}{\partial x}(t,x)=0}$

(2) $\tilde{F}(t,x)=e^{-rt}F(t,xe^{rt})$ .

Show that ${\displaystyle \frac{\partial\tilde{F}}{\partial t}(t,x)=-\frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2}\tilde{F}}{\partial x^{2}}(t,x)}$.

Although to derive Black-Scholes we might need PDE and Ito's lemma, I think here we only need calculus. However I cannot seem to get it right. I guess my problem has something to do with the partial derivative of $F(t,xe^{rt})$. Thank you!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

First notice that (2) implies $$F(t,x) = e^{r t} \tilde F(t,x e^{-r t}).$$ Plug this directly into the Black-Scholes partial differential equation (1). You should find, for example, $$\frac{\partial F(t,x)}{\partial t} = r e^{r t} \tilde F(t,x e^{-r t}) + e^{r t} \tilde F^{(1,0)}(t,x e^{-r t}) - r x \tilde F^{(0,1)} (t,x e^{-r t})$$ where $\tilde F^{(0,1)}(t,x e^{-r t})$ indicates a first derivative on the second argument, i.e., $$\tilde F^{(0,1)}(t,x e^{-r t}) = \frac{\partial \tilde F(t,X)}{\partial X}|_{X=x e^{-r t}}.$$ Notice that the partial derivative acts on the factor $e^{r t}$, the first argument of $\tilde F$, as well as its second argument. If you can get this term, you should be able to work out the others.

Working out the rest of the terms we find Black-Scholes implies $$e^{r t} \tilde F^{(1,0)}(t,x e^{-r t}) = -\frac{1}{2} \sigma^2 x^2 e^{-r t} \tilde F^{(0,2)} (t,x e^{-r t}).$$ Change variables. Let $X = x e^{-r t}$. We immediately find
$$\frac{\partial\tilde F(t,X)}{\partial t} = -\frac{1}{2} \sigma^2 X^2 \frac{\partial^2 \tilde F (t,X)}{\partial X^2},$$ which is the required equation.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.