1
$\begingroup$

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!

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .