# question on PDE with chain rule

We have PDE for $v(t,x)$:

$$\partial_tv+\frac{1}{2}\sigma^2x^2\partial_{xx}v=0$$

Then it says for $\tilde{v}$, with $x=S_t$ and $\tilde{x}=e^{rt}x$:

I understand the steps in between, but how do they then finally get to this PDE?

Apparently one can use the middle equality to recalculate the partial derivatives in the first PDE, but I didnt get to the resulting PDE then.

• Use the product rule to find the derivatives of $v$ in terms of the derivatives of $\tilde v$. Plug the results into the original PDE.
– user147263
Aug 21, 2014 at 16:53
• @900sit-upsaday Can you please do this ;) ? Aug 21, 2014 at 16:54
• Ah the black scholes equation. :) Aug 21, 2014 at 16:54
– user147263
Aug 21, 2014 at 16:54
• @900sit-upsaday ..then you dont get the answer vote you could get Aug 21, 2014 at 16:56

Let me answer a different more general question $$v(t,S) = \mathrm{e}^{-rt}U\left(t,g(x,t)\right).\tag{1}$$

$$v_t = -r\mathrm{e}^{-rt}U\left(t,g(x,t)\right) + \mathrm{e}^{-rt}U_t.\tag{2}$$ now you must be happy with Eq.(2)?

so lets compute $U_t$,

$$U_t = \dfrac{\partial U}{\partial t} + \dfrac{\partial U}{\partial g}\dfrac{\partial g(x,t)}{\partial t}$$ so putting it all together we have

$$v_t = \mathrm{e}^{-rt}\left[\dfrac{\partial U}{\partial t} + \dfrac{\partial g(x,t)}{\partial t}\dfrac{\partial U}{\partial g}\right]-r\mathrm{e}^{-rt}U\left(t,g(x,t)\right)$$

To go further.

$$U = \tilde{v},\\ g(x,t) = \tilde{x} = x\mathrm{e}^{rt}$$

therfore

$$v_t = \mathrm{e}^{-rt}\left[\dfrac{\partial \tilde{v}}{\partial t} + rx\mathrm{e}^{rt}\dfrac{\partial \tilde{v}}{\partial \tilde{x}}\right]-r\mathrm{e}^{-rt}\tilde{v}$$

now lets look at $$v_x = \mathrm{e}^{-rt}\dfrac{\partial \tilde{v}}{\partial \tilde{x}}\dfrac{\partial \tilde{x}}{\partial x} = \mathrm{e}^{-rt}\dfrac{\partial \tilde{v}}{\partial \tilde{x}}\left(\mathrm{e}^{rt}\right) = \dfrac{\partial \tilde{v}}{\partial \tilde{x}},\\ v_{xx} = \frac{\partial}{\partial x}\dfrac{\partial \tilde{v}}{\partial \tilde{x}} = \mathrm{e}^{rt}\dfrac{\partial^2 \tilde{v}}{\partial \tilde{x}^2}$$

remembering $x = \tilde{x}\mathrm{e}^{-rt}$

we find

$$v_t + \frac{1}{2}\sigma^2v_{xx}$$ is equivalent to

$$\mathrm{e}^{-rt}\left[\dfrac{\partial \tilde{v}}{\partial t} + rx\mathrm{e}^{rt}\dfrac{\partial \tilde{v}}{\partial \tilde{x}}-r\tilde{v}\right] + \frac{1}{2}\sigma^2 \tilde{x}^2\mathrm{e}^{-2rt}\left[\mathrm{e}^{rt}\dfrac{\partial^2 \tilde{v}}{\partial \tilde{x}^2}\right]$$ or finally

$$\mathrm{e}^{-rt}\left[\dfrac{\partial \tilde{v}}{\partial t} + r\tilde{x}\dfrac{\partial \tilde{v}}{\partial \tilde{x}}-r\tilde{v} + \frac{1}{2}\sigma^2 \tilde{x}^2\dfrac{\partial^2 \tilde{v}}{\partial \tilde{x}^2}\right] = 0$$

now all you have to do is a trivial re-arrangement.

• But please also take @900 note on board and update the question ( you may use this hint to further the approach you have taken and if you get stuck then an answer would be more forthcoming :) ) Aug 21, 2014 at 17:35
• Can you please write it in the form as I had? I really cannot follow sry. Aug 21, 2014 at 20:21
• Nice and complete answer. I added another answer, on the other typical form to derive the BS equation. Aug 21, 2014 at 21:48
• @Chinny84 Thank you, can you please explain why there is $\partial_xv$ in $\partial_tv$? Aug 21, 2014 at 22:11
• It is due to the transformed variable $\tilde{x}$ having a variable $t$ in there. Aug 21, 2014 at 22:19

In order to derive the Black-Scholes equation, you can do it in a more intuitive way (with economics). Also, this needs understanding of Brownian Motions and stochastic differential equations (SDEs), Îto's Lemma, and some basical assumptions that I will not list here (they're already available in Wikipedia and other references).

Let's see it in two steps:

Let's assume the asset $x$ evolves according to $$dx=\mu x\,dt+\sigma x\,dW_t$$

Let $\Pi$ denote the valie of a portfolio of one long option position and a short position in a quantity $\Delta$ of the underlying: $$\Pi=V(x,t)-\Delta x$$ The value of the portfolio changes partly because the change in the option value, and partly because of the change in the underlying: $$d\Pi=dV-\Delta dx$$

From Îto (because $W_t$ is a Brownian Motion), $$dV=V_tdt+V_xdx+\frac{1}{2}\sigma^2x^2V_{xx}dt$$

Then $$d\Pi=V_tdt+V_xdx+\frac{1}{2}\sigma^2x^2V_{xx}dt-\Delta dx$$

Deterministic terms are those terms with $dt$. The random terms are those with $dx$, and are the risk in the portfolio.

Then, using Delta hedging ($\Delta=V_x$), to perfectly eliminate risk (dynamically), you have a portfolio whose value changes as $$d\Pi=V_tdt+\frac{1}{2}\sigma^2x^2V_{xx}dt\qquad(1)$$

$(1)$ is a completely riskless situation.

This is the first part to derive the PDE of BS-PDE. Note that your fisrt PDE is wrong (you assume the porfolio value change is $0$).

If, from $(1)$, we have a completely risk-free change $d\Pi$ in the value $\Pi$, then it must be the same as the growth we would get if we put the equivalent amount of cash in a risk-free intereset-bearing account: $$d\Pi=r\Pi dt=r\bigl(V-\Delta x\bigr)dt=r\bigl(V-V_xx\bigr)dt\qquad(2)$$ an example of the no arbitrage principle.

So, make $(1)=(2)$ and $$\begin{array}{rcl} d\Pi&=&d\Pi\\ V_tdt+\frac{1}{2}\sigma^2x^2V_{xx}dt&=&r\bigl(V-V_xx\bigr)dt\\ \end{array}$$

Finally, $$V_t+\frac{1}{2}\sigma^2x^2V_{xx}+rxV_x-rV=0$$

Hope this helps understading the problem.

• I think it was just the transformation. However, nice answer for deriving the B-S equation. Aug 21, 2014 at 21:49
• Also, I like to add that if you are interested in these quant related subjects check out quant stackexchange :). Aug 21, 2014 at 22:22
• I'll give it a look, thanks! Aug 21, 2014 at 22:29