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I want to obtain the PDE for the Black-76 model. I believe it has to be the following PDE:

$$\left(\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^{2}F^{2}\frac{\partial^{2} V}{\partial F^{2}}\right)dt-rV=0.$$

I know that the PDE of Black-Scholes model is given by: $$\left(\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^{2}S^{2}\frac{\partial^{2} V}{\partial S^{2}}+rS\frac{\partial V}{\partial S}\right)dt-rV=0.$$

Here you start with geometric Brownian motion process, $dS_{t}=\mu S_{t}dt+\sigma S_{t}dW_{t}$. Further you consider a trading strategy under which one holds one option and continuously trades in the stock in order to hold some $\Delta$ shares.

What I have done so far:

Now I start with the process: $dF=\sigma F_{t} dW_{t}$. By Ito's lemma I have

\begin{align} dV &= \frac{\partial V}{\partial F}dF+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}dF^{2} \\ &= \frac{\partial V}{\partial F}\sigma F dW+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}(\sigma F dW)^{2} \\ &= \frac{\partial V}{\partial F}\sigma F dW+\frac{\partial V}{\partial t}dt+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2} dt \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW. \end{align}

Now I make use of the same hedging strategy as in the Black Scholes case so I have: $$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF.$$

So I have: \begin{align} d\Pi &= dV-\Delta dF \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW-\Delta dF \\ &= \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt+\frac{\partial V}{\partial F}\sigma F dW-\Delta (\sigma F dW) \end{align} Thus: $$\Delta=\frac{\partial V}{\partial F}$$ Also $d\Pi=r\Pi dt$, so:

$$r\Pi dt = \left(\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}\right)dt.$$

Integrating gives:

$$\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}-r\Pi = 0$$

Now $$\Pi=V-\Delta F=V-\frac{\partial V}{\partial F}F,$$ so we have

$$\frac{\partial V}{\partial t}+\frac{1}{2}\frac{\partial^{2} V}{\partial F^{2}}\sigma^{2} F^{2}-rV+\frac{\partial V}{\partial F}rF = 0$$

This is the Black-Scholes PDE not Black PDE. What am I doing wrong? How do I get the correct PDE?

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Nobody with any suggestion? –  Lech121 May 5 '13 at 0:42

1 Answer 1

You should apply a transform to the time derivative, remember that $F = \exp(r(T-t))\,S_t$

So $$\frac{\partial V}{\partial t}_s = \frac{\partial V}{\partial t}_t - \frac{\partial V}{\partial F}_t \frac{\partial F}{\partial t}_t$$

Which should remove the spare term.

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Welcome to Math.SE! Please put dollar signs like $...$ around your $\LaTeX$-code so it gets rendered properly. I edited your answer. You can click edit to see the code. Theres also a changelog when you click on the edit. –  Christoph Apr 23 at 13:07
    
Thank you Christoph, I didn't know. I'll be more proper in the future. –  MrGullaksen Apr 23 at 15:10

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