4
$\begingroup$

There are many ways to derive the Black Scholes PDE. The Martingale way would be to demand the option price is driftless according to particular measures. Below I derive the correct PDE using the bank account as the numeraire but fail to get the correct PDE when using the stock as a numeraire. I am hoping someone will be able to point out what I am doing wrong.

Deriving Black Scholes PDE using bank account as numeraire

One of the ways to derive the Black-Scholes equation is to take the bank account $B_t$ as a numeraire and then demand that $d\frac{C_t}{B_t}$ be driftless. Below I keep the subscript denoting time implicit.

Concretely, under this numeraire $W_B$ (where $B$ stands for bank account)

$$ dS=S r dt + S \sigma dW_B \\ dB=B r dt $$ so we simply get $$ \begin{eqnarray} d\frac{C}{B} &=& \frac{\partial_t C dt + \partial_S CdS + \frac{1}{2} \partial_{S,S} CdS^2 }{B}-\frac{CdB}{B^2} \\ &=& \frac{\partial_t C + r S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC}{B} dt + \frac{\sigma S \partial_S C}{B} dW_B + \mathcal O({dt}^{3/2}) \end{eqnarray} $$ and demanding that $\frac{C}{B}$ be a Martingale requires the vanishing of the drift term and we get the Black Scholes PDE: $$ \partial_t C + r S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC=0 $$

Trying to derive Black Scholes PDE using stock as numeraire

Now I try to do the same while taking the Stock as a numeraire. I will demand, as usual, that $d \frac{C}{S}$ is a Martingale under this measure. Under this measure we have $$ dS = S(r+\sigma^2) dt + S \sigma dW_S $$ so we get $$ \begin{eqnarray} d\frac{C}{S} &=& \frac{\partial_t C dt + \partial_S CdS + \frac{1}{2} \partial_{S,S} CdS^2 }{S}-\frac{CdS}{S^2} + \frac{CdS^2}{S^3} \\ &=& \frac{\partial_t C + (r+\sigma^2) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C}{S} dt + \frac{\sigma S\partial_S C dW_S}{S} - \frac{C}{S}\big((r+\sigma^2) dt + \sigma dW_S \big)+\frac{C}{S}\sigma^2 dt +\mathcal O(dt^{3/2}) \\ &=& \frac{\partial_t C + (r+\sigma^2) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC}{S} dt + \frac{\sigma S \partial_S C-C}{S} dW_S +\mathcal O({dt}^{3/2}) \end{eqnarray} $$

NOW demanding the drift term to be zero gives me an extra term $$ \partial_t C + (r+\color{red}{\sigma^2}) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC=0 $$

$\endgroup$
5
$\begingroup$

You miss the cross-derivative term in the Ito formula you use to express $d\left ( \frac {C_t}{S_t} \right)$. More specifically (see [Remark] below),

$$d\left ( \frac {C_t}{S_t} \right) = \frac {1}{S_t} dC_t - \frac {C_t}{S_t^2} dS_t + \frac {C_t}{S_t^3} d\langle S_t, S_t \rangle {\color{green}{- \frac {1}{S_t^2} d\langle C_t, S_t \rangle}}$$

This last term evaluates to $$-\partial_S C_t \sigma^2 dt $$

Meaning that one can write:

$$d\left( \frac{C_t}{S_t} \right) = \frac {1}{S_t} (\partial_t C_t dt + \partial_S C_t dS_t + \frac {1}{2} \partial_{SS} C_t \sigma^2 S_t^2 dt) - \frac {1}{S_t} \left( (r+\sigma^2) C_t dt + \sigma C_t dW_t \right) + \frac {1}{S_t} \sigma^2 C_t dt - \frac {1}{S_t} \partial_S C_t \sigma^2 S_t dt$$ or equivalently after re-arranging some terms $$d\left( \frac{C_t}{S_t} \right) = \frac {1}{S_t} (\partial_t C_t + r S_t \partial_S C_t + \frac {1}{2} \partial_{SS} C_t \sigma^2 S_t^2 - rC_t ) dt + (.) dW_t$$

Hence the Black-Scholes pde from the martingale representation theorem.

[Remark] This result simply comes from applying the bidimensional version of Ito's lemma $$df = (\partial_t f) dt + (\partial_X f) dX_t + \frac {1}{2} (\partial_{XX} f) d\langle X_t \rangle + (\partial_Y f) dY_t + \frac {1}{2} (\partial_{YY} f) d\langle Y_t \rangle + (\partial_{XY} f) d\langle X_t, Y_t \rangle$$

To the function $f (t, X_t,Y_t) = \frac {X_t}{Y_t} $

$\endgroup$
  • $\begingroup$ Note that these are Ito expansions not Taylor expansions, thus I don't get your Big oh notation. $\endgroup$ – Quantuple Mar 13 '16 at 21:48
  • $\begingroup$ Thanks a lot! That's it of course. About the notation, I meant dW is like the square-root of dt so in the Ito expansion I understand we neglect terms higher than dt. $\endgroup$ – Borun Chowdhury Mar 13 '16 at 22:03
  • $\begingroup$ Fair enough! Glad I could help $\endgroup$ – Quantuple Mar 13 '16 at 22:07
  • $\begingroup$ (Ps if the answer suits you could you please mark it as the accepted answer so that the topic is closed. Cheers mate) $\endgroup$ – Quantuple Mar 13 '16 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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