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 $$ \partial_t C + (r+\color{red}{\sigma^2}) S\partial_S C + \frac{1}{2} \sigma^2 S^2\partial_{S,S} C -rC=0 $$


1 Answer 1


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} $

  • $\begingroup$ Note that these are Ito expansions not Taylor expansions, thus I don't get your Big oh notation. $\endgroup$
    – Quantuple
    Mar 13, 2016 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. $\endgroup$ Mar 13, 2016 at 22:03
  • $\begingroup$ Fair enough! Glad I could help $\endgroup$
    – Quantuple
    Mar 13, 2016 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, 2016 at 22:26

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