# Vanila Option self financing under Stock as numeraire

I am trying to see how the vanilla call option can be seen as self financing using the Stock as the numeraire. The case with the Bond as numeraire is quite simple and can be found in Wilmott's FAQ for instance.

We assume that under the real world probability measure

$$dS=S(\mu dt + \sigma dW)\\ dB=B rdt$$

It is well known that under the Stock as as numeraire we can get the call option as

$$\frac{C_t}{S_t} = E_{\mathbb S,t} [ \frac{S_T-K}{S_T} \Theta(S_T-K) ]$$

where under measure $\mathbb S$ we have

$$dS= S( (r+ \sigma^2) dt + \sigma dW_S) \\ dB= B r dt$$ which gives $dW_S = dW + \frac{\mu - r- \sigma^2}{\sigma} dt$. Specifically under this measure $\frac{B}{S}$ is a Martingale

$$d(\frac{B}{S})= \sigma \frac{B}{S} dW_S$$

My question is that I should be able to see this as a self financing portfolio but I cannot. I am sure that I am making a mistake but cannot find it. I do the following (I suppress the time subscript in the following)

From the Martingale representation theorem we have $$d(\frac{C}{S})=\alpha d( \frac{B}{S})$$ for some pre-visible process $\alpha$.

This can be written as

$$dC = \alpha dB + \frac{ C-\alpha B}{S} dS - \frac{ C-\alpha B}{S^2} dS^2$$

It's the last piece that I don't understand. Without it we would have a self-financed portfolio. I could try to write it in terms of $dB$ but it still doesn't give me a self-financed portfolio. What am I doing wrong?

• I'm trying to follow your argument. You say 'from the martingale property' then you specify the dynamics of C/S. How do you get those specific dynamics ? – dm63 Dec 15 '17 at 14:37
• If you mean just the line after 'from the Maringale property', the fact that it has not drift in the $W_S$ measure is the fact that its defined to be a Martingale under that measure. The rest is just a process that I am trying to find and for convenience I write it as $\alpha_t C_t/S_t$ absorbing the unknown part in $\alpha_t$. – Borun Chowdhury Dec 15 '17 at 14:45
• The self financing property is not related to a specific measure. – Antoine Conze Dec 15 '17 at 16:30
• I know. That's why I am asking what am I doing wrong. Said differently, if I had never derived the self financing in the Bond numeraire and just did it this way I would be making a mistake. What is it? – Borun Chowdhury Dec 15 '17 at 16:41

## 1 Answer

I found the mistake I was making. Its the same mistake I made a variant of this problem about two years ago and Quantuple answered it. I am giving the correct answer here for completeness.

From the Martingale representation theorem we have

$$d(\frac{C}{S}) = \alpha d (\frac{B}{S})$$

Now the LHS is

$$LHS = \frac{dC}{S} - \frac{CdS}{S^2} + \frac{CdS^2}{S^3} - \color{red}{\frac{dCdS}{S^2}}$$

I was missing the red term in my derivation. We are keeping terms only upto order $dt$ and the piece that contributes from the $dC$ in the last term will be $\mathcal O(dW)$.

The RHS is

$$\alpha ( \frac{dB}{S} - \frac{BdS}{S^2} + \frac{BdS^2}{S^3})$$

and there is no equivalent term for the red one because $dB$ has not stochastic piece so such a term would not appear at order $dt$.

From these one gets

$$dC= \alpha dB + (C-\alpha B) \frac{dS}{S}$$

making $C$ a self-financed strategy.