Im having a question about this standard derivation of the Black-Scholes formula:


The paper states

$$C=\Delta S+B$$

and finally $\Delta = C_s$. But what is the value of $B$? Is there a functional form for $B$? I could only think of an implicit solution as $B=C-\Delta S$ (where $C$ and $\Delta$ represent the solved Black-Scholes price and delta).

  • $\begingroup$ You are correct. $B$ is computed as $B = C - \Delta S$ and represents the cash amount required to fund the option and its delta hedge. $\endgroup$ – Antoine Conze Jan 24 '19 at 13:31
  • $\begingroup$ @AntoineConze If one can derive $C$ and $\Delta S$, there should also be a way to derive $B$. If we think of $B=C-\Delta S$ in the Black-Scholes model, this would be a pretty complicated formula so I would be interested in its justification $\endgroup$ – emcor Jan 26 '19 at 14:13
  • 1
    $\begingroup$ Very simple formula in the BS case $B = -e^{-rT}K N(d_2)$. You can also note that $B=-K\frac{\partial C}{\partial K}$, which is quite intuitive since you can view the Call option $(S_T - K)^+$ as being the option of exchanging the stock $S_T$ for a quantity $e^{-rT}$ of risk free asset $A_T = Ke^{rT}$, so that $B$ represents the option delta hedge with respect to $A_0$. $\endgroup$ – Antoine Conze Jan 28 '19 at 8:20
  • $\begingroup$ @AntoineConze Very nice argument, thank you! $\endgroup$ – emcor Jan 29 '19 at 11:06

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