# Black Scholes Replicating Portfolio Riskfree Asset

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

http://www.soarcorp.com/research/BS_hedging_portfolio.pdf

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).

• 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. – Antoine Conze Jan 24 '19 at 13:31
• @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 – emcor Jan 26 '19 at 14:13
• 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 $A_0$. – Antoine Conze Jan 28 '19 at 8:20
• @AntoineConze Very nice argument, thank you! – emcor Jan 29 '19 at 11:06