On the derivation of Black–Scholes

Question

While going through Steven Shreve's Stochastic Calculus for Finance II, I encountered a problem in section 4.5.3. Equating the evolutions, why can't we use $d(X(t)) = d(c(t, S(t))) $ instead of $ d(e^{-rt} X(t)) =d (e^{-rt} c (t, S(t))) $ to derive the Black-Scholes equation?

Any help or suggestion would be appreciated.

Answer 1

You can if you want - but it corresponds to a different hedging portfolio. You may compare the following two portfolios:

• Put part of your asset into a money market with interest rate $r$, and the other part into a stock market with drift $\mu$ and volatility $\sigma$.
• Keep part of your asset by hand (or equivalently, put it into a money market with interest rate $0$), and put the other part into a stock market with drift $\mu$ and volatility $\sigma$.

For the first portfolio, you need to include the $r$ term, because you will earn/pay interest if your risk-free asset is positive/negative. For the second portfolio, however, you need to exclude the $r$ term, because there is no interest-related issue.

Beyond the money market, you may also consider other portfolios, e.g., put part of your asset into one stock market with drift $\mu_1$ and volatility $\sigma_1$, and the other part into another stock market with drift $\mu_2$ and volatility $\sigma_2$.

Different portfolios lead to different pricing models. And the model that gives a lower price should be regarded as more optimal (if you are an option buyer, you will definitely choose the cheapest if other conditions, e.g., strike price and maturity, are identical). You may compare the price predicted by the original Black-Scholes model, and by the model in which you ignore $r$ (or you set $r=0$ in the original Black-Scholes model). You will find that the model that includes $r$ gives a lower price. Therefore, taking $r$ into account is a more optimal choice.