Understanding the relationship between the Black-Scholes formula and a replicating portfolio

Question

I'm self-studying and I'm considering the below example. The specific example is not especially relevant, but I included it for reference.

I'm trying to understand the relationship between a replicating portfolio and the Black-Scholes equation.

It is my understanding that a replicating portfolio for a put involves short selling stock and lending money. There would be a positive cash flow of $Se^{-\delta T} N(-d_1)$ and a negative cashflow of $Ke^{-rT} N(-d_2).$

However, the Black-Scholes formula for a put is: $P = Ke^{-rT}N(-d_2) - Se^{-\delta T} N(-d_1)$.

This formula suggests that a long position in a put is a positive cashflow of $Ke^{-rT} N(-d_2)$ and a negative cashflow of $Se^{-\delta T} N(-d_1)$.

So wouldn't the replicating portfolio create a short position put, while the Black Scholes formula provides a long position put (since the signs are swapped comparing the replicating portfolio to the Black Scholes formula for a put)?

Should examples like the one below clarify what position the put is in?

Basically I'm looking for clarification on the signs of the terms in the formula.

Answer 1

It is my understanding that a replicating portfolio for a put involves short selling stock and lending money.

You cannot statically replicate an option. So this is not true in general, you'll need to re-balance your replicating portfolio (underlying + cash) dynamically if you want to replicate the option. This will imply sometimes buying stock and borrowing money.

Basically I'm looking for clarification on the signs of the terms in the formula

Take the case $S=0$ and you'll see that the signs $$ P(K,T) = Ke^{-rT}N(-d_2) - Se^{-\delta T} N(-d_1) $$ are indeed correct since in that case you need to find $P(K,T) = K e^{-rT} > 0$, because the payout of a put option is: $\max(K-S_T,0)$.

Extending this idea:

• Call (Payoff $\to$ Price): $$\max({\color{blue}{+S}}_T{\color{blue}{-K}},0) \to C(K,T) = {\color{blue}{+S}}e^{-\delta T}\phi(d_1) {\color{blue}{-K}} e^{-rT}\phi(d_2)$$
• Put (Payoff $\to$ Price): $$\max({\color{blue}{+K}}{\color{blue}{-S}}_T,0) \to P(K,T) = {\color{blue}{+K}}e^{-rT}N(-d_2) {\color{blue}{-S}} e^{-\delta T} N(-d_1)$$