Black–Scholes Call Price Equivalence

Article Quant Q&A · Author: Bob

Summary

The note examines whether a European call priced at rate r and strike K has the same value as a zero-rate call with strike reduced by discounting K over the option’s life. It derives the relationship by substituting the adjusted strike into the Black–Scholes d terms and call-price formula. The adjusted strike makes the zero-rate d values match those for the original call, while the discounted strike term in the original formula also matches directly.

The resulting equality is between the original call price and the zero-rate call price at the adjusted strike; there is no additional discount factor multiplying the latter. The derivation assumes the standard Black–Scholes setup for a European plain-vanilla call, with the same underlying, volatility, and maturity. It does not establish the equivalence for other option styles or models.

Key ideas

Discounting the strike by the risk-free rate aligns the Black–Scholes d terms with those of a zero-rate call.
The call price at rate r equals the zero-rate call price with strike K discounted to maturity.
Applying another discount factor to the adjusted call price would give the wrong relationship.
The derivation concerns European plain-vanilla calls under Black–Scholes assumptions.

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Options Derivatives pricing Statistics

Full text

# Modeling the price of an option with another option using Black Scholes


# Modeling the price of an option with another option using Black Scholes












I have been told that:

The price of a call with strike price $k$ and risk-free interest $r$ is identical to the price of a call with strike price $ke^{-r}$ and risk-free interest $0$.

Here is what I claim. Let $c_1$ be the price of an option with interest rate $r$ and strike price $k$. Let $c_2$ be the price of an option on the same underlying security, with strike $ke^{-rt}$, same duration as the first call option and an interest rate of $0$. My claim is that $c_1 = (c_2)e^{-rt}$.

Do I have that right?

Bob

## Answer by LocalVolatility (score 3, accepted)

https://quant.stackexchange.com/a/30766

This is easy to show by just rearranging the Black Scholes solution for European plain vanilla calls. We make the dependence on the strike and rate explicity in what follows and set $\hat{K} = K e^{-r T}$. First,

\begin{eqnarray} d_\pm(K, r) & = & \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K} \right) + \left( r \pm \frac{1}{2} \sigma^2 T \right) \right)\\ & = & \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K e^{-r T}} \right) \pm \frac{1}{2} \sigma^2 T \right)\\ & = & \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{\hat{K}} \right) \pm \frac{1}{2} \sigma^2 T \right)\\ & = & d_\pm \left( \hat{K}, 0 \right). \end{eqnarray}

Thus

\begin{eqnarray} C_0(K, r) & = & S_0 \mathcal{N} \left( d_+(K, r) \right) - K e^{-r T} \mathcal{N} \left( d_-(K, r) \right)\\ & = & S_0 \mathcal{N} \left( d_+ \left( \hat{K}, 0 \right) \right) - \hat{K} \mathcal{N} \left( d_- \left( \hat{K}, 0 \right) \right)\\ & = & C_0 \left( \hat{K}, 0 \right) \end{eqnarray}

So yes, what you have been told is correct. And no, your formula is wrong as you seem to claim that

\begin{equation} C_0(K, r) = e^{-r T} C_0 \left( \hat{K}, 0 \right). \end{equation}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

Black–Scholes Call Prices with Adjusted Strikes and Rates

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