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

Question

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 1

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}