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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

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  • $\begingroup$ This is only true if the risk-neutral distribution is lognormal, of course. It is not true in the marketplace where the lognormal assumption is only an approximation. $\endgroup$
    – dm63
    Oct 28, 2016 at 10:42

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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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