$$ c_{f,\Sigma} = f N(d_1) - k N(d_2)$$

with

$$ d_1 = \frac{\log(f/k)+\Sigma^2/2}{\Sigma} $$

We have simply:

$$ c'(k) = - N(d_2)$$

He also proves that we can always find those two parameters for any \( k_0 > 0, c_0 > 0, -1 < c_0' < 0 \)

Then I had the silly idea of trying to match with a put instead of a call for the left wing (as those are out-of-the-money, and therefore easier to invert numerically). It turns out that it works in most cases in practice and produces relatively nice looking extrapolations, but it does not always work. This is because contrary to the call, the put value is bounded with \(f\).

$$ p_{f,\Sigma} = k N(-d_2) - f N(-d_1)$$

Inverting \(p_0'\) is going to lead to a specific \(d_2\), and you are not guaranteed that you can push \(f\) high and have \(p_{f,\Sigma}\) large enough to match \(p_0\). As example we can just take \(p_0 \geq k N(-d_2)\) which will only be matched if \(f \leq 0\).

This is slightly unintuitive as put-call parity would suggest some kind of equivalence. The problem here is that we would need to consider the function of \(k\) instead of \(f\) for it to work, so we can't really work with a put directly.

Here are the two different extrapolations on Kahale own example:

Extrapolation of the left wing with calls (blue doted line) |

Extrapolation of the left wing with puts (blue doted line) |

I find it interesting that some smiles can not be extrapolated by displaced diffusion in a C1 manner except if one allows negative volatilities in the formula (in which case we are not anymore in a pure displaced diffusion setting).

Extrapolation of the left wing using negative displaced diffusion volatilities (blue dotted line) |

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