2
$\begingroup$

I came across the Vola Dynamics page about the W-shaped vol before an event: https://voladynamics.com/marketEquityUS_AMZN.html

I'm a bit confused by "this term does not have any butterfly arbitrage". I thought butterfly arbitrage suggests that price against strike is convex, i.e., $\partial^2 C/\partial K^2 > 0$. But the W-shape around the forward is clearly not convex.

I guess it may be because the y-axis is not price but vol, but then I thought roughly as vol is higher the price is higher too.

Any formula to check the butterfly arbitrage in the vol space? I mean, with some rule to check including for example $\partial \sigma^2 / \partial K^2$.

$\endgroup$
3
  • 1
    $\begingroup$ Hint: $C(K) = C^{BS}(K,IV(K))$. Differentiate both sides twice wrt to $K$ and then find the equivalent condition on IVs for no butterfly arbitrage. $\endgroup$
    – user34971
    Commented Oct 8, 2022 at 10:52
  • $\begingroup$ @Frido, Thanks. I found the answer in this paper: arxiv.org/pdf/1204.0646.pdf $\endgroup$
    – Michael
    Commented Oct 8, 2022 at 14:26
  • $\begingroup$ Yes, so in that paper equation 2.1 is I think what you're looking for, i.e $g(k) \geq 0$. $\endgroup$
    – user34971
    Commented Oct 8, 2022 at 16:28

2 Answers 2

2
$\begingroup$

Given the call option price $C$ as a function of strike $K$ and (strike-)implied volatility $\sigma(K)$, we have $C(K,\sigma(K))$. No-arbitrage requires the total derivative of the call option price w.r.t. the strike to be $\geq 0$, i.e.:

$$ \begin{align} \frac{\mathrm{d}^2C}{\mathrm{d}K^2}&\geq 0\\ \Rightarrow \quad\quad 0&\leq\frac{\partial^2C}{\partial K^2}+2\frac{\partial^2C}{\partial K\partial\sigma }\frac{\partial \sigma}{\partial K}+\frac{\partial^2C}{\partial\sigma^2 }\left(\frac{\partial \sigma}{\partial K}\right)^2+\frac{\partial C}{\partial\sigma }\frac{\partial ^2\sigma}{\partial K^2} \end{align} $$

$\endgroup$
0
$\begingroup$

Local volatility (LV) must be positive. The expression for LV in vol space can be found here on page 10.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.