8
$\begingroup$

In the book 'Dynamic Hedging', Nassim Taleb writes:

    All operators in options learn that a rise in 
volatility would cause the delta of an out of the money
 call to rise and that of an in the money call to drop,
thereby bringing deltas closer to 50%.

Why is this? Why would the OTM delta rise? It can only rise if the underlying rallies. But if underlying rallies, the ITM delta will also rise, approaching 1. Why would it drop?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
11
$\begingroup$

Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know).

If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases).

If the option is already ITM, most(many) of its payoffs are already positive. Hence an increase in volatility will increase the likelihood for zero payoffs instead from a shift in the underlying price (i.e. delta decreases).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ How about delta of an at the money option when volatility increases? $\endgroup$
    – Idonknow
    Commented Mar 19, 2020 at 16:17
  • $\begingroup$ @Idonknow, you can look here for ATM. Generally, this answer and the book is incomplete. ITM call deltas will eventually also rise if vol is significantly high. $\endgroup$
    – AKdemy
    Commented Jul 10, 2021 at 15:28

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.