According to the formula for pricing binary options with a volatility skew, it appears that the value of the binary option for a given strike gets lower, the higher the volatility skew at that strike. Why is this, intutively? I (think I) understand the derivation, but intuitively it seems like a positive volatility skew should make far OTM binary call options more expensive, not cheaper, because it should create "fatter tails" in the implied PDF of the price of the underlying at that expiry.

  • $\begingroup$ Yes, get your scepticism... how exactly are you defining "volatility skew"? $\endgroup$
    – demully
    Mar 11, 2021 at 23:49
  • $\begingroup$ the derivative of IV with respect to strike price $\endgroup$
    – KD89042
    Mar 12, 2021 at 4:20

2 Answers 2


You can replicate the payout of a binary with a put spread with strike prices which are very close to one another. Higher skew makes the further out of the money put more expensive, which makes the put spread cheaper.

  • $\begingroup$ My question is really why the put spread is cheaper with higher skew, since the put spread should replicate the payout of a binary option as the spread goes toward 0. Intuitively it seems like the skew should increase the chance of the price being above some large number, or below some small number, at expiry, and the value of put spreads far away from the spot price should reflect that? $\endgroup$
    – KD89042
    Mar 11, 2021 at 6:49
  • $\begingroup$ I tried to adress that in my answer trying to develop intuition behind P- and Q-probability $\endgroup$
    – Mats Lind
    Mar 11, 2021 at 22:34

Just on pure intuition (close to a guess); the binary does not pay you for the fat tail. All you want is high volatility locally around the strike to add to the chances for the option to expire in the money. Just narrowly in the money counts as much as far out on the tail.

Edit: The put spread in the other answer above could provide a good starting point to deepen the intuition. It buys volatility nearer the current underlying price and sells it back farther out on the tail. Narrow it down to (an out of the money, OTM) binary: it is long volatilty OTM and short volatility (vola) in the money all (ITM) the way out on the tail. The fat-tail distribution has relatively low vola when we are OTM and want vola to put us ITM. It has high vola when we are ITM and want low vola to stay there.

The takeaway here I guess is not to look at the expected return (P-probability): even though the chances for the very far OTM binary to expire ITM may increase with the fatness of the tail, its fair price will not! The fair price is what you would pay to break-even when you carry the option risklessly with a dynamic delta-hedge in the underlying until expiry (risk neutral Q-probability).

And where you are OTM you are long vola and long gamma and reap high rebalancing profits when vol is high. Where you are ITM and out on the tail it is the reverse, you lose in rebalancing when the large high vola moves producing the fatness even further out occur. Forget the (non-risk adjusted P-probability) expected return and look at how the vola skew matches where you will be long and short vola and gamma!

  • 1
    $\begingroup$ Yes. To expand on that, skewed distributions have a fat tail to the right. But to create that, while keeping the expected value the same, you have to move the peak of the distribution to the left. Hence the ATM digit call worth less that 50%. $\endgroup$
    – dm63
    Mar 4, 2021 at 12:33
  • $\begingroup$ I'm wondering specifically about binary options with a strike price far out on the tail. It seems like they should be priced higher when accounting for skew, relative to using constant volatility. Instead they are lower, and I don't understand why. $\endgroup$
    – KD89042
    Mar 11, 2021 at 6:50

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