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Comment by skzv | Hacker News Reader

> Delta is not a linear function, meaning it will not change proportionately with the stock price.

Someone please correct me: can't gamma (the second derivative) be a constant, implying delta (first derivative) is linear? Then the price itself changes quadratically.

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No, because delta can't ever go below 0 or above 1, but it will be close to 0 for low prices (OTM options) and close to 1 for high prices (ITM options). So it'll always form a kind of S-curve.

Oh okay, makes sense. Can you share a resource where I can read up on this?

Natenberg wrote the classic tome on the subject (Option Volatility and Pricing) but the burned hand teaches best. Trade some options, lose some money, and you'll learn what to do and what not to do.

Delta for put options is typically below 0.

Sure, for put options it ranges between 0 and -1 for the same reasons, and the same argument holds for why gamma can't be constant.

Gamma can occasionally be constant for a single contract. It doesn't work out that way mathematically, but in practice it is the result of IV being spot dependent (spot up vol down, spot down vol up are two common phenomena) or reset lower due to a passing catalyst. This tends to be rare and localized, but it's important to think about charm vs speed if you trade weeklies around earnings because your true gamma will affect your decisions about how many shares to use to hedge delta. Let's look at an example. Say that you're long an OTM call into earnings, around 40d. The earnings reaction is muted compared to expectations, amounting to a small gain. You're now closer to your strike but still OTM, and IV has reset lower due to the catalyst having passed. In this case it's entirely possible that the speed of your gamma was counteracted by the charm of the passing catalyst (event theta), leaving you with delta that corresponds to what your first-order gamma would have indicated. Aka constant gamma for that particular contract in that specific event.

Yeah that part wasn't worded very well.

The process is nonlinear, necessitating a quadratic taylor expansion, at least.

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