0
$\begingroup$

I was reading a lot about the idea that long positions (call or put) have positive gamma, and short positions (call or put) have negative gamma. But I couldn't understand why. In "Bunds and Bund Futures" book, the author argues as: "The reason for this is that as the futures price rises the delta of a long call (or long put) option will become more positive (or less negative). For short call (or short put) options, the deltas will become more negative (or less positive) as the futures price increases." But why!? Why "as the futures price rises the delta for a long put option will become less negative"? I would really appreciate some feedback about this.

$\endgroup$
  • $\begingroup$ I assume this refers to European call/put. Best way to understand would be to draw the price as a function of the underlying, and then draw tangents at three different points (tangent being the equivalent of delta and change in slope, steepening/flattening, the convexity). This is similar to the concept of convexity in the bond price as a function of yield. $\endgroup$ – Magic is in the chain Jun 21 at 0:33
  • $\begingroup$ Yes I was referring to European options. I know that the tangent (first derivative) is delta, and the second derivative (or first derivative of delta) is gamma which is convexity/concavity. What I'm not getting is: How did the author know that "futures price rises the delta of a long put option will become less negative" without looking at a specific function? Like if you give me a specific function, I can show that gamma is positive for this function, etc. But how can he know that gamma is positive simply because there's the word "long"? Is there a conceptual idea behind it? $\endgroup$ – Guess601 Jun 21 at 2:07
  • $\begingroup$ I also read on some blogs that they use a rule-of-thumb, "long" implies positive gamma, and "short" implies negative gamma. But there's no conceptual explanation why. $\endgroup$ – Guess601 Jun 21 at 2:09
  • 1
    $\begingroup$ If $f(S)$ has positive slope at $\hat{S}$ then $−f(S)$ has negative slope there, yes? Same for curvature. This is Curve for Long Call encrypted-tbn0.gstatic.com/… and this for Short Call tradersfly.com/wp-content/uploads/2019/06/… Verify the signs of the first and second derivative $\endgroup$ – noob2 Jun 21 at 9:42
  • 1
    $\begingroup$ I think now I get it. The graph of long (short) call or put looks like a convex (concave) curve. $\endgroup$ – Guess601 Jun 21 at 14:12
1
$\begingroup$

It works in the Black-Scholes world assuming other factors are constant. For the Black-Scholes, you can check online the formula for Gamma. It is positive obviously if you are long. In reality this may not hold. A simple example would be the case when the implied vol is correlated with spot price. In this case, the "Gamma" would not be as simple as the plain Black-Scholes since there will be "mixed-derivative" terms when calculating second derivatives.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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