Integrating Gamma

Let C x C_x be the call option with a strike price of x x .
Let γ C x \gamma_{C_x} be the gamma of C x C _ x .
What is

0 γ C x d x ? \int_0^{ \infty } \gamma_{ C_ x } \, dx ?

1 1 2 - \frac{1}{2} -1 Cannot be determined 0 1 2 \frac{1}{2}

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1 solution

Dionys Nipomici
May 30, 2020

Since γ \gamma is the derivative of Δ \Delta , we have:

0 γ d x = Δ ( ) Δ ( 0 ) = 1 0 = 1 \int_0^{\infty} \gamma dx = \Delta (\infty) - \Delta (0) = 1-0=1

Remember that Gamma is the derivative of the delta with respect to a change in the underlying stock.

What you're thinking of is "For a fixed x = P x = P , 0 γ C P d S = [ δ C P ] 0 \int_0^\infty \gamma_{C_P} d S = [\delta_{C_P} ]_0^\infty .

However, in this question, note that we're integrating with respect to the Strike price.
There's a "change of variables" that's going on.

Calvin Lin Staff - 1 year ago

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