I recently met the claim that for standard put and calls the gamma of the options are always positive. Is this a general result?
I am hoping not to assume any model, especially not Black-Scholes.
Answer
I'll use a European call option as an example, I think you can easily generalize it for a put option.
Given underlying S(t)=St, maturity T, strike K and risk-free rate r, the price of a call option as time t under the rik-neutral measure Q is
C(t,St)=EQ[e−r(T−t)max(ST−K,0)]=EQ[e−r(T−t)(ST−K)⋅1ST≥K]=EQ[e−r(T−t)ST⋅1ST≥K]−EQ[Ke−r(T−t)⋅1ST≥K]=e−r(T−t)EQ[ST⋅1ST≥K]−Ke−r(T−t)EQ[1ST≥K]
where 1ST≥K is a function values 1 when ST≥K and 0 otherwise. For the first expectation, we can change the probability measure to make it more manageable. Call P the new measure; the Radon-Nikodym derivative between the P and Q is
dQ=StSTer(T−t)dP
Therefore you get
C(t,St)=e−r(T−t)EQ[ST⋅1ST≥K]−Ke−r(T−t)EQ[1ST≥K]=e−r(T−t)EP[ST⋅1ST≥K⋅StSTer(T−t)]−Ke−r(T−t)EQ[1ST≥K]=e−r(T−t)EP[1ST≥KSter(T−t)]−Ke−r(T−t)EQ[1ST≥K]=StEP[1ST≥K]−Ke−r(T−t)EQ[1ST≥K]
Expanding the expectations as integrals you get:
C(t,St)=St∫∞KfP(ST)dST−Ke−r(T−t)∫∞KfQ(ST)dST=StP1−Ke−r(T−t)P2
where P1,P2 highlight that the integrals are probabilities.
Now the Greeks:
Δ=∂C∂St=∫∞KfP(ST)dST=P1Γ=∂2C∂St=∂Δ∂St=fP(St)∂fP(St)∂St
The derivative in Γ is the key. I don't think you can prove Γ to be positive for any probability density (i.e. any model).
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