Convex Payoffs and Nonnegative Gamma in European Claims

Summary

The document asks why a European contingent claim with a convex payoff in the underlying should itself be convex in the underlying, which corresponds to nonnegative gamma when the value is twice differentiable. The attempted Jensen argument is confused: it switches to the logarithm and derives the arithmetic-geometric mean inequality, which does not establish the claim’s gamma property.

The answer gives a direct finite-difference argument. Convexity implies that the payoff at a point is no greater than the average of its values at equally spaced points on either side. Rearranging yields a nonnegative centered second difference; dividing by the squared spacing and taking the limit gives a nonnegative second derivative. This establishes the result for smooth convex functions. The reasoning is local and does not itself derive the full claim value from a pricing model or address nonsmooth payoffs, where convexity can instead be understood through generalized derivatives.

Key ideas

• A smooth convex payoff has a nonnegative second derivative.
• Convexity bounds the value at the midpoint by the average of values at symmetric points.
• The centered second difference is nonnegative and converges to the second derivative after scaling.
• The logarithm and arithmetic-geometric mean argument in the attempted proof does not establish the gamma result.
• Nonsmooth convex payoffs require a generalized interpretation of curvature.

Tags

# Convexity in Markovian contingent claim


# Convexity in Markovian contingent claim












Background information:

I believe we can use Jensen's Inequality here

Show that if the payoff function $V(S_T)$ is a convex function on $S_T$, then the Markovian European contingent claim with payoff $V(S_T)$ has non-negative $\Gamma$, i.e. $V(\tau,S)$ is convex on $S$ for all $\tau$.

Attempted proof: Suppose we have a function $V(S_T)$ that is convex on $S_T$, then if $p_1,\ldots,p_n$ are positive numbers that sum to 1, then $$V\left(\sum_{i=1}^{T}p_i S_i\right) \leq \sum_{i=1}^{T}p_i V(S_i)$$ Now, let $p_i = 1/n$, then $\ln S$ gives, $$\ln\left(\frac{1}{T}\sum_{i=1}^{T} S_i\right) \geq \frac{1}{T}\sum_{i=1}^{T}\ln S_i$$ Through exponentiation we have the arithmetic mean-geometric mean inequality, $$\frac{S_1 + S_2 +\ldots + S_T}{T} \geq \sqrt[T]{S_1 S_2,\ldots S_T}$$ Which is non-negative, thus the result follows.

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/24539

Form a smooth convex function, the second derivative is always non-negative. In particular, for any $\varepsilon >0$, \begin{align*} V(S) &= V\Big(\frac{1}{2}(S+\varepsilon ) + \frac{1}{2}(S-\varepsilon )\Big)\\ &\le \frac{1}{2}\Big(V(S+\varepsilon )+ V(S-\varepsilon) \Big). \end{align*} That is, $$V(S+\varepsilon )+ V(S-\varepsilon) - 2 V(S) \ge 0 $$ Then \begin{align*} \frac{\partial^2 V}{\partial S^2} = \lim_{\varepsilon \rightarrow 0}\frac{V(S+\varepsilon )+ V(S-\varepsilon) - 2 V(S)}{\varepsilon^2} \ge 0. \end{align*}

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