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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.

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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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  • $\begingroup$ Was I completely wrong with my attempted proof? $\endgroup$ – Wolfy Feb 25 '16 at 21:36
  • $\begingroup$ Nothing is wrong, but it does not lead to the second derivative. $\endgroup$ – Gordon Feb 25 '16 at 21:37
  • $\begingroup$ I see is there a way to lead to the second derivative from what I did? $\endgroup$ – Wolfy Feb 25 '16 at 21:52
  • $\begingroup$ It is difficulty, as you only described what does convexity mean. $\endgroup$ – Gordon Feb 25 '16 at 22:09
  • $\begingroup$ Gordon, I need your help with a question on two-period binomial model. Could you provide an answer? $\endgroup$ – Wolfy Mar 16 '16 at 0:50

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