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The book Advanced Equity Derivatives Volatility and Correlation page 22 said

To preclude arbitrage we must at least require:

  1. No call or put spread arbitrage : $\dfrac{\partial c}{\partial K}\leq 0,\ \dfrac{\partial p}{\partial K}\geq 0.$

  2. No butterfly spread arbitrage : $\dfrac{\partial^2 c}{\partial K^2}\geq 0,\ \dfrac{\partial^2 p}{\partial K^2}\leq 0.$

  3. No calendar spread arbitrage : $\dfrac{\partial c}{\partial T}\geq 0,\ \dfrac{\partial p}{\partial T}\geq 0.$

I know $\dfrac{\partial c}{\partial K},\dfrac{\partial^2 c}{\partial K^2},\dfrac{\partial c}{\partial T}$ are the limitation of their corresponding option strategies, but why $\leq 0$ and $\geq 0$ can represent the sufficient conditions of no arbitrage?

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Let's focus on a European call option for the sake of the argument. Assume deterministic rates to keep notations uncluttered. Define $\Bbb{Q}$ as the probability measure associated to the money market numéraire $B_t$. $$ C(K,T) = \frac{1}{B_T} \Bbb{E}^\Bbb{Q} \left[ (S_T-K)^+ \right] = \frac{1}{B_T} \int_K^\infty (S - K) q(S) dS $$ Whence (Leibniz rule) $$ \frac{\partial C}{\partial K}(K,T) = - \frac{1}{B_T} \int_K^\infty q(S) dS = -\frac{1}{B_T}\Bbb{Q}(S_T \geq K)$$ since $B_T$ is a numéraire (traded asset with positive value at all times) and since by definition of a probability $$\Bbb{Q}(\omega) \geq 0 , \forall \omega \in \Omega$$ the resulting no static arbitrage condition is indeed $ \frac{\partial C}{\partial K}(K,T) \leq 0 $

Similarly, $$ \frac{\partial^2 C}{\partial K^2}(K,T) = \frac{1}{B_T}q(K) $$ where the same argument applies for the sign of $B_T$ and again by definition of a p.d.f. $$ q(\cdot) \geq 0 $$ so that we we get $ \frac{\partial^2 C}{\partial K^2}(K,T) \geq 0 $

As you can see the weak inequalities directly come from the positivity on the c.d.f. and p.d.f.

As far as calendar arbitrage is concerned, the relation you wrote holds for American options and is a direct consequence of the latter's definition $$ C^{AM}(K,T) = \sup_{\tau \in [0,T]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] $$ where $\tau$ is a stopping time and obviously $$ \sup_{\tau \in [0,T_2]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] \geq \sup_{\tau \in [0,T_1]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] $$ for any $T_2 \geq T_1$.

You have a similar inequality for the absence of calendar arbitrage for European options though, see this related question for further info.

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  • $\begingroup$ Why do $\mathbb{Q}<0$ lead a arbitrage? I think both the $Q>0$ and $q>0$ are the definition of probability. $\endgroup$ – A.Oreo Nov 2 '17 at 13:38
  • $\begingroup$ See Mark Joshi's answer here: quant.stackexchange.com/questions/32607/…. The gist is that if there is a strictly positive probability $q(S_T=K)>0$ then the price of the infinitesimal butterfly spread should be strictly positive as well. If not, i.e. $q(S_T=K)=0$, then the price should be zero as well. $\endgroup$ – Quantuple Nov 2 '17 at 13:56
  • $\begingroup$ Why the price should be zero when $q(S_T = K) = 0?$ I think if $q$ is positive for some other $S_T=K'>K,$ the price can still be positive? $\endgroup$ – A.Oreo Nov 3 '17 at 1:43
  • $\begingroup$ Yes but that was is not the definition of an Arrow Debreu security, which is non-negativite everywhere but positive only at $S_T=K$. You need such kind of security to isolate the influence of $q(K)$ only hence deriving the butterfly arbitrage condition. You do something similar with digitals to obtain the vertical spread condition. $\endgroup$ – Quantuple Nov 3 '17 at 8:03
  • $\begingroup$ you mean the $q(S_T = K)<0$ will lead a negative price of Arrow Debreu security, but non-negative payment at $T$ for the Arrow Debreu security? $\endgroup$ – A.Oreo Nov 3 '17 at 8:48

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