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I have read somewhere that we can show by using arbitrage argument the following relationship for call option :

$$\frac{\partial{C_t(T,K)}}{\partial{K}}\leq0$$

$$\frac{\partial^2{C_t(T,K)}}{\partial{K^2}} \geq0$$

$$\frac{\partial{C_t(T,K)}}{\partial{T}} \geq0$$

where $C_t(T,K)$ is a call price at time t, for a strike K, with maturity T, obviously we can show these relation by taking the derivation for example with respect to the maturity for the last relation, or for showing the second relation (i.e. the butterfly spread relation, we just need to notice that the payoff is a convex function the strike and then the second derivative is always positive or null,but I am looking for arbitrage argument to show these relations. If someone has any idea to show them I would be grateful.

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  • $\begingroup$ First inequality: if not true, a lower strike call would be cheaper than a higher strike call. So sell the higher strike, buy the lower strike for a credit spread, and note that, regardless of closing price, the lower strike will never be worth less than higher strike. The third one is probably only true for American-style options and would involve creating a calendar spread. Not sure after the middle one. $\endgroup$ – barrycarter Apr 18 '16 at 3:11
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$$\begin{array}{rcl} (1) & \partial_KC_t(T,K) & \leq 0 \\ (2) & \partial^2_KKC_t(T,K) & > 0 \\ (3) & \partial_T C_t(T,K) & \geq 0 \\ \end{array}$$

If $(1)$ doesnot hold, it exists $K_1<K_2$ such that $C_t(T,K_1)<C_t(T,K_2)$. Then as barrycarter said in his comment, you sell $C_t(T,K_2)$ and you buy $C_t(T,K_1)$, so your cash position is $C_t(T,K_2)-C_t(T,K_1)>0$, at maturity you receive $(S_T-K_1)^+-(S_T-K_2)^+\geq 0$. There is an arbitrage.

If $(2)$ doesnot hold, it exists $\epsilon>0$ and $K>\epsilon$ such that $C_t(T,K-\epsilon)+C_t(T,K+\epsilon)\leq 2 C_t(T,K)$. Then you buy $C_t(T,K-\epsilon)$ and $C_t(T,K+\epsilon)$ and you sell $2C_t(T,K)$, your cash position is $2 C_t(T,K) - C_t(T,K-\epsilon)+C_t(T,K+\epsilon)\geq 0$. at maturity you get $(S_T-(K+\epsilon))^++(S_T-(K-\epsilon))^+-2(S_T-K)^+\geq 0$ which is the butterfly spread you mention. Note that with non-null probability, this payoff is positive. There is again an arbitrage.

Assuming you talk about american call options. If $(3)$ doesnot hold, it exists $T_1<T_2$ such that $C_t(T_1,K)>C_t(T_2,K)$. Then you buy $C_t(T_2,K)$, you sell $C_t(T_1,K)$, your cash position is $C_t(T_1,K)-C_t(T_2,K)>0$. At any time $\tau\leq T_1$, the buyer of $C_t(T_1,K)$ can exercise its right, and then you owe him $(S_\tau-K)^+$, but since you buy an american option $C_t(T_2,K)$, you can also exercise your right at time $\tau$ and you net position. Again since you build a positive cash position leading to a non-negative positive (here equal to $0$), you build an arbitrage.

Note that this third relationship will hold with any american options. And note that the two first relationships will also work with american calls by adapting the $T$ to $\tau$ being the exercise time of the agent who buys call options.

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  • $\begingroup$ Nice answer. Note that (3) also holds for European options provided the derivative is computed for a fixed forward moneyness level instead of a fixed strike. $\endgroup$ – Quantuple Apr 18 '16 at 17:39

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