Relevant definition:

Assumption 2.1 (No dominance). If the payoff $P$ of a financial instrument is nonnegative, then the price $p$ of the financial instrument is nonnegative.


$T$ - the maturity

$K_1$,$K_2$ - Strike prices

$S$ - stock price

$t$ - current time

$B_t(T)$ - price of bond

It is well known that a convex function has right and left derivatives at all points. From the above exercise it follows that (exercise) $\delta_{K_{\pm}}C(T,K,S,t)$ exists. Use no dominance to show $$-B_t(T) \leq \delta_{K_{\pm}}C(T,K,S,t)\leq 0$$. (hint: Consider a portfolio made of a long position in a call with strike $K_2$ and short positions in a call option with strike $K_1$ and $K_2 - K_1$ bonds.

Attempted solution: Suppose we have a portfolio made up of a long position in a call with strike $K_2$ and a short position in a call with strike $K_1$ and $K_2 - K_1$ bonds. The price of the long call is defined by $$C(T,K_2,S,t)$$ the price of the short position in the call is defined by $$-C(T,K_1,S,t)$$ and the price of the $K_2 - K_1$ bonds is defines as $$B_t(T)$$ The portfolio value is thus $$P_v = C(T,K_2,S,t) - C(T,K_1,S,t) + (K_2 - K_1)B_t(T)$$ (Taking the hint provided by the user: barrycarter) Suppose the return on $P_v$ is less than the risk-free rate, then one would have to be compensated to take this position. Therefore by the no dominance assumption the price of this position would be negative which is a contradiction. Does the result follow from exercise 2.2?

I am not sure where to go from here, any suggestions is greatly appreciated.

  • $\begingroup$ Quibble with Assumption 2.1: if the payoff is less than the risk-free interest rate, could the price be negative (you'd have to pay someone to take it to make up for the interest they would lose)? $\endgroup$
    – user59
    Commented Jan 16, 2016 at 4:06
  • $\begingroup$ @barrycarter Is the start of my solution on the right track though? I assume by your hint I am going to prove this by contradiction? $\endgroup$
    – Wolfy
    Commented Jan 18, 2016 at 1:39

1 Answer 1


Note that, for $K_1 < K < K_2$, \begin{align*} -(K_2-K_1) \le (S_T-K_2)^+ - (S_T-K_1)^+ \le 0. \end{align*} Taking the conditional expectation with respect to information set $\mathcal{F}_t$, \begin{align*} -(K_2-K_1)B_t(T) \le C(T, K_2, S, t) - C(T, K_1, S, t) \le 0. \end{align*} That is, \begin{align*} -B_t(T) \le \frac{C(T, K_2, S, t) - C(T, K_1, S, t)}{K_2-K_1} \le 0. \end{align*} Since \begin{align*} \lim_{K_2-K_1\rightarrow 0}\frac{C(T, K_2, S, t) - C(T, K_1, S, t)}{K_2-K_1} = \frac{\partial C(T, K, S, t)}{\partial K}. \end{align*} The result follows immediately.


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