I need to prove that vertical spread is bounded, by using no arbitrage condition.

  0 > (C(T,K1 )- C(T,K2))/(K1- K2 )  >-e^(-r*T )

I have documented my solution below. Can you kindly review it and comment on my approach.

In order to prove vertical spread condition, I am using following inequality, which is lower bound on option value:

 C  ≥ S(0) − Ke^(−rT)

Vertical spread is created by going long (1) call at strike (K1) and going short (-1) call at strike (K2), where K2 ≥ K1.

C(T,K1) ≥ S(0) – K1e^(−rT) and C(T,K2) ≥ S(0) – K2e^(−rT) , when I calculate difference in call price I get:

C(T,K1) - C(T,K2) ≥ K2e^(−rT) - K1e^(−rT)             --(1)

C(T,K1) - C(T,K2) ≥ e^(−rT)* (K2 - K1), and as K2 ≥ K1 , I get:

C(T,K1) - C(T,K2) ≥ 0  --(2)

If I rearrange following inequality K2 ≥ K1 , I get:

K1 - K2  ≤ 0    -- (3)

If I divide (2) by (3), I get:

(C(T,K_1 )- C(T,K_2))/(K1- K2 )  <0   --(4) 

As dividing positive numerator with negative numerator gives a negative number. You may notice that I have removed equals to symbol from the inequality. Reason being, when K1= K2 , then the numerator is equal to zero and so is the denominator, leading to undefined state of solution. (4) satisfies the upper bound of inequalities. By rearranging inequality (1), I get:

C(T,K1 )-C(T,K2 )≥ -e^(-rT ) (K1- K2) 

Dividing above inequality by (3.3), I get:

 (C(T,K1 )- C(T,K2))/(K1- K2 )  >-e^((-r)T )          --(5)

Combining inequality (4) and (5), gives me required vertical spread condition

0 > (C(T,K1 )- C(T,K2))/(K1- K2 )  >-e^((-rT )
  • $\begingroup$ Seems like you and stud91 are enrolled in the same class: quant.stackexchange.com/questions/32607. The solution approach is almost identical to the one in that question as well. And again: you generally can't simply subtract inequalities. $\endgroup$ Feb 23, 2017 at 11:38
  • $\begingroup$ Thanks. I reviewed your reply for butterfly spread. Can you kindly elaborate on how can I apply no arbitrage condition on vertical spread. I can assume V to be value of vertical spread and use your explanation of V<0 and V=0 to prove first condition of V being less than 0. However, there is delta K in denominator and there is the other side of inequality that I can't resolve. Your help is much appreciated. $\endgroup$ Feb 23, 2017 at 12:56

2 Answers 2


We want to show that

\begin{equation} 0 \leq C_0 \left( K_2 \right) - C_0 \left( K_1 \right) \leq e^{-r T} \left( K_2 - K_1 \right) \end{equation}

where $K_1 < K_2$. For the moment we don't worry about strict inequality but I will get back to that later. We also need to assume that there are no holding returns to the underlying asset (e.g. dividends, repos, ...).

To show the first inequality, it is sufficient to note that the portfolio $C_0 \left( K_2 \right) - C_0 \left( K_1 \right)$ has a non-negative payoff at maturity

\begin{equation} V_T = \begin{cases} 0 & \text{if } S_T < K_1\\ S_T - K_1 & \text{if } S_T \in \left( K_1, K_2 \right)\\ K_2 - K_1 & \text{otherwise} \end{cases}. \end{equation}

Thus, if the current value of this portfolio would be negative, we'd have an arbitrage of the type "free lunch".

For the second inequality, note that the payoff is bounded from above by $\Delta = K_2 - K_1$. If the current value $V_0$ of the spread was greater than $e^{-r T} \Delta$, then we would sell it now, invest the proceeds in the risk-free asset and have a terminal value of strictly more than $\Delta$ for an obligation of at most $\Delta$. Again, this is a free lunch.

Regarding the strictly inequality: In the first case if $V_0 = 0$ and there is a non-zero probability of $S_T > K_1$, then we'd have a "free lottery" arbitrage. I.e. you pay nothing for something that gives you a non-negative payoff with probability one and a strictly positive payoff with a strictly positive probability. In the second case if the probability of $S_T < K_2$ is also non-zero, then you can make the same argument for receiving $e^{-r T} \Delta$ now for a future obligation that is never bigger than $\Delta$ but has a strictly positive probability of being smaller.


Let $x^+ = \max(x, 0)$. Note that, for any two real numbers $x$ and $y$, \begin{align*} (x+y)^+ \le x^+ + y^+. \end{align*} Then, for $K_1 < K_2$, \begin{align*} (S_T-K_1)^+ &= (S_T-K_2+ K_2-K_1)^+\\ &\le (S_T-K_2)^+ + K_2-K_1, \end{align*} and, consequently, \begin{align*} (S_T-K_1)^+ -(S_T-K_2)^+ \le K_2-K_1. \end{align*} Therefore, \begin{align*} C(T, K_1) -C(T, K_2) \le (K_2-K_1)e^{-rT}. \end{align*}


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