# Prove that the butterfly condition is always greater than zero

I need to prove that the butterfly condition is always positive under no arbitrage theorem. We are constructing a long butterfly using European call options

C(T,K+∆K) - 2C(K) + C(T,K-∆K) > 0 where ∆K < K


I have managed to prove for greater than or equal to zero using the following steps:

Lower bound of a European call option of a non-divided paying stock is as follows:

C(T,K) >= S(0) - Ke^-rT
K = strike price, T = time to maturity, r = interest rate, S(0) = stock price at time=0


Hence for options in the butterfly this evaluates to

C(T,K+∆K) >= S(0) - (K+∆K)e^-rT  --- (Eq:1)
C(T,K) >= S(0) - (K)e^-rT --- (Eq:2)
C(T,K-∆K) >= S(0) - (K-∆K)e^-rT --- (Eq:3)


Doing (Eq:1) - 2*(Eq:2) + (Eq:3), I get the following

C(T,K+∆K) - 2C(T,K) + C(T,K-∆K) >= 0


However, how do i go further and prove that the above inequality is not equal to zero under no arbitrage.

You generally can't simply subtract two inequalities as you did in your attempt. Here are two approaches to solve your problem:

No-Arbitrage Argument

Assume that the initial value of the Butterfly spread was strictly negative $V_0 < 0$. Buying the butterfly spread would thus yield a strictly positive cash-flow at time $t = 0$. Next note that the terminal payoff $V_T$ is non-negative. It is zero when $S_T \in [0, K - \Delta] \cup [K + \Delta, \infty)$ and strictly positive when $S_T \in (K - \Delta, K + \Delta)$. This is a free lunch (you get cash now and have a non-negative payoff in the future) and thus contradicts the absence of arbitrage.

Now assume that $V_0 = 0$. In this case the initial cash-flow from buying the butterfly spread is zero. In the future you have a non-negative cash-flow. If this cash-flow has a non-zero probability of occurrence, then this represents a free lottery and again contradicts no-arbitrage.

State Price Density

From the Breeden-Litzenberger result, we know that

$$C_0(K) = e^{-r T} \int_K^\infty (x - K) f(x) \mathrm{d}x \qquad \Leftrightarrow \qquad \frac{\partial^2 C_0}{\partial K^2} = e^{-r T} f(K).$$

I.e. the compounded second derivative of the call price w.r.t. the strike is equal to the risk-neutral probability density. For $f$ to be valid, we thus require that the second derivative is non-negative everywhere. Now consider a finite difference approximation:

$$\frac{\partial^2 C_0}{\partial K^2} = \lim_{\Delta \downarrow 0} \frac{C_0(K - \Delta) - 2 C_0(K) + C_0(K + \Delta)}{\Delta^2}.$$

The numerator is just the butterfly spread and it follow that is has to be non-negative as well. The argument for it being strictly positive again depends on whether you allow for regions with zero probability mass or not.

• The second argument is valid only for sufficient small Delta. It doesn’t have to hold for any Delta. – quallenjäger Jun 25 at 22:59

I think the proof should go like this:

The payout of the option butterfly is always non-negative. See the graph in wikipedia.

The no-arbitrage condition is equivalent to that the probability density is always non-negative for any state.

Therefore, the (discounted) expected payout, i.e., the option premium, should be non-negative.

• Non-negative implies that it is equal to zero. How can i prove that it is not equal to zero – stud91 Feb 22 '17 at 13:51
• Given there is no assumption on the distribution, it's not possible. If the stock price has a upper bound and K is higher than the bound, the butterfly price can be zero. – Jaehyuk Choi Feb 22 '17 at 14:21
• The payoff diagram makes it clear. If V < 0 it means you receive some money now, looking at the payoff diagram you see that at expiry you either receive some more money or you receive nothing. This is clearly an arbitrage: you receive money now and never have to pay out anything in the future... – noob2 Feb 22 '17 at 14:49

Alternatively, note that the dual gamma (see here) for a vanilla call is given by \begin{align*} \frac{\partial^2 C}{\partial K^2} = e^{-r(T-t)}\frac{\phi(d_1)}{K\sigma \sqrt{T-t}}, \end{align*} where $\phi$ is the density function of a standard normal random variable. That is, $C(K)$, as a function of $K$, is strictly convex. Then \begin{align*} C(K) &= C\left(\frac{K+\Delta K + K-\Delta K}{2}\right)\\ &<\frac{1}{2}\Big(C(K+\Delta K)+ C(K-\Delta K)\Big). \end{align*} In other words, \begin{align*} C(K+\Delta K)-2C(K)+ C(K-\Delta K)>0 \end{align*}

it's a model-free result. The pay-off is non-negative everywhere and positive somewhere.

Since it's non-negative everywhere, if its price was negative there would be a clear arbitrage.

We have to show positive. We assume that there is a positive probability that the stock lands in the region where the pay-off is positive.

Now, if the contract is worth zero then it has zero value today , and positive value with positive probability and negative nowhere so it defines an arbitrage.

Hence, no arbitrage implies the value is positive today, provided there is a positive probability of reaching the area where the pay-off is positive.

(see my book concepts etc for more discussion)