options - 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.

Answer

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

\begin{equation} 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). \end{equation}

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:

\begin{equation} \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}. \end{equation}

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.