Prove that the butterfly condition is always greater than zero

Question

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 1

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.

Answer 2

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)

Answer 3

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.

Answer 4

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*}