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I want to prove why there are no arbitrage opportunities when a long butterfly is strictly positive. I know there is a similar topic out there, but it seems it doesn't solve my question: Prove that the butterfly condition is always greater than zero.

By call-put parity, I know

$C(T,K+∆K)-2C(T,K)+C(T,K-∆K) \geq 0$

is valid, but how to prove strict positivity? I know it makes sense for a long butterfly to be strictly positive in no arbitrage condition. But I just don't know how to get it mathematically. Hope I can get some idea here.

enter image description here

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  • $\begingroup$ Does my answer clarify things? $\endgroup$ Feb 8 at 17:14
  • $\begingroup$ @DaneelOlivaw Thanks for your answer. I just left you a comment. Sorry if the question sounds stupid, I am new to arbitrage . $\endgroup$
    – will_www
    Feb 8 at 17:17
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I am not sure why the question you link to does not provide an answer. I’ll try to answer it but it is really similar to what has already been said there. Bottom line is: if the value $K$ is reachable by the underlying asset $S$, that is $K$ belongs to the domain of process $S$, then the butterfly should be strictly positive.

First note that the butterfly is actually an approximation of the second derivative w.r.t. to strike: $$\lim_{h \rightarrow 0}\frac{C(t,K+h)-2C(t,K)+C(t,K-h)}{h^2}=\frac{\partial^2C}{\partial K^2}(t,K)$$ where obviously $h^2>0$. Yet by the Breeden-Litzenberger formula, we know that: $$\frac{\partial^2C}{\partial K^2}(T,K)=e^{-rt}q(t,K)\geq 0$$ where $q$ is the risk-neutral density of the underlying $S$ and $r$ the risk-free rate. You now see that if $K$ is a value which $S$ can reach, that is $K$ belongs to the domain of $S$, then the density of $S$ at $K$ must be strictly positive, that is: $$C(t,K+h)-2C(t,K)+C(t,K-h)\approx h^2e^{-rt}q(t,K)>0$$

Going further, let us introduce the Dirac delta function $\delta$, which is characterized by the following property for any real-valued function $f$: $$\int_{-\infty}^{+\infty}\delta(x)f(x)dx=f(0)$$ Hence the density can be expressed as: $$q(t,K)=\int_{-\infty}^{+\infty}\delta(s-K)q(t,s)ds=E^Q\left(\delta(S_t-K)\right)$$ That is, the risk-neutral density corresponds to the price of a payoff which is non-negative everywhere and strictly positive for one state the world, i.e. if $S_t=K$ $-$ informally the payoff would be infinite if $S_t=K$, see the definition of the Dirac delta. Hence to avoid arbitrage the price of this claim, $e^{-rt}q(t,K)$, must be strictly positive.

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  • $\begingroup$ Thanks for detailed explain, but I am a bit confused now. So, no arbitrage actually means the value is positive today and in future right? From the graph I added above, consider the upper line, even today the zero is 0, the arbitrage opportunities in future still exist since its possible that V >0 in future? $\endgroup$
    – will_www
    Feb 8 at 17:15
  • $\begingroup$ @will_www as you rightly point out, the payoff of the butterfly (grey line) is positive. More specifically, you see it is null below $K-\Delta K$ and above $K+\Delta$, and strictly positive in between. Hence, if you are long a butterfly, there will be some scenarios where you receive 0, other where you will receive a positive sum. [cont'd] $\endgroup$ Feb 8 at 17:29
  • $\begingroup$ [cont'd] @will_www suppose now the price of the butterfly was 0. Then you could be infinitely as many. This would amount to free money, that is arbitrage: if the price of the underlying ends up below $K-\Delta K$ or above $K+\Delta K$, you made zero PnL because you do not receive any payoff but anyway the butterfly cost you zero. But if you end up between those 2 values, then you are actually making a positive profit. $\endgroup$ Feb 8 at 17:31
  • $\begingroup$ In conclusion, as you point out, arbitrage implies that, if the value is positive in the future (that is, non-negative always and at least strictly positive sometimes), then the value must be strictly positive today. Otherwise, you can make free money. $\endgroup$ Feb 8 at 17:33
  • $\begingroup$ Thank you so much! Got it now! $\endgroup$
    – will_www
    Feb 9 at 2:33
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as it was stated correctly in the question all long butterfly options have to have a non-negative premium in order for No-arbitrage to hold. So we can say that:

No-Arbitrage holds implies All Butterfly spreads have a non-negative premium.

However, the reverse is not true. Just because all butterfly spreads have non-negative premiums does not mean that there is No-Arbitrage. There is a well-known paper by Davis and Hobson. They state necessary and sufficient conditions for option prices to fulfill No-Arbitrage. Their notation is a bit hard to get used to. So here is quick (and a bit sloppy) summary of theorem 3.1.

There is no Arbitrage in the market if and only if

  1. All Butterfly spreads have a non-negative premium
  2. All Call-spread (i.e Long Option with Strike K1 and Short Option with Strike K2>K1) have a non-negative premium.
  3. Call spreads are not too expensive (so there is a upper bound on call spread premiums).
  4. The premium of a call spread can only be zero if both call options have zero premium.

The authors distinguish between model independent arbitrage and weak-arbitrage. They show that if conditions 1., 2. or 3. fail there is model independent arbitrage. If condition 4. fails there is weak arbitrage.

Coming back to you question, here is an easy example of why positive butterfly premiums do not guarantee No-Arbitrage:

Let $K_1 = 1, K_2 = 2, K_3 = 3$ be three strikes and $p_1 = 1, p_2 = 4, p_3 = 8$ be the premiums where $p_i$ is the premium for the call option with strike $K_i$. Also, we assume that the risk-free rate is equal to zero. The corresponding butterfly option $$ \frac 12 C_1 - C_2 + \frac 12 C_3 $$ has a premium of $$ p_{\text{butterfly}} = \frac 12 p_1 - p_2 + \frac 12 p_3 = \frac 12 > 0. $$ But there is an arbitrage opportunity. Simply going long in $C_1$ and short in $C_2$ results in an arbitrage opportunity.

I hope that this was helpfull.

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  • $\begingroup$ Thanks! yes its helpful $\endgroup$
    – will_www
    Feb 9 at 2:34

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