Why Arbitrage Does Not Force Every Derivative to Have Zero Value
Summary
This discussion asks whether the existence of arbitrage in a theoretical market necessarily makes every derivative worthless. The accepted answer gives a counterexample in a discrete market with several traded securities and states. Some underlying assets have positive prices despite the arbitrage, so calls with zero strike, which reproduce those assets’ payoffs, also have positive prices. A further example considers a claim paying the maximum of the securities’ terminal values and notes that its price is positive in the constructed model.
The main lesson is that arbitrage does not by itself imply zero prices for all contingent claims; conclusions depend on the market structure and the claim being priced. The examples rebut the universal claim but do not provide a general classification of which derivatives retain positive value, nor do they establish a unique arbitrage-free pricing measure. The original question also raises completeness, frictionlessness, short selling, and other derivative types, but the answer’s examples address only particular claims in one model. Treat this as a conceptual counterexample, not a general pricing recipe.
Key ideas
- The presence of arbitrage does not imply that every derivative has zero value.
- A zero-strike call can have the same payoff and price as a positively priced underlying asset.
- A claim on the maximum of several terminal asset values can also have positive value in the example.
- The examples disprove a universal assertion but do not characterize all claims or market models.
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Full text
# Does presence of arbitrage necessarily make all derivatives have zero value?
# Does presence of arbitrage necessarily make all derivatives have zero value?
Spin-off from: Pricing when arbitrage is possible through Negative Probabilities or something else
I mean in a theoretical sense: If we have a particular market model with some fancy assumptions such as completeness, frictionlessness, permission of shorting and fractional purchases, etc, does presence of arbitrage necessarily make all kinds of derivatives have zero value?
It seems that in the previous question, such market models might make all European call options have zero value, but what about Eur puts? Am calls/puts? Forwards? Options? Futures? Other Derivatives? Even if the answer is yes to all those, what about in other kinds of market models?
I mean in a theoretical sense: If we have a particular market model (which I guess we may assume is complete or frictionless if need be) where shorting and fractional purchases are allowed, does presence of arbitrage necessarily make all kinds of derivatives have zero value? It seems that in the previous question, such market models might make all European call option
Update: Is it fair in an introductory stochastic calculus/derivatives pricing class to ask for the price when absence of arbitrage is violated?
## Answer by mbison (score 3, accepted)
https://quant.stackexchange.com/a/22559
To answer your question:
> I mean in a theoretical sense: If we have a particular market model (which I guess we may assume is complete or frictionless if need be) where shorting and fractional purchases are allowed, does presence of arbitrage necessarily make all kinds of derivatives have zero value?
The answer is no. See example below. Went over your original link and I took the model you constructed from there.
S_0 =\begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}, S_1 = \begin{bmatrix} S_1^0\\ S_1^1\\ S_1^2 \end{bmatrix}, D = \begin{bmatrix} 1 & 2 & 3\\ 2 & 2 & 4\\ 0.8 & 1.2 & 1.6 \end{bmatrix}
First of all, consider the the trivial derivatives. The trivial call option with strike K = 0 are the ones with the same payoff as the stocks. In your model the stocks have positive price:
S_0 =\begin{bmatrix} 2 \\ 3\\ 1 \end{bmatrix}
Therefore the K=0 calls have the same price and therefore the price are bigger than zero (answering your question).
If you want an example that is less trivial consider the option that pays out the $max(S_1^0,S_1^1,S_1^2 )$. This derivative is worth more than each of the original products, which had positive price. Thus this derivative also has positive price.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.