# Does presence of arbitrage necessarily make all derivatives have zero value?

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

• What do you mean with zero value? The price is not zero for all derivatives. – mbison Dec 31 '15 at 11:30
• @mbison Could you give an example of a derivative in that market with positive value? – BCLC Jan 1 '16 at 2:29
• Given your question, most likely i am completely misunderstanding you. But is the price of a call option not bigger than 0? i.e. you have to pay something to enter the contract? Or a bond, you got to pay something to now to get the payoff at the end. Maybe when you are talking about "value" you mean something else then when I am talking about "price" – mbison Jan 1 '16 at 13:50
• @mbison value = price. In the question linked, the option price is zero – BCLC Jan 1 '16 at 21:41

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.