Risk-neutral pricing and statistical arbitrages

Question

I'm studying the martingale approach to asset pricing. Dealing with the concept of risk-neutral probability, I came up with a question about the possibility of "arbitrages in expectation". I'll be more precise with a (maybe too simplistic) example:

Consider a discrete-time framework with only two points $t=0$ (today) and $t=1$ (tomorrow). In this context we consider a market composed by

• a risk-free asset $B$ whose price is given by: $B(0)=1, B(1)=1+r$
• a risky asset $S$ whose price at today is given by: $S(0)=1$ and tomorrow price will be determined by a fair-coin toss: $S(1)=10$, if head, $S(1)=0$ otherwise.

Consider that the market sets the value of $r$ equal to $0.06$.

The risky asset dynamics is given under a "physical probability" $P$.

To have no-arbitrage pricing we have to find a martingale measure of probability $Q$, i.e $Q$ s.t.:

$\frac{1}{1+r}\mathbb{E}^Q[S(1)|\mathcal{F}_0]=S(0)$

The sample space in this simple context is $\Omega=\{head, tail\}$.

• Under the physical probability $P(head)=P(tail)=0.5$.
• Under the martingale probability $Q(head)=0.106$ and $Q(tail)=0.894$

Now want to price a derivative on S, for example a EU call option with strike $5\$$. The payoff of this option will be $\Phi(s_1)=(s_1 - 5)^+$.

According to the physical probability $P$, the expected value of this contract today is

$\mathbb{E}^P_0[\Phi(S(1))]=0.5 \cdot 5 + 0.5 \cdot 0=2.5\$$.

Thus the price that I would give to the contract is $\frac{2.5}{1.06} \$ \approx 2.36\$ $

However this will not be the market price of this contract: under the arbitrage-free condition it's price will be given by the discounted expectation w.r.t. $Q$, i.e.:

$\frac{1}{1.06}\mathbb{E}^Q_0[\Phi(S(1))]= \frac{0.106 \cdot 5 + 0.894 \cdot 0}{1.06} =0.5\$ $

This sounds strange to me. It seems that I have the opportunity to make a sort of "arbitrage in expectation" in the sense that my expected return on the investment is much higher then what I have to invest.

I know that this example is very simplistic, but this phenomenon seems to hold in general. Reading many resource about risk-free measure I understood that this new measure on the sample space of possible outcomes (the "set of potential states of the world") take in account the risk-aversion of markets, in the sense that people want to pay less very risky assets. However this opens the possibility to the aforementioned "statistical arbitrage". Abstractly if there's a market with an infinity of very very risky stocks, then a rich trader should buy one call option on each stock and for the Law of Large Numbers make money for sure (it's obviously an abstraction but this catch is what I mean by "statistical arbitrage").

I cannot figure out where this reasoning fails. The question is: am I misunderstanding the real meaning of arbitrage-free pricing or the only reason for which this it seems to be strange to me is that I miss some economical / "real markets world related" point of view?

Answer 1

"Cannot figure out where this reasoning fails. The question is: am I misunderstanding the real meaning of arbitrage-free pricing or the only reason for which this it seems to be strange to me is that I miss some economical / "real markets world related" point of view?"

In opposite to the first comment, I think you do miss something a bit subtle here. First the reason why one assumes arbitrage-freedom in order to price is very simple: Because if you do not, you cannot determine a price of an asset in a consistent and senseful meaning. Assume you would allow arbitrage opportunities in your mathematical model. Then the price of a position that does arbitrage must be arbitrarily high. And the price of a position without arbitrage would have the price 0 because of opportunity costs. Therefore in order to be able to price in senseful way you need to assume arbitrage-freedom. So in other words: The possible physical probabilites determined by your model are given by the martingale measures! Economically one could argue that is due to the fact that real probabilites are implied in the price determined by the market.

Obviously the assumptions used in mathematical models fail in many ways in real life. And therefore the calculated probabilites of the model are not correct or at least not precise. However estimating the probabilites with statistical methods is not an easy way out because of two reasons:

1.) There is no way of testing whether your statistically estimated probability is precise or "correct". You can simply invest and see whether you make money but ex post you do not know whether you just had luck or did a good estimation. Power laws should hold to a certain degree but again this can be risky.

2.) Transaction fees and the spread of interest rates can keep real-world-arbitrage-freedom in place in spite of mispricing of assets and options.

Also a risky asset does not necessarily have a better expected return because the risk is priced. When doing arbitrage you usually try to leverage the arbitrage opportunity and therefore also the return itself does not play such a big role but rather the ratio of yield and risk, since you can reach the same return by leveraging.

Answer 2

What you say is perfectly true and there is no contradiction. Arbitrage means risk free profit , so your ‘statistical arbitrage’ is not arbitrage at all. It just says that if you take risk, your expected returns can be higher than the risk free rate. How much higher depends in the risk aversion of market participants.

Answer 3

Ok I was having the same question while studying Financial Mathematics. After some careful thinking, I came to the obvious conclusion that this is nothing weird. To explain, the risk-neutral price of 0.5$ is actually the cost of the replicating portfolio. That is, you can completely hedge your option by buying 0.5 of the risky asset for 0.5\$ and have the same payoff as the option in both future cases. The subtle thing is in the assumptions of the model itself: it is assumed that the market is frictionless and you can sell short and go long as much as you want, when you want. Additionally, the model assumes that all information is shared equally among all participants.
For the option issuer to hedge the option in question, she would need to buy 0.5 shares of the risky asset. However, if everyone in the market knew the underlying probability distribution of the stock price, probably no one would sell you the asset for 1\$ per share, since it has positive expected value and it would make more sense to keep the stock (assuming some risk-taking), which would drive the price up.
So in short, if someone is able to buy the risky asset at 1\$, she can sell you the option for 0.6\$ (and therefore earn 0.1\$ of riskless profit) and you will have an expected value of 1.9\$. The only one who lost here (comparing expected values) is the one who sold the risky asset to the option issuer used for hedging, since he gave up an expected 2.5\$ for an expected 0.5\$ (100% * 0.5).
Hope this was helpful to someone :)