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.:


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?


3 Answers 3


"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.

  • $\begingroup$ I agree with you with the idea that the risk neutral measure is the only feasible in order to price. The question should be why the approach of every book involves "physical probability". I mean, does in real world exists a known "physical probability"? If so, a totally rational market would price assets in such a way that physical and risk neutral probability would coincide. (e.g. in my previous example if the market has only those two asset, if the risky asset is given, it doesn't make sense to have such a risk free asset because the law is assigned (and not modeled!) to the risky asset) $\endgroup$
    – Kinderone
    Commented Jul 15, 2019 at 18:25
  • $\begingroup$ Whether the physical probability exists or not is almost a philosphical question. But let me try to answer it that way: Under the assumption that the markets behave in a certain way random. At every time point there has to exist a probability distribution that describes that random behaviour. But there is no way to find out that implicit distribution because it can change at any time. If you assume though that this distribution does not change over time you can estimate that implicit probability, i.e. the "physical probability". $\endgroup$
    – Max
    Commented Jul 15, 2019 at 18:39
  • $\begingroup$ But this assumption as all assumptions has to be justified logically or economically, e.g. using empirical distributions. So you really have to differentiate between the real world, the mathematical model and the applications of these models. $\endgroup$
    – Max
    Commented Jul 15, 2019 at 18:41
  • $\begingroup$ So regarding your example: If that physical probability you mentioned in your model is indeed the implicit prob. and is known to all market participants, then indeed the market should immediately adjust the prices s.t. the Martingale Prob. and physical prob. coincide. In practice though there is no way to figure out the implicit prob. Further not all participants agree on a good estimation. Therefore with more information/data extracted of public data or gathered somewhere else than other participants, you can do statistical arbitrage. $\endgroup$
    – Max
    Commented Jul 15, 2019 at 18:45
  • 1
    $\begingroup$ That's the best explanation I've ever found. Much more clear then many books I read about. Thank you very much $\endgroup$
    – Kinderone
    Commented Jul 15, 2019 at 18:50

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.

  • $\begingroup$ Thanks for your reply. So this means that a perfectly rational and "infinitely rich" trader should always prefer risky assets because the expected returns are always higher, right? (I know I have made an abuse using the word "arbitrage", I just meant "the market admits strategies (the risky ones) more remunerative then others in expectation") $\endgroup$
    – Kinderone
    Commented Jun 24, 2019 at 8:04
  • $\begingroup$ Indeed. It seems that wealthy people do largely employ your strategy. $\endgroup$
    – dm63
    Commented Jun 24, 2019 at 10:17
  • $\begingroup$ Oh nice freaky world the capitalism... when I'll be rich I'll call you to thank you! $\endgroup$
    – Kinderone
    Commented Jun 24, 2019 at 10:32

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 :)


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