I just want to be sure if my thinking is correct and does not have any flaws.

Let's define stock as a process $S$ (see the picture below) with real-world measure $\mathbb{P}$, where $p=0.9$ and Bonds with zero interest rate $r=0$.

process S

Market Maker (MM) wants to price an options with $T=1$ and $K=100$. So the payoff of the options is same as claim $X_T$ (see the picture). To price this option, he finds a $\mathbb{Q}$ measure such that the process $S$ is martingale with respect to $\mathbb{Q}$. MM derives that the $\mathbb{Q}$ measure has $q=0.5$ and the price of the options is $10$ and the hedge is $(0.5, -40)$ (buying $0.5$ of stock and $40$ cash loan).

Now comes the idea: Random trader realise that $\mathbb{Q} \neq \mathbb{P}$ and $E_{\mathbb{P}}[X|F_0] = 18$. So the trader buys the option, but he will not hedge it.

If we simulate this process multiple times for trader, we will see that the trader generated a profit $E_{\mathbb{P}}[X|F_0] - E_{\mathbb{Q}}[X|F_0] = 8$ on average per simulation.

Does this means, if every trader sees $\mathbb{Q} \neq \mathbb{P}$, he will buy the option, which forces the MM to hedge it by buying the stock? So then this forces $S_0$ price to rise (to $116$) until there does not exist any profit: $E_{\mathbb{P}}[X|F_0] = E_{\mathbb{Q}}[X|F_0]$, which occurs at $q=p=0.9$ ($\mathbb{Q} = \mathbb{P}$).

So my conclusion is, if $\mathbb{Q} \neq \mathbb{P}$ then there exists some strategy, which can generate you positive expected yield.

Trading this positive expected yield strategy will force $S_0$ and $\mathbb{Q}$ to change to reflect $\mathbb{P}$.


Replaced arbitrage in title to trading strategy

  • 2
    $\begingroup$ All traders know $\mathbb Q\neq\mathbb P$. The ``difference’’ between these two measures is defined by risk aversion of traders. Thus, the expected return of an asset $\mathbb{E}^\mathbb{P}[X]$ is very different from its price $e^{-rT}\mathbb{E}^\mathbb{Q}[X]$. Just because you have an strategy that has a positive expectation doesn't mean it's an arbitrage. $\endgroup$
    – Kevin
    Commented Dec 2, 2022 at 13:17
  • $\begingroup$ @Kevin yea, you are right is more trading strategy. But I was amazed that you can sell option and do not lose any money and an buyer can generate expected positive return. $\endgroup$
    – lukas kiss
    Commented Dec 2, 2022 at 15:02
  • 1
    $\begingroup$ It should be at least slightly positive though. The party who net buy options transfer risks away. If the net buy does not result in expected loss, there is only benefit, and no cost. Why would the other party who net sell options be willing to transact? Conceptually, that cannot work. $\endgroup$
    – Argyll
    Commented Dec 2, 2022 at 15:23
  • $\begingroup$ @lukaskiss There are many (simpler) strategies that have a positive expected return (buy equities for example). All of these positive expected returns come with their appropriate risk. But you're right $\frac{\mathbb{E}^\mathbb{P}[\max\{S_T-K,0\}]}{e^{-rT}\mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}]}$ is how you calculated the expected return of a European call option held until maturity. You can also calculate Sharpe ratios etc. of your option trading strategies. $\endgroup$
    – Kevin
    Commented Dec 2, 2022 at 15:37
  • $\begingroup$ @Kevin yea it sounds good, but it is very difficult to find $\mathbb{P}$ measure. It is also very interesting that when you hedge, you suddenly appear under $\mathbb{Q}$ with "no risk", but if you do not hedge, your returns are under $\mathbb{P}$ with risk. $\endgroup$
    – lukas kiss
    Commented Dec 2, 2022 at 18:19

1 Answer 1


As people pointed out, P and Q are different because of risk aversion. Under the P measure, the payoff in bad states are worth more; whereas Q measure is risk neutral. This is similar to why people don't invest all their money into stocks instead of risk-free bonds, although it is well known that stocks have higher expected returns.


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