Why Risk-Neutral and Real-World Option Values Can Differ
Summary
The document considers an option priced under a risk-neutral probability measure and compares that price with the payoff's expected value under the real-world measure. It asks whether the gap represents a positive expected-profit opportunity and whether traders buying the option would force the underlying price and risk-neutral probabilities to adjust until the two measures coincide.
The answer explains that the measures can differ because investors value states differently: bad-state payoffs can command a premium, while risk-neutral valuation incorporates that risk pricing. A higher physical expected return therefore does not by itself establish an arbitrage or a positive expected return after accounting for risk. The example gives hypothetical probabilities, prices, and a hedge, but the response does not fully analyze the proposed trading strategy, market impact, or how prices would adjust. Its main lesson is the distinction between risk-neutral pricing probabilities and real-world outcome probabilities; the measures need not become identical for markets to be consistent.
Key ideas
- Risk-neutral probabilities are used to price claims in an arbitrage-free framework, while real-world probabilities describe outcome frequencies.
- Differences between the measures can reflect compensation for bearing risk.
- A payoff's larger real-world expected value than its risk-neutral price does not alone prove an arbitrage.
- The response does not establish that trading pressure would make the two probability measures coincide.
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Full text
# $\mathbb{Q}$ measure and $\mathbb{P}$ measure, trading strategy
# $\mathbb{Q}$ measure and $\mathbb{P}$ measure, trading strategy
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 $r=0$.
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 $\mathbb{P}$.
EDIT:
Replaced arbitrage in title to trading strategy
## Answer by ml-guy (score 1, accepted)
https://quant.stackexchange.com/a/73961
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.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.