Using Option Prices to Infer Return Distributions, Not Spot Trade Signals
Summary
The discussion explains why Black–Scholes–Merton is primarily an option-pricing framework, not a recipe for generating entry and exit signals in spot markets. Its geometric Brownian motion assumptions imply lognormal stock prices, but the answer notes that this is a poor description of equity and index returns. The model’s pricing role comes from changing from the physical probability measure to a risk-neutral one, under which discounted asset prices satisfy a martingale condition.
For information about the underlying, the response points to extracting a risk-neutral return density from the curvature of call prices across strikes, using finite differences in practice. Such a density reflects both expectations and risk compensation, so it is not a pure forecast of actual returns. Option-implied parameters might inform a model-based estimate, but the source stresses that results depend on assumptions and do not guarantee profitable trades. A second answer similarly frames implied volatility as a market-price-derived measure and notes that apparent mispricing still requires a viable counterparty or arbitrage opportunity.
Key ideas
- Black–Scholes–Merton prices European options under assumptions that do not closely describe actual equity returns.
- Risk-neutral pricing changes the probability measure so discounted asset prices satisfy a martingale condition.
- The curvature of call prices across strikes can be used to infer a risk-neutral density for the underlying at expiration.
- Risk-neutral distributions combine expectations with risk compensation and should not be treated as direct physical forecasts.
- Option pricing models can identify relative mispricing only under their assumptions and do not by themselves create profitable spot-trading signals.
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# How to use black scholes for spot trading?
# How to use black scholes for spot trading?
Since I am very new to this topic, i am struggling to connect the option pricing with Spot Trading.
Is there a way to use the Black-Scholes model to derive entry and exit Events? Any blogs or papers?
I've just started to research the Black-Scholes model of using online courses on coursera and somehow fail to find the reasoning for this approach besides option price calculation.
Thanks!!
## Answer by Stéphane (score 2, accepted)
https://quant.stackexchange.com/a/53087
I am going to assume that you meant to ask how we can use option prices to inform trading of underlying stocks.
The Black-Scholes-Merton model says that under the physical measure, the underlying stock obeys a geometric Brownian motion: \begin{equation} dS_t = \mu S_t dt + \sigma S_t dZ_t. \end{equation} You can apply Ito's Lemma and integrate to obtain \begin{equation} S_t = S_0 \exp \left( \left( \mu - \frac{\sigma^2}{2} \right)t + \sigma Z_t \right). \end{equation} Now, $(Z_t)_{t \geq 0}$ is a standard Brownian motion under the physical measure. By definition, this means $Z_t \sim N(0, t)$. In other words, gross returns $R_{0,t} := \frac{S_t}{S_0}$ are distributed log normally; or, if you prefer, returns $r_{0,t} := ln S_t - ln S_0$ are distributed normally. The first issue here is that this is a bad model for the behavior of returns on equity and equity indexes. Whether you test for the normality of returns or for the log normality of gross returns, you will find that it is a bad model. Or, if you prefer, it's easy to find more flexible models that will be prefered from a Bayesian perspective, even if you use priors that favor this model.
However, the interest of the Black-Scholes-Merton model was not to describe the behavior of returns on equity as much as it was to provide a way to price European options. My favorite way of approaching this is to say that Girsanov's theorem tells you that the pricing kernel you should use is \begin{equation} m_t = m_0 \exp \left( - \int_0^t \theta_s ds - \frac{1}{2} \int_0^t \theta_s^2 ds \right) \end{equation} where $\theta_t := \frac{\mu - r}{\sigma}$ is your Sharp ratio. To be clear, this variable variable is chosen so that $m_t S_t$ and $m_t B_t$ (where $B_t$ would be the price of a riskless bond) are martingales under the physical measures. It so happens that $\exp(r \tau) m_t$ is a proper Radon-Nikodym derivative and you can use it to change to the risk-neutral measure: \begin{equation} C(t, t+\tau,, K, S_t) = E^P \left(m_{t+\tau} \left( S_{t+\tau} - K \right)_+ \right) = \exp(-r \tau) E^Q \left( \left( S_{t+\tau} - K \right)_+ \right). \end{equation} From those equations, you can derive the Black-Scholes-Merton formula. The martingale restriction in particular will impose that $S_t$ grows at a rate of $r$ rather than $\mu$ under $Q$ and Girsanov's theorem will tell you that a translated and scaled version of $Z_t$ is a standard Brownian motion under $Q$. With a bit of algebra, you would see that stock prices follow a geometric Brownian motion under $Q$ and you'd get an expression similar as that above for prices. From there, you'd exploit the log-normality of prices to compute the expectation under $Q$ and some tedious algebra later you would land on the famous Black-Scholes formula.
The problem with all of the above is that it relies on a lot of assumption, one of which says that the arbitrages you think you can find by trading stocks cannot exists... But it might not be an insurmontable problem.
If you assume that the Black-Scholes-Merton argument is "almost" correct in the sense that markets deviate temporarily from this and come back to it, you might be able to do something interesting. In particular, Black-Scholes-Merton gives you a direct link between option prices and the conditional density of returns under either measures. Even if the model is wrong, nothing forbids you to use the model to estimate the parameters $\mu, \sigma$ using option prices and plug that into a log-normal distribution to have a guess of where prices are likelier to land "sort of according to the derivative market."
But, if you want to go in that direction, there might be better alternatives. In particular, Breeden and Litzenberger (1978) gave us a way to related the risk-neutral density with option prices. Specifically, \begin{equation} f^Q(S_{t+\tau}, K) = \frac{\partial^2 C(t, t+\tau, K, S_t)}{\partial K^2}. \end{equation} This can approximated using a finite difference on a grid of strikes prices of options at time $t$, with all options maturing at time $t + \tau$. You pick the moment they mature according to whatever horizon you'd like to "forecast." Now, the twist here is that you have risk-neutral densities: this will mix information about changing risk compensations and changing expectations. However, nothing says you couldn't bet a machine learning algorithm with a suitable loss function couldn't use these as inputs and learn to provide you the information you need like a point forecast, an interval forecast, or quantile forecasts. The major advantage of the BL(1978) result is that it is as model-free as it will get and it allows you to use large cross-sections of options to say something potentially useful about the underlying.
## Answer by Andreas (score 1)
https://quant.stackexchange.com/a/53089
In short, no, you can not use the Black-Scholes model to derive entry and exit events. It is a bad approximation with several flaws and therefore nowadays only used to obtain a measure of volatility called "implied volatility". As the only latent variable in the models equation volatility is used to obtain the "correct" market price. Implied volatility therefore indicates what the volatility should be in order for the BS-model to come to the current market price.
Black-Scholes is, due to it's relative simplicity, often taught to give people a basic understanding of the dynamics of options prices. Several more sophisticated models have since been proposed, fixing the shortcomings of the BS model, however, none will guarantee you to make money.
Option pricing models can be used to figure out, if an option is over or underpriced (relative to your model and it's assumptions such as input parameters) but even if the model indicates the option should be worth more or less, you need a counterparty to trade with or be able to make an arbitrage profit.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.