r/wallstreetbets post alleges that because ~$1.9 Trillion USD of SPY call options expire on 3/20, the price of SPY will skyrocket on 3/20.

Is this correct? What can be deduced? I'm not expecting to predict the price of SPY on 3/20, but can we at least predict its price direction?

  • $\begingroup$ No, the person has no basis for claiming that "most of these options are on the short side". Where did he get that info? $\endgroup$ – noob2 Mar 19 at 7:14
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    $\begingroup$ That's complete nonsense. Imo, reading Reddit posts for financial insight is not wise. $\endgroup$ – amdopt Mar 19 at 20:17
  • $\begingroup$ Note that if there are many options of a certain strike outstanding with very short time to maturity, and the spot price happens to be near the strike already, there is a possibility of stock pinning due to hedging activity. This is not the same though as 'predicting' price direction. $\endgroup$ – ilovevolatility Mar 21 at 12:24

Background information

You need to put some structure on this to get a sense of what is going on. Skip down if you're acquainted with asset pricing. Define $p_t$ as the price of a security at time $t$. Its payoff at time $T$ is given by $x_T$. In the absence of arbitrage, there exists some positive random variable $m_T$ such that: $p_t = E_t^P(m_T x_T)$. Moreover, we can show that this equation implies $exp(-r_f (T-t)) = E_t^P(m_T)$ where $r_f$ is the risk-free rate of return if we use it to price a riskless bond.

Another way to write the pricing equation above is by using the fact that \begin{equation} E_t^P(exp(-r_f(T-t))m_T) = \int_{\omega \in \Omega} exp(-r_f(T-t))m_T(\omega) dP(\omega)= 1 \end{equation} and the fact that $\forall \omega \in \Omega \; exp(-r_f(T-t))m_T(\omega) > 0$. In plain English, that thing is always positive and it adds up to 1, so for all intent and purposes you can treat it as a new, "twisted" probability distribution which incorporates BOTH risk aversion concerns AND the uncertainty captured the real, physical distribution. We write it this way: \begin{equation} p_t = E_t^P(m_T x_T) = exp(-r_f(T-t))E_t^Q(x_T) \end{equation} it should apply to all assets, including options and their underlying, and we call this the risk-neutral distribution because it discounts at at the risk-free rate.


So, when you are looking at option prices, what do you learn? Up to some interest rate discounting, you're gauging information about the risk-neutral (or Q) distribution, not the physical (or P) distribution. To go from option prices to information about the physical distribution of the underlying, you have to make some assumptions about $m_T$ -- or, stated differently, you have to pick a market price process for all sources of risks (which impact the payoff $x_T$).

Is the direction of the price of a stock market index predictable based on option prices? Well, theoretically, setting aside dividend payments, $(m_tp_t)_{t \geq 0}$ is a martingale: it's not the change in price that is unpredictable, but the stochastically discounted price. What the model DOES say is that betting on something like the direction of the market is profitable to the extent that it EXPOSES you to some risk that you cannot diversify away.

It's possible that you manage to use option prices to predict direction a little better than a coin toss would. It's not entirely crazy, especially since using options greatly expands your information set for modeling. However, it's very likely that any profit you'd get from that would be priced risk and not arbitrage. You can seemingly make a fortune and think to yourself that the efficient market hypothesis is BS, until a series of large jump drains every penny you made in the last decade and you finally learn something: you were being paid to assume that risk and, sometimes, it blows back in your face.

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