1
$\begingroup$

From the option chain of a security, we can calculate the implied probability density at the maturity $T$ (assume the options are European. Now suppose we have our own view/prediction on the probability density of the underlying price at time $T$. What is the best way to trade this view?

For example, say the implied probability of the underlying will end above $150$ is $5\%$ and our own view of that is $10\%$. We would think that the upside calls are underpriced. However, if we express our view by simply buying an upside call, the chance for us to have profit is still very small even if our view is totally correct. Trading volatility does not seem to be quite relevant also since the view is about the terminal probability distribution.

To be more specific, in the situation described above, what are the good ways trading the view? Buying a 150 call only gives you a $10\%$ chance of winning even if our view is $100\%$ correct. Are there any other ways so that the winning probability is higher but still give us a positive expectation?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Note that option prices contain distributional information about the risk-neutral measure $\mathcal{Q}$, not the physical one.

That being said, per the well-known Breeden-Litzenberger formula, the undiscounted price $\widehat{C}$ of a European call option on a security $S$ with strike $K$ and expiry $T$ is: $$\widehat{C}(t,K,T)=\int_K^\infty(S_T-K)\text{d}\mathcal{Q}(S_T)$$ Differentiating with respect to $K$: $$\frac{\partial \widehat{C}}{\partial K}(t,K,T)=-\int_K^\infty \text{d}\mathcal{Q}(S_T)$$ That is: $$\mathcal{Q}(S_T\leq K)=1+ \frac{\partial \widehat{C}}{\partial K}(t,K,T) $$ Approximating by finite central differences: $$\mathcal{Q}(S_T\leq K)\approx 1+ \frac{\widehat{C}(t,K+\delta,T)-\widehat{C}(t,K-\delta,T)}{2\delta} $$ The second term is a static position in a bull spread, which give you direct exposure to the cumulative probability density.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Thank you for your answer. I know this result however it does solve the problem. Let's say we know the call spread is underpriced from our view. But the problem is the call spread won't make us any money if S ends below, say $K-\delta$. Yes the expectation profit of this trade is positive but it is very asymmetric, say with 90% probability losing money and 10% probability winning, which is not very good from my perspective. $\endgroup$
    – MainCom
    Commented Mar 10, 2021 at 9:17
  • 1
    $\begingroup$ You don't need to hold the options to maturity. When you "trade the probability distribution" you are betting that the implied distribution converges toward your view, normally over a relatively short time frame. You could buy options with 1Y to maturity but plan to exit the position in two weeks. $\endgroup$
    – Chris Taylor
    Commented Mar 10, 2021 at 9:35
  • $\begingroup$ not really. the implied density may converge to its historical average, but definitely need not converge to my density (which I think though is correct). $\endgroup$
    – MainCom
    Commented Mar 10, 2021 at 9:48
  • $\begingroup$ @MainCom not sure I follow. You have a view on the "price" of $\mathcal{Q}$, and a bull spread is a very close proxy to that. If you believe the market is mispricing $\mathcal{Q}$, then you must be thinking the value of bull spreads should adjust to your view, right? $\endgroup$
    – Daneel Olivaw
    Commented Mar 10, 2021 at 10:03
  • 1
    $\begingroup$ @MainCom I don't think what you are looking for is possible. When you have a view that the market is wrong, you're always assuming it will revert to what you believe is the fair level. Thus Keynes aphorism: "the markets can remain irrational longer than you can remain solvent". Unless I am missing something? But you seem to be saying: for a given strategy $\theta$, I want to find a strategy $\theta^\star(\theta)$ with same dependence on risk factors $R_1,\dots,R_n$ as $\theta$ but with 1) higher probability of positive payoff and 2) higher payoff. That would constitute an arbitrage. $\endgroup$
    – Daneel Olivaw
    Commented Mar 10, 2021 at 10:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.