# Does option delta predict expiration probability?

I've read that delta is a probability of option assignment. For example, if you have a put with a delta of -.70, there is a 70% chance the option will be assigned. If the delta is -.04, there is only a 4% chance the option will be assigned.

I can't remember where I read that reference. But everything I've read lately on delta simply regurgitates the standard definition. Can anyone confirm the above definition (with a reference)?

Googling "delta probability expiration" will offer lots of references on the web that discuss delta as a proxy for the probability of expiring in-the-money.

Delta is an approximation of the likelihood that the option will expire ITM. Since delta is affected by the various pricing variables, it will vary during the life of the option. Dramatic change in implied volatility can change delta significantly on a day to day basis so it's an estimate of probability rather than a guarantee.

Also, in a bull market such as we've had since 2009, calls expiring ITM exceeded the probability forecast by delta whereas puts under performed.

I consider Sheldon Natenberg to be an authority on options. In "Option Pricing And Volatility Strategies", he wrote:

There is one other definition of the delta which is of less practical value to the trader but which may be of interest from a theoretical standpoint. If we assume that futures prices are lognormally distributed, then the delta of an option is approximately the probability that the option will finish in-the-money ...

Since most option strategies depend not on the probability of an option finishing in-the-money but rather on the total expected return, this definition will not be of much use to the serious trader. A trader who is courageous enough (or foolish enough) to simply sell naked options might perhaps use this probability concept to help assess risk.

Delta is a very poor approximation of the risk neutral probability that the option will expire in the money.

N(d2) is the actual probability, which eventually becomes zero for very long dated options and / or when there is very high implied vol. On the other hand, delta (N(d1) if undiscounted forward delta), becomes 1 (100%) in this case.

You can find some details here.

Edit

N(d2) simply is the measure for the risk neutral probability of exercise. It becomes zero because of the way the model is built. The higher σ (Vol / sigma), the more the global maximum of the probability density function (the mode) shifts towards the lower bound of the lognormal distribution.

The reason the option value of the option itself is approaches the discounted value of the forward is that more and more contribution of the expected value come from values of ST when ST≥K as σ (sigma) grows, although the probability for those values to occur goes to zero. So, we have a sort of competition of limits, where the values of ST above K increase faster than their probability to occur goes to zero. It is mainly a theoretical result based on basic probability theory because very high vol and very long tenors are unlikely in option pricing.

Nonetheless, Warren Buffett explained his take on the Black Scholes formula for long-dated options (which has a similar effect as large IVOL) in his 2008 letter to the Shareholders of Berkshire Hathaway.

The Black-Scholes formula has approached the status of holy writ in finance, and we use it when valuing our equity put options for financial statement purposes. ... If the formula is applied to extended time periods, however, it can produce absurd results. In fairness, Black and Scholes almost certainly understood this point well.

Below is a gif I made in Julia showing this impact (you need to compare the mode vs the mean). Edit 2

That vol is not constant and returns are not normally distributed has very little to do with this question. If you get the math, you get the result. Delta is the probability of the option being ITM under the stock measure (this is yet another equivalent martingale measure which uses the stock as numeraire). N(d2) is the probability of the event {ST≥K} in the risk-neutral world. That is always the case.

Insofar, for the question at hand, N(d2) is the only correct answer, and N(d1) only approximates it (which the answer referencing Natenberg points out correctly) since time to maturity and volatility are typically small numbers, i.e., d1=d2+ σ*sqrt(T) ≈ d2.

Regarding the concern that vol is not constant, and returns are not normal; option markets correct for this with the IVOL skew - which affects N(d2). See for example this answer for an interactive chart showing how implied vols correct for skewness and kurtosis in the return distribution and how to fit vol smiles with SVI (one way to do it).

• So why is the probability of exercise zero for long-dated or high-vol options? Why does this make N(d2) a better measure than N(d1)? Jan 3 at 15:43
• Thanks for the edit - I get the math but my question was more practical - if the probability of exercise goes to zero for long-dated of high-vol options, does that make it a "better" measure of probability than delta? Why would long-dated options have virtually no probability of exercise under the B-S model? Jan 3 at 16:16
• To be fair, that's a loaded question... My assertion is that N(d2) is not a true measure of probability of exercise because of flaws in the B-S model (namely that vol is not constant and returns are not perfectly normally distributed) Jan 3 at 16:18