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In an interview I was once asked what the delta of an option was and my answer started from the fact that it is the first derivative of the option with respect to the price, and then I concluded saying that it is practically used as probability of the option to end In The Money at Maturity. The interviewer, very bother by this conclusion, replied:

That is absolutely not true from a mathematical point of view

Why is that so?

Mining
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  • Can you please detail how did you draw your conclusion? I.e. how did you motivate that it is probability to end ITM from the fact that it is the first derivative? – Yoda And Friends Sep 16 '21 at 15:43
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    Check out this question: https://quant.stackexchange.com/q/46836/848 – Bob Jansen Sep 16 '21 at 15:43
  • @YodaAndFriends it is simply how it is known to be used. If you study option greeks on books, you can basically read everywhere that the delta is used by traders as the probability of the option ending ITM, in fact, as the option is at the money the delta is 50%. When option gets in the money the delta is above 50% and as the maturity approaches the delta increases until getting 100% the day before maturity. – Mining Sep 16 '21 at 15:46
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    Marco, make sure you're not confusing $N(d1)$ and $N(d2)$. To cut to the chase, I show in this answer here that $N(d2)$ is the (risk-neutral) probability of the Option ending up in the money. $N(d1)$ is the Delta: sensitivity of option price per one unit move in the underlying. Sorry if this sounds harsh, but confusing the two concepts would deff be a red flag in a quant interview: these are considered relatively basic and you need to make sure you can nail these before applying. – Jan Stuller Sep 16 '21 at 15:51
  • Hi @JanStuller. Thanks for your answer. You are right. Can I ask you then why the Delta is widely known to be interpreted as the probability of the option ending in the money? – Mining Sep 16 '21 at 16:00
  • It's a vague "rule of thumb" and nothing else. It comes from the fact that a deep ITM call option has a delta of approx 1 and is very likely to expire in the money, whereas a (deep) OTM call option has a delta of approx 0, and are likely to expire OTM. I do not believe it has any mathematical basis. There's an answer here detailing the rule of thumb. In the answer he also emphasizes that delta and probability of expiring in the money are not the same. – Pleb Sep 16 '21 at 16:10
  • thank you guys! – Mining Sep 16 '21 at 16:12
  • $N(d_1)$ and $N(d_2)$ are probabilities of the exercise event ${S_T\geq K}$. However, these are artificial probability measures that have nothing to do with what the underlying is doing in the real world. None of these two numbers tells you something about the dynamics of the underlying stock in reality. – Kevin Sep 16 '21 at 16:16
  • This article explains it nicely in detail: https://www.globalcapital.com/article/28mwtvkodfvd0968sq6m8/derivatives/option-delta-versus-probability-to-exercise – Jan Stuller Sep 16 '21 at 16:18
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    Who wants to combine all these useful comments into one answer? – Bob Jansen Sep 16 '21 at 16:24
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    Feel free, @BobJansen: :P :P :P :) – Jan Stuller Sep 16 '21 at 16:34
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    this answer shows the difference between delta and the prob of ending up in the money. Here is a related topic. – AKdemy Sep 16 '21 at 16:40

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The comments already give links to many top answers and articles outlining the answer. Here's the summary:

The Black-Scholes formula for European-style call options is $$C = Se^{-qT}\Phi(d_1)-Ke^{-rT}\Phi(d_2).$$ The option delta (sensitivity to changes in the stock price) is $$\Delta=\frac{\partial C}{\partial S} =e^{-qT}\Phi(d_1).$$

Firstly, the delta of an option cannot be the probability of anything: it can exceed one, depending on the cost of carry $q$ (think of long-dated deep ITM currency options).

You can show that $\Phi(d_2)$ is the risk-neutral probability of the event $\{S_T\geq K\}$. Thus, a few people call $\Phi(d_2)$ the probability of ending up in the money. It couldn't be further from the truth. This number doesn't tell you where the asset will likely be at maturity. Risk-neutral valuation is a beautiful and very convenient pricing tool, but it makes no predictions about the future distribution of stock prices.

You can show that $\Phi(d_1)$ is the (risk-neutral) probability of the event $\{S_T\geq K\}$ associated to a different numéraire. That's even more technical and even less related to where the underlying is going to end up in the real world.

Because $\Delta$ is easily observable on any trading platform, and because $d_1=d_2+\sigma\sqrt{T}\overset{?!}{\approx} d_2$, some people may suggest that delta proxies the probability of exercise. As you now know, this is just wrong and bad. As @Jan said this is a "red flag in a quant interview".

What you can do: You can interpret $Se^{-qT}\Phi(d_1)$ as price of an asset-or-nothing and $e^{-rT}\Phi(d_2)$ as price of a cash-or-nothing option, or as the aforementioned risk-neutral probabilities.

Kevin
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  • +1. Btw: would it be fair to say that the risk neutral Probability of exercise (i.e. $N(d2)$), over-estimates the real-world probability of exercise? The reason I am saying that is that the real-world drift $\mu$ cannot be lower than the risk-neutral drift associated with the money-market numeraire, i.e. $r$ (for no arbitrage reasons): then, $\mathbb{E}^Q[S_t]\leq \mathbb{E}[S_t]$ (where the second expectation is under the real-world measure). – Jan Stuller Sep 17 '21 at 14:23
  • It’s possible that $\mu<r$, it the asset loads negatively on positively priced risk factors (acts as insurance in some sense). Granted, for pretty much all equity options, that’s not the case. But it can very well be the case for other assets. It’s fair to say that $\Phi(d_2)$ is the exercise probability to people who know what is meant by this. But I hear it too often interpreted in a sense that the option will actually expire ITM with probability $\Delta$, which is just nonsense. Not everyone appreciates what “risk neural probability” means. – Kevin Sep 17 '21 at 15:08
  • You're right: in the discrete binomial model that I discussed in one of my earlier answers here, the no arbitrage condition is that $d \leq e^r \leq u$, but that doesn't necessarily mean that $\mathbb{E}[S_1] > S_0 e^r$. For the vast majority of risky assets though, I assume we're safe to say that the risk-neutral probability over-estimates the real-world probability of exercise: would you agree? – Jan Stuller Sep 18 '21 at 09:59
  • Yeah, I most certainly agree!:) Most assets (certainly pretty much every stock) have a higher expected return than a risk-free bond. I still warn to equate delta with the exercise probability and to make any suggestion that the risk-neutral exercise probability carries any significant information for real world trading. – Kevin Sep 18 '21 at 10:11
  • Agreed 100% that linking risk-neutral probability to real world likelihoods is a "no-go". To me, the risk-neutral probability on stocks at least puts an " upper bound" on the real world probability though: still pretty useless for decision making. I wonder if there's any good research on trying to extract real-world probabilities from risk-neutral ones via some method: I guess you could pick your individual "real-world" drift and compute the market price of risk and extract the real-world probability via radon-nikodym? – Jan Stuller Sep 18 '21 at 10:32
  • It’s not necessarily an upper bound. Remember that the Black Scholes model is completely misspecified. There are many other modelling aspects $\Phi(d_2)$ neglects. For real applications, I’d say this number is pretty useless (but then, I have never traded an option in real life either haha). And you’re again absolutely righty: There’s certainly a ton of research about extracting real world probabilities from option prices! – Kevin Sep 18 '21 at 10:41
  • Cool, always great to discuss these things and get views from others. If you know any good (or at least decent) papers on extracting real world probabilities from option prices, pls do share a link: I'd be curious to see at least the approach they take... – Jan Stuller Sep 18 '21 at 10:52
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    @JanStuller To get $\mathbb{P}$ dynamics from $\mathbb{Q}$ dynamics, you need to make a stance on the SDF, see this answer for a textbook approach. Current research on recovery theory (starting from Ross (2015, JF)) tries to extract $\mathbb{P}$ dynamics from market prices, but it's empirically not quite successful. The inequalities following Ian Martin's seminal 2017 QJE paper are extremely interesting and promising. [P.S. Sorry for the late response. I only had access to the internet via my phone during the last days.] – Kevin Sep 21 '21 at 19:03