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?

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    $\begingroup$ Check out this question: quant.stackexchange.com/q/46836/848 $\endgroup$
    – Bob Jansen
    Commented Sep 16, 2021 at 15:43
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    $\begingroup$ 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. $\endgroup$ Commented Sep 16, 2021 at 15:51
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    $\begingroup$ Who wants to combine all these useful comments into one answer? $\endgroup$
    – Bob Jansen
    Commented Sep 16, 2021 at 16:24
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    $\begingroup$ Feel free, @BobJansen: :P :P :P :) $\endgroup$ Commented Sep 16, 2021 at 16:34
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    $\begingroup$ this answer shows the difference between delta and the prob of ending up in the money. Here is a related topic. $\endgroup$
    – AKdemy
    Commented Sep 16, 2021 at 16:40

1 Answer 1


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.

  • $\begingroup$ +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). $\endgroup$ Commented Sep 17, 2021 at 14:23
  • $\begingroup$ 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. $\endgroup$
    – Kevin
    Commented Sep 17, 2021 at 15:08
  • $\begingroup$ 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? $\endgroup$ Commented Sep 18, 2021 at 9:59
  • $\begingroup$ 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. $\endgroup$
    – Kevin
    Commented Sep 18, 2021 at 10:11
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    $\begingroup$ @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.] $\endgroup$
    – Kevin
    Commented Sep 21, 2021 at 19:03

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