This seems to be another easy question but I am a bit confused. I know delta is a proxy for an option finishing ITM. Delta also happens to be N(d1) in the BSM pricing model. N(d1) usually is pretty close to N(d2) but not exact and deviates as time to expiration increases. Some sources say that N(d2), is actually the probability of the option expiring in the money.

However, if you look at the equation for N(d1), below, you'll see that it involves "r" which is the result of risk neutral pricing.

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A final source mentions that the above d1 equation, involving "r" is actually not accurate for the probability of an option expiring ITM. In fact, this source claims that "r" should be replaced by mu, or the mean return of the underlying. Also the subsequent + sign should be replaced by a - sign. Basically claims that we should examine probabilities in a risk natural world.

So now I am confused. What am I missing? If I really want to calculate the probability of an option finishing ITM, what equation should I use? Is every source right and there are just small caveats I am missing?



1 Answer 1


You got to be careful with $\mathbb{P}$ and $\mathbb{Q}$. Indeed, $N(d_2)$ is the probability of the event $\{S_T\geq K\}$ in the risk-neutral world. Note that $r$ (or $r-q$) is the drift in the risk-neutral world and hence this variable occurs in $d_2$. Since time to maturity and volatility are typically small numbers, i.e. $d_1=d_2+\sigma\sqrt{T-t}\approx d_2$, i.e. Delta approximates the ITM probability.

By the way, Delta may be seen as a probability as well: 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).

This is important: If you want to compute the probability of your stock being above a certain threshold $K$ on day $T$, then please don’t use any of these formulae!!! You could go back to $\mathbb{P}$ and replace $r$ by $\mu$ but you have at least two big problems:

1) how do you estimate $\mu$? There is low autocorrelation in log-returns and estimating the expected drift of a stock is quite difficult.

2) the formula is only true if the stock price follows a geometric Brownian motion but we have plenty of evidence that the real world is (much) more complicated: fatter tails, skews, stochastic volatility etc.

So, the Black Scholes model (and it’s related probabilities) is a good way of starting to learn about financial models but you should not apply them in real life, they are too simplified.

That being said, you can attempt to estimate real world probabilities, for instance one can get the distribution under $\mathbb{Q}$ from traded option prices (Breeden Litzenberger 1978) and then transform this distribution into a real world distribution, see Chapter 16 in Stephen Taylor’s book (Asset Price Dynamics, Volatility, and Prediction, 2005).

  • $\begingroup$ Thanks! So to clarify, we are talking about two different models here is that correct? Basically two different distributions, the P vs. the Q distributions. And if so, for more complicated option models like SV and SVJ, this is where the "market price of risk" comes in, to translate one world to the other? $\endgroup$
    – confused
    Jul 29, 2019 at 15:51
  • $\begingroup$ Yeah precisely. Unfortunately, we do not know the distribution of $(S_t)$ under $\mathbb{P}$. So you model them under $\mathbb{Q}$ (i.e. assume some model, BS, SV, SVJ, SVJJ, etc) and price derivatives using change of measure techniques. As you say, IF every real-world market participant was risk-neutral, $\mathbb{P}=\mathbb{Q}$ but, of course, people ask for an exyta price for bearing risk. Thus, the distributions of $\mathbb{Q}$ and $\mathbb{P}$ are different. $\endgroup$
    – Kevin
    Jul 29, 2019 at 16:33

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