Everyone says $N(d_2)$ is the probability of the option being exercised but stocks that have really high volatility have really expensive options indicating a high likelihood of expiring in the money. If volatility is very high we see that s $N(d_2)$ goes to 0, so doesn't that contradict our first statement? Isn't $N(d_1)$ the true probability?

  • $\begingroup$ Please add definitions for N, d_1 and d_2. What is sN(d_2)? What is a true probability? What kind of option are you talking about? $\endgroup$
    – user1157
    Jan 31 '14 at 17:20
  • $\begingroup$ Given the context of the question, N(.) is probably the CDF of the standard normal, with d1 and d2 defined as in the standard Black-Scholes formula. $\endgroup$
    – crcvd
    Jan 31 '14 at 20:36

This is indeed a very basic question. Please have a look at the following paper which will answer all your questions in a very intuitive and step-by-step fashion:

Understanding $N(d_1)$ and $N(d_2)$: Risk-Adjusted Probabilities in the Black-Scholes Model by Lars Tyge Nielsen


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