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?
1
-
$\begingroup$ 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. $\endgroup$ – Corcovado Jan 31 '14 at 20:36
2
$\begingroup$
$\endgroup$
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:
