Thanks for that f/back! I myself can't intuit an explanation for d1, d2 since they follow from a differential. But, collectively, the N(d1) and N(d2) are probability-adjusting the minimum value, where minimum value (MV) = Stock - (Strike)EXP[(-rate)(t)]. That MV = Black-Scholes value in the special case where the volatility is zero [if vol = 0, then both N(d1) and N(d2) = 1]. Now, as we "add volatility" the spread between N(1) and N(2) increases (i.e., N(1) - N(2) which I like to think of as a volatility wrapper around the MV) such that the BS value increases with greater volatility.
If you have the John Hull text, he provides another interpretation, based on rearranging (see Chapter 13). e.g., N(d2) is the probability the option will be struck.
But, for FRM purposes, our more relevant point is probably that N(d1) is the call option's delta (for a Euro call) and N(d1) - 1 is the put option's delta (for Euro put). (for detail, see John Hull p. 346, Options 6th edition).
It won't give intuition for d1, d2 but it is the best i've been able to do concerning an intuition. Taking Hull's example (p. 296) chapter 13, it shows the BS formula for a given call option but at different volatility levels; e.g., 0, 10%, 20%, ...
You can see at almost 0 volatility, the BS = the minimum value. That would be like purchasing the share with borrowed cash (i.e., long share + short bond, bond = discounted strike price). If the stock grows at riskless rate, the Min Value is the Present Value of the payoff if the stock indeed grows at the riskless rate. And, here is the key idea, this is the (dynamic) hedge underling the formula. So, as you look down the table (as the volatilities go up), note the "spread" between N(d1) and N(d2) is increasing. This is "plussing up" the value; in fact, what this means refers the same dynamic hedge. At 50% vol, for example, the hedge is long N(d1) = 0.68 shares and short N(d2) = 0.54 bond units. Higher volatility is requiring less bond units to maintain a hedge.
I hope that sheds additional light, but I'm always interested in better ways to intuit the formula since I think few of us really manage that!
I was looking at the connection among the option delta, N(d1) and N(d2). Is the connection "exact"? i.e. if the delta of a call option = 0.25, does that mean N(d1) = 0.25? Also, we know the delta of a put option is always negative. The question is, if N(d2) = 1 - N(d1) and the maximum value that N(d1) can assume in any situation is 1, how do we get the delta of the put option being negative? What am I missing?
* For a non-dividend Euro call, yes, the relationship is exact. That's because if we take the 1st (partial) derivative of the Black-Scholes with respect to the stock price (i.e., what's the small change in call given small change in stock), we are solving for the option delta (by definition) and it reduces mathematically to N(d1).
* I goofed above (I just fixed it), sorry to mislead. The delta of non-dividend Euro is N(d1) - 1. So, your confusion was warranted!
Hi, sorry I have an unrelated question about N(d1) and N(d2).
I am using the Black Scholes model to try and value some performance rights. As such, they will have an exercise price of zero.
Is it true to say that N(d1) reflects the cumulative probability related to the current value of the financial security (between 0 and 1), and N(d2) reflects the cumulative probability related to the exercise price of the financial security (between 0 and 1)?
If so, then in the case of the performance rights with zero exercise price, both N(d1) and N(d2) should have values of zero, shouldn't they?
If this is the case, the value then becomes C = S doesn't it? Because the multiplier is 1, and the present value of the exercise price is zero.
I've been plugging the data into a valuation model in excel and the value comes out as being less than the spot price. Also, the value is entirely independent of the level of volatility!
Any advice or assistance would be greatly appreciated.
These must be calls-yes?
If so, your model must be wrong: N(d1) is the option's delta. If strike near 0, these rights issue deep "in the money" (analogous to restricted stock grants) and delta is near to 1.0 not 0.0 (0 if they were deeply out of the money).
Both N(d1) and d(2) should approximate 1.0. You might check to see if somehow you are reversing stock/strike and inputing a deeply out of money call? That would get your zeros and a model that it is insensitive to volatility changes
That's absolutely correct, in the extreme you are left with (S)[N(d1)] - Discounted[exercise]. And if strike = 0, then just S*N(d1).
The impact of volatility is vega (change in option given change in volatility). Vega approaches zero as option goes, in both directions, in and out-of-the money.
Intuitively, if the option is far out of the money, adding volatility just doesn't help you in any reasonable timeframe (e.g., 1 year). If your option is deep in the money, you are near the value of the underlying (the share itself) and the option can't get be worth more than it's underlying - you are already at the upper bound.
if you want to attach the XLS, i'll take a look..thanks, David
Dave--doing research on N(d1) and N(d2) and ran across this thread... do you happen to know of a formula that estimates the probability that the underlying stock price will reach the strike price at any point during the term (as opposed to N(d2), which I believe estimates the probability at the end of the term)? Not N(d1) is it?
I don't. Maybe such a closed-form formula exist, but I have not seen it nor can I reason to it. You are correct that N(d1) is not a probability, it is delta (for a Euro call on non-dividend paying stock). And yes it is true also that N(d2) is the probability that the European option EXPIRES in the money. I only have two thoughts:
* I wonder if it might be somehow embedded in the formula for a knock-in barrier option where barrier = strike. Under this assumption, you are looking for the probability that the option is knocked-in?
* This is a natural application for the binomial - build the tree and literally add up the probabilities to all paths that cross the "in the money" barrier. It's numerical but seems straightforward.
Hi, Could you give me a hint how to prove this equation? Where P stands for put option, not probability. Does it involves proving this equation with integrals, such in the video below or the appendix of the book of John Hull, chapter 14 8th edition or chapter 15 newer edition?