Questioning the meaning of Delta Delta for Calls is in the range 0 to 1, and for Puts in the range 0 to -1. The abs(Delta) is the probability that the option closes ITM (usually multiplied by 100 to get a range of 0% to 100%). See also https://en.wikipedia.org/wiki/Greeks_(finance)#As_a_proxy_for_probability BUT, take a look at this BSM calculation: S=100, K=100, s=0.3, t=1, r=0, q=0 : C: Value=11.923538 Delta=0.559618 P: Value=11.923538 Delta=-0.440382 It means the Call option has about 56% chance to close ITM, whereas the Put option has about 44% to close ITM. How come? Can this be right at all? Shouldn't both be 50%? Is Delta itself wrong, or is its interpretation wrong, or are both 'kinda wronga'?
Nope, they should not. In a lognormal world the stock can go to infinity yet it can't fall below zero - I am pretty sure you can see how that's reflected by the delta. As a take-home assignment, try shocking vol to 300% and see what happens to deltas for the put and the call. PS. Also, while delta is a proxy for probability, it's not exactly probability
I understand, but even in the lognormal-world it has to be 50% for both, IMO. Vola 300%: C: Value=86.638560 Delta=0.933193 P: Value=86.638560 Delta=-0.066807 Hmm. yes, as said, really very funny. IMO this can't be the probability for closing ITM at exp date. So, what is it? Probability or not?
Well, think of it this way - if volatility is very high and an asset can not go below zero, than the asset is "cheap" and it's more likely to be above the current level at time T. As I said, delta is a proxy, while N(d2) is the actual risk neutral probability.
This diagram would be better if peaks were aligned. You will see a similar 50/50 distribution in a small interval that bisects the top of the curves. Mostly ATM and near ATM. A higher volatility which flattens the curves will mildly broaden the tops but also magnify the effect of the skew.
Using this lognormal formula S_t = S * e^(z * s * sqrt(t)) one can calculate the upper and lower bounds for 1 SD: S=100, s=3.0, t=1 : The upper 1 SD (ie. z=1) price: 2008.55 The lower 1 SD (ie. z=-1) price: 4.98 But they are both equal distance (ie. 1 SD) from the mean S=100. Therefore the probability for both has to be equal, ie. 50%. Is maybe the Black-Scholes formula wrong? I know, millions of people have studied it and using it daily, but does there exist a proof for its correctness? What is the easiest method to prove its correctness?
Probably by getting a PhD in Applied Mathematics. It is a well-trodden path but all the kids want Pure Mathematics nowadays. You will calculate the very similar and elementary Heat Equation in your sleep. Then, you can translate your closed form solutions to Numerical Methods in your computer programs.
I think there could be a better formula out there to be discovered, which also would allow to compute the IV directly (without iterating)...
Well, you are missing an vol drift correction term if you really want to simulate where the asset is going to end up and yes, that's where the difference between N(d1) and N(d2) comes into play, as d2 includes the vol drift. Derive the option pricing from BSM differential equation, if you so desire. Or you can go further and derive the BSM PDE from the basics, i.e. Ito lemma etc (several ways to do that). A light version would be to derive BS option pricing from a binomial tree. PS. BS option pricing has some assumptions that are questionable from the market perspective (and the market corrects for it), but the actual mathematics are correct.
I've intentionally omitted the drift (the risk-free-rate) and dividend when I set them r=q=0 in the initial posting to keep it simple.