[original post removed]

2018 Level II CFA curriculum, vol. 5, p. 406, §6.1:

Because:

Δ_{c} = e^{−δT}N(d_{1})

One reason is because there may not be enough time to get to the strike if you are far out of the money If X is 100, and S is 90 with one day to go to expiration, a $2 dollar movement in the stock probably won’t move the call option value very much, if at all. If you had a month to go, a $2 dollar movement int he stock may move the option more.

I am wondering why that is the case, as in why in any case would the call option’s delta increase as the option itself gets closer to maturity?

I understand how the call option changes with the change in stock price (since the call option probability to be exercises changes at different levels of the stock price). But I am wondering the same thing in the context of how the call option delta would increase as the call option moves closer to maturity.

Thanks,

The delta might decrease; it doesn’t have to increase.

Sorry, do not recall, but what is so unique about the formula:

e^{−δT}N(d_{1})?

I understand that N(d1) is the random component of the formula (for the random movement of the stock price) but how does this formulas as a whole relate to option delta that are in the money moving closer to matury (call option in this formula above)?

N(d1) isnt random… thats basically the call delta before the range calcuation… the e^{T} is your time component.

I already said this in my original post but here’s straight from the book. "When the option gets closer to maturity, the delta will drift either toward 0 if it is out of the money or drift toward 1 if it is in the money. "

This makes perfect sense because you’re running out of time, so if you’re out of of the money, your delta is going to dwindle to zero because it will likely not hit the strike. Opposite is true of you’re deep in the money.

There’s nothing unique about it. It simply shows that the delta is a function of the time left to maturity. As that time changes, the delta of the option changes.

(Note that d_{1} is also a function of time to maturity, but I didn’t want to add unnecessary complications to the explanation. What you see there is clearly a function of time to maturity.)

Ok it makes sense now.

Thanks for all the help so far!

My pleasure.