# Black model: Delta - strike relationship regardless of expiry?

While wandering through some QuantLib experimental classes for FX trading, I've found this Black Delta Calculator.

By reading its .cpp, it seems that no use of options time to expiry is made at all.

Usual Black, Scholes & Merton $d_{1}$ argument of cumulative Normal distribution has time to expiry as argument: on the contrary, Black Delta Calculator makes a weird use of $d_{1}$ and $d_{2}$, expressing them like at line 222 or 250:

d1_ = std::log(forward_/strike)/stdDev_ + 0.5*stdDev_; // .cpp line # 222
d2_ = std::log(forward_/strike)/stdDev_ - 0.5*stdDev_; // .cpp line # 250


This is quite different than what usual Itö correction produces over classic GBM process (squared variance and of course time adjustment for annual basis).

By reading that code it seems that going from Delta to strike and back can be made regardless of expiry date, while common sense says that there are a lot of options whose Delta can match any strike if you can search all over implied volatility surface without an expiry boundary.

Questions

• It must be I am missing some important relations which allow for such a simplification: could you show me which one?
• Is this possible relation viable just for FX options or can it be extended to any use of Black model (e.g. interest rates)?
• Time to expiry is already included in the inputs (the two discounts $e^{-rt}$ and $e^{-qt}$ and the standard deviation $\sigma \sqrt{t}$). – Luigi Ballabio Mar 31 '15 at 8:59
• You should give it an answer, Luigi... something like: «As far as I know, there's no way to go straight from Delta to strike without knowing $T$... and you're a lazy ass who doesn't read my codes from the beginning» ...Ahahah, always thank you for your help :) – Lisa Ann Mar 31 '15 at 9:39
• Ok. I'll skip the lazy ass part, though. :) – Luigi Ballabio Mar 31 '15 at 9:42

The time to expiry is required, but it's included in the inputs: the two discounts $e^{-rT}$ and $e^{-qT}$ and the standard deviation $\sigma\sqrt{T}$. You might argue it could be documented more clearly, and I might agree with you.