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When the Black scholes formula is derived, $T$ is just some time in the future. We don't specify what it is.

So why is it that if you go to an option pricing calculator, it asks specifically for days until expiration? When these calculators then use the formula, do they just plug in these number of days remaining for $T$?

If they do this, then what if instead of saying 10 days, I said 1440 hours. That's still 10 days, yet plugging in 1440 for T obviously gives a different answer than if we plugged in 10 .....

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  • $\begingroup$ T = 1 most often denotes a year in the literature, and we often assume 252 trading days for a year. Thus a 10 day expiry is achieved by T = 10/252. I'm not sure that 1440 hours is 10 days. $\endgroup$ – Forgottenscience Feb 8 '17 at 19:26
  • $\begingroup$ 10*24 = 240. Not sure why I wrote 1440, maybe I was thinking of 60 days and wrote 10. $\endgroup$ – Morgenman Feb 8 '17 at 19:41
  • $\begingroup$ Don't worry, I make this mistake all the time... Welcome to QuantStackexchange! $\endgroup$ – noob2 Feb 8 '17 at 20:04
  • $\begingroup$ @noob2 your comment is an answer. Please post it as such. $\endgroup$ – SRKX Feb 9 '17 at 1:44
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You describe an online calculator that takes input in days. Days is just convenient for some people, although I have seen other calculators using other units for time to maturity. Internally the calculator is translating the number of days into years before plugging it into the BSM formula. That is because mathematically the units used in the formula have to be consistent for 3 variables $r$,$\sigma$ and $T$. It just turns out $r$ and $\sigma$ are usually expressed on a yearly basis.

I would add that the calculators used professionally allow you to enter exact dates (e.g. 2017 02 08 to 2017 05 19)(maybe even the hour) rather than days or years when specifying $T$.

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  • $\begingroup$ The units can be anything, as long as they are consistent. If you use $T$ in years, you must use $r$ in years and $\sigma$ in years. As an example, imagine $r=0$, $T=1$ year and $\sigma=0.20$ standard deviations per year. If you wanted to use months as the unit, then you would need everything in units per month. In this case, you would have $T=12$ months (1 year), and $\sigma=0.2/\sqrt{12}$ (standard deviations per month). Both should give you the same result. $\endgroup$ – Guilherme Salomé Jan 10 '18 at 20:34

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