Does the Black (1976) model require the existence of the relation $F(t,T)=S(t)e^{r(T−t)}$?

I studied the derivation of the Black-Scholes formula. However, although I know the Black formula, I've never studied its entire derivation process. And probably the easiest way to justify the formula is using that relationship (something that motivates the previous question).


1 Answer 1


No, the Black model does not require this relationship to hold. For example, futures on currencies exhibit a different relationship between the future and spot price because of the interest debit/credit nature of currency borrowed/lent. However for the Black model on European futures options to hold the following condition has to be met:

The product of the asset price probability distribution and the pricing kernel has to be log-normal.

It can be shown that the asset price probability distribution does not necessarily have to be log-normal but the above has to hold to price, for example specific stock options (such as European spot options), bond options (where bond prices are log-normal), and some of the interest rate options (where such interest rates follow a MSS-BGM process). The Libor Market Model (LMM) is actually just a special case of the Black model.

  • $\begingroup$ when you mention the asset price probability distribution, do you mean, in this case, the underlying futures contracts or the the underlying asset of the futures?? $\endgroup$ Commented Aug 16, 2013 at 11:44
  • $\begingroup$ @JoaoSerafim, I meant the asset on which the option is written, which could be a foreign exchange rate, a futures price or an interest rate,...please note that the risk-adjusted probability distribution of the above mentioned product has to be lognormal. $\endgroup$
    – Matt Wolf
    Commented Aug 16, 2013 at 14:00
  • $\begingroup$ I see. I have another question relted to this subject. Let's suppose we have a future contract F in a market where the relation F(t,T)=S(t)er(T−t) doesn't hold. What are the the boundary conditions for the drivation of the Black formula?? $\endgroup$ Commented Aug 19, 2013 at 14:17
  • $\begingroup$ I've created a new post for this question. $\endgroup$ Commented Aug 20, 2013 at 4:31

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