# What is the Rho of an option on a futures contract priced using the Black 76 model?

I wanted to quickly confirm some simple calculations for the Black 76 greeks and was making use of the formulas on this website:

http://riskencyclopedia.com/articles/black_1976/

I have an issue with their statement of Rho and I wanted to check with you guys. Their version for a call has:

$$\rho = \tau e^{-r\tau}K\Phi(d_2)$$

Whereas I feel it should be:

$$\rho = \tau e^{-r\tau}(K\Phi(d_2) - F \Phi(d_1))$$

i.e.

$$\rho = - \tau C$$

Where $C$ is the Black76 call option price:

$$C = e^{-r\tau}(F \Phi(d_1) - K\Phi(d_2))$$

Would really appreciate your input as the one person to post on my comment disagrees with me!

I agree with you. In the setting of Black 76 we consider an option on a forward/futures price $F$. $F$ is modelled as geometric Brownian motion and in contrast to the usual BM model the futures price does not grow with the risk-less rate.
Then you formula for the call (C) is correct. Note that in general neither $F$, $d_1$ nor $d_2$ contain the term $r$. Then taking the derivative w.r.t. $r$ we arrive at your formula.
In the case where the forward price can be found by cost-of-carry pricing (e.g. options on equity indices things simplify because then $$F_t = S_0 \exp((r-q)t),$$ which simplifies further if $q=0$ Then the $r$ terms cancles out but then in fact we are back in the BS model. Thus I would say: if cost-of-carry in the sense of the above formula holds, then Black-Scholes and Black 76 are just the same thing.