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Futures Option Rho Includes Forward Rate Sensitivity

Article Quant Q&A · Author: Xiaohuolong

Summary

The question examines why the interest-rate sensitivity of a European option on futures can differ from the rho formula for an option on an asset with a continuous yield. The explanation identifies two rate exposures: discounting the expected payoff and the interest-rate dependence of the forward or futures price used to value the option.

It applies the chain rule to option value as a function of both the rate and the forward price. The direct rate derivative captures discounting, while the forward-price derivative, multiplied by the forward’s rate sensitivity, captures the financing effect associated with replicating the option. This resolves the apparent mismatch when the underlying forward is treated as rate-dependent. The note gives a conceptual decomposition rather than a numerical example, and its formulas presume the stated option-pricing setup.

Key ideas

  • Option rho can include both discounting effects and changes in the forward price.
  • The chain rule separates direct rate sensitivity from sensitivity transmitted through the forward.
  • The option’s forward-price sensitivity is its delta with respect to that forward.
  • Forward rate sensitivity reflects the financing cost embedded in the underlying forward price.

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Full text
# Hull's book - Futures option's rho


# Hull's book - Futures option's rho












In Hull's book (9th edition), on page 420, in table 19.6, it says rho of a European call on an asset with yield $q$ is $$KTe^{-rT}N(d_2)$$ Below it says we can compute greeks of European options on futures by setting $q=r$. But then it says the rho for a call futures option is $-cT$. I am a bit confused here. The price of a futures call option is $$c=e^{-rT}(F_0N(d_1)-KN(d_2))$$ where the futures price $F_0=e^{rT}S_0$ when there is no dividend and interest rate is constant. Then wouldn't the rho of this option be $$\frac{\partial c}{\partial r}=KTe^{-rT}N(d_2)$$ rather than $$-cT=KTe^{-rT}N(d_2)-TS_0e^{-rT}N(d_1)?$$ Are we not treating $F_0$ as a function of $r$? What am I missing here?

## Answer by Soumirai (score 1)

https://quant.stackexchange.com/a/60009

Your option has exposure to interest rates for two different reasons:

- The discounting of (expected) terminal payoff.

- The forward (~cost of financing of the delta hedging).

Mathematically, if you note your call as a function of rates and of the forward (itself a function of rates) $C(r, F(r))$, you have by chain rule:

$$\frac{dC}{dr} = \frac{\partial C}{\partial r} + \frac{\partial C}{\partial F}\frac{\partial F}{\partial r}$$

The first RHS term $\frac{\partial C}{\partial r}$ comes from discounting the expected terminal payoff (i.e. the call present value). The second RHS term comes from financing the call's replicating strategy. To replicate the call, you must hold a certain quantity of forward (which is the delta of the call $\frac{\partial C}{\partial F}$. And that forward has a sensitivity to interest rates $\frac{\partial F}{\partial r}$ which is the cost of funding the underlying asset.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.