Stochastic Short Rates and Forward Rates Under the Forward Measure

Summary

The document explains why an instantaneous forward rate is generally not the ordinary risk-neutral expectation of a future short rate when rates are stochastic. It corrects a derivation that differentiates the expected discount factor as though the result were an unweighted expected rate. Differentiating the bond price and dividing by that price instead produces an expectation weighted by the stochastic discount factor.

That weighting corresponds to expectation under the maturity-specific forward measure, whose numeraire is the zero-coupon bond maturing at the forward rate’s horizon. The answer gives the measure change explicitly and observes that when rates are deterministic, the change of measure is trivial, so the forward rate equals the future short rate. The explanation assumes the relevant derivatives and expectations can be interchanged and focuses on the pricing identity rather than the dynamics or calibration of a particular interest-rate model.

Key ideas

• With stochastic short rates, the instantaneous forward rate is not generally the risk-neutral expected future short rate.
• Differentiating the bond price gives a discount-factor-weighted expectation divided by the bond price.
• This weighted expectation is taken under the maturity-specific forward measure.
• When rates are deterministic, the forward and risk-neutral measures coincide, recovering the deterministic rate identity.

Tags

# why $f(t,u) \neq E_t^Q [r(u)]$ when $r$ is random?


# why $f(t,u) \neq E_t^Q [r(u)]$ when $r$ is random?












If I suppose the short rate $r$ deterministic, and the risk neutral measure $Q$, I can write the following :

$$f(t,u) = -\frac{d}{du}\ln P(t,u) = -\frac{d}{du} E_t^Q \left[ e^{-\int_t^{u}r_sds} \right] = E_t^Q \left[ \frac{d}{du} \int_t^{u}r_sds \right] = E_t^Q \left[ \frac{d}{du} (R_u - R_t) \right] = E_t^Q [r_u]$$

with $f$ the instantaneous forward rate and $P$ the price of a zero coupon bond.

Now I wonder, which one of the equalities here that doesn't hold when $r$ is a random process? Any help please?

## Answer by Antoine Conze (score 5, accepted)

https://quant.stackexchange.com/a/43845

Your equations are flawed. Also there is no expectation if the process $\{r_s\}$ is deterministic.

The correct reasoning is, assuming $\{r_s\}$ is stochastic: $$ f(t,u)=-\frac{d}{du}\ln P(t,u)=-\frac{\frac{d}{du}P(t,u)}{P(t,u)}\\ =-\frac{\frac{d}{du}E^Q_t[e^{-\int_t^u r_s ds}]}{P(t,u)} =\frac{E^Q_t[e^{-\int_t^u r_s ds} r_u]}{P(t,u)} =E^Q_t\left[\frac{e^{-\int_t^u r_s ds}}{P(t,u)} r_u\right]\\ =E^{Q^u}_t[r_u] $$ where $Q^u$ is the $u$-forward measure (the measure associated with $P(.,u)$ as numeraire) defined as $$ \frac{dQ^u}{dQ}=\frac{e^{-\int_t^u r_s ds}}{P(t,u)} $$

If $\{r_s\}$ is deterministic then $\frac{dQ^u}{dQ}=1$, i.e. the two measures are identical.

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