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

Question

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 1

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.