 Solution

Let $F_0(T)$ be the observed initial instantaneous forward rate curve. Since ${\hat{W}}_0=0$, part (i) gives $f_0(T)=g(T)-{\sigma }^2{T}^2/2$. Thus the forward-curve calibration of a shifted Brownian short rate is achieved by the deterministic choiceSubstitution gives $f_0(T)=F_0(T)$ at every maturity where the curve is defined classically. It also reproduces the corresponding initial zero-coupon bond curve:There is no shape restriction on the initial curve beyond the regularity needed to define these integrals and forward derivatives; for continuous curves the match is pointwise, and for locally integrable curves the forward identity has its almost-everywhere interpretation.