= Solution
Let $q(T)$ denote the original <zero-coupon bond> price $P(0,T)$ computed above. A deterministic shift $h_t$ of the <short rate>, with $h_0=0$, gives $\widehat r_t=r_t+h_t$. Subtracting the two recursions shows that the required added drift is
$$
\alpha_t=h_t-\beta h_{t-1}.
$$
Since the shift is nonrandom, the new <discount factor> differs by $\exp(-\sum_{s=1}^Th_s)$, and hence $\widehat P(0,T)=q(T)\exp(-\sum_{s=1}^Th_s)$.
For <deterministic calibration of an autoregressive short rate>, define
$$
d_T=\log q(T)-\log p(T),\qquad d_0=0,\qquad h_T=d_T-d_{T-1}\quad(T\geq1).
$$
All these quantities are finite. The telescoping sum $\sum_{s=1}^Th_s=d_T$ gives $\widehat P(0,T)=p(T)$. Explicitly,
$$
\boxed{\alpha_1=d_1,\qquad
\alpha_T=d_T-(1+\beta)d_{T-1}+\beta d_{T-2}\quad(T\geq2).}
$$
These constants fit every maturity simultaneously. Each successive maturity fixes $h_T$ uniquely, so the deterministic calibration is also unique with the prescribed $r_0$.
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