Nearest-grid probability in phase estimation

Put $n=\lfloor Q\theta \rfloor$ and $r=Q\theta -n$, so $n\in \{0,\dots ,Q-1\}$ and $0\le r<1$ by [F1]. If $r\le 1\mathrm{/}2$, choose $y=n$; if $r>1\mathrm{/}2$ and $n+1<Q$, choose $y=n+1$; if $r>1\mathrm{/}2$ and $n+1=Q$, choose $y=0$. The resulting circular distance is respectively $r\mathrm{/}Q$ or $(1-r)\mathrm{/}Q$, at most $1\mathrm{/}(2Q)$.

Every integer $k$ has a lift representation $k=y+sQ$ with $s\in \mathbb{Z}$ and $y\in \{0,\dots ,Q-1\}$: apply [F1] to $k\mathrm{/}Q$ and set $s=\lfloor k\mathrm{/}Q\rfloor$, $y=k-sQ$, so $0\le y<Q$. Thus grid labels and their lifts range over all integers, and minimizing circular phase distance is the same as choosing an integer nearest to $Q\theta =n+r$. If $r\le 1\mathrm{/}2$, every integer $k\le n$ is at distance at least $r$ and every $k\ge n+1$ is at distance at least $1-r\ge r$, so $n$ is nearest; when $r=1\mathrm{/}2$, $n+1$ is also nearest. If $r>1\mathrm{/}2$, every integer $k\le n$ is at distance at least $r>1-r$ and every $k\ge n+1$ is at distance at least $1-r$, so $n+1$ is nearest. If a nearest integer is $Q$, its label is $0$: this is the wrap-around choice in step 1.1 when $r>1\mathrm{/}2$, and the tied wrap-around label when $r=1\mathrm{/}2$. Thus every nearest label has distance $d\le 1\mathrm{/}(2Q)$.

Fix a nearest label and set $x=\theta -y\mathrm{/}Q$, $d=d_{\mathbb{T}}(x)$. Since $x\in (-1,1)$, the definition of $d$ gives an integer $\mathrm{\ell }\in \{-1,0,1\}$ and sign $\epsilon \in \{-1,1\}$ with $x=\mathrm{\ell }+\epsilon d$. By [F6], $\exp (2\pi ix)=\exp (2\pi i\epsilon d)$. If $\epsilon =-1$, Euler's formula and the evenness of cosine and oddness of sine in [F8] give $\exp (-2\pi id)=\overline{\exp (2\pi id)}$, so [F7] shows its distance from $1$ has the same modulus as for $d$. Euler's formula gives $1-\exp (2\pi id)=\exp (\pi id)(\exp (-\pi id)-\exp (\pi id))=-2i\exp (\pi id)\sin (\pi d)$. Since $\mathrm{\mid }\exp (\pi id)\mathrm{\mid }=1$ and $\sin (\pi d)\ge 2d\ge 0$ by [F5], $\mathrm{\mid }1-\exp (2\pi ix)\mathrm{\mid }=2\sin (\pi d)$. The same calculation with $Qx=Q\mathrm{\ell }+\epsilon Qd$ gives $\mathrm{\mid }1-\exp (2\pi iQx)\mathrm{\mid }=2\sin (\pi Qd)$ when $d>0$, because $Qd\le 1\mathrm{/}2$.

When $d>0$, [F4] and step 3.1 yield $P_{\theta }(y)=Q^{-2}(\sin (\pi Qd)\mathrm{/}\sin (\pi d){)}^2$. If $d=0$, then $x=0$ and [F3] gives $P_{\theta }(y)=1$.

If $d>0$, step 2.1 gives $0<Qd\le 1\mathrm{/}2$. By [F5], $\sin (\pi Qd)\ge 2Qd$ and $0<\sin (\pi d)\le \pi d$; hence step 4.1 gives $P_{\theta }(y)\ge Q^{-2}(2Qd\mathrm{/}(\pi d){)}^2=4\mathrm{/}{\pi }^2$. If $d=0$, step 4.1 gives $P_{\theta }(y)=1\ge 4\mathrm{/}{\pi }^2$. Thus every nearest label, including either label at a tie, has the claimed probability.