## 4 Theorems and Proofs
: Let \(U_{j}=\sum_{i=0}^{d-1}U_{j,i}\) from (100), \(C\left(X_{j}\right)=\sum_{i=0}^{d-1}C\left(X_{j,i}\right)\) and \(C\left(\Delta_{j}\right)=\sum_{i=0}^{d-1}C\left(\Delta_{j,i}\right)\). Exploiting the result of Theorem 1, in the scalar reconciliation process [PERSON] decides on the scalar quantity \(U_{j}^{\prime}=a\), if:  
\[\Pr\left(U_{j}=a\big{|}U_{j}^{\prime}\right)\geq\Pr\left(U_{j}=b\big{|}U_{j}^ {\prime}\right). \tag{108}\]  
Similarly, she decides on \(U_{j}^{\prime}=b\), if:  
\[\Pr\left(U_{j}=b\big{|}U_{j}^{\prime}\right)\geq\Pr\left(U_{j}=a\big{|}U_{j}^ {\prime}\right). \tag{109}\]  
Conditioned on \(a\) or \(b\), the received \(U_{j}^{\prime}\) has mean \(\mu_{a}=a\) or \(\mu_{b}=b\), with \(\mathbb{N}\left(\mu_{a},\eta^{2}\right)\) and \(\mathbb{N}\left(\mu_{b},\eta^{2}\right)\). Applying the maximum-likelihood-based correction rule [15; 16; 17; 18; 19], [PERSON] calculates with the following inequalities:  
\[\frac{1}{\sqrt{2\pi\eta^{2}}}e^{\left(-\frac{\left(U_{j}^{\prime}-a\right)^{2} }{2\eta^{2}}\right)}\geq\frac{1}{\sqrt{2\pi\eta^{2}}}e^{\left(-\frac{\left(U_{ j}^{\prime}-b\right)^{2}}{2\eta^{2}}\right)} \tag{110}\]  
and:  
\[\frac{1}{\sqrt{2\pi\eta^{2}}}e^{\left(-\frac{\left(U_{j}^{\prime}-b\right)^{2} }{2\eta^{2}}\right)}\geq\frac{1}{\sqrt{2\pi\eta^{2}}}e^{\left(-\frac{\left(U_{ j}^{\prime}-a\right)^{2}}{2\eta^{2}}\right)}, \tag{111}\]  
which then leads to (for a comparison see (77) and (78)):The received \(U^{\prime}_{j}\) has mean \(\mu_{a}=a\) or \(\mu_{b}=b\), hence one obtains the following conditional probability for an error event, conditioned on [PERSON] has sent \(U_{j}=a\):  
\[\Pr\!\left(U^{\prime}_{j}=\frac{U_{j}}{C\big{(}X_{j}\big{)}}C\big{(}\Delta_{j} \big{)}<\frac{\mu_{a}+\mu_{b}}{2}\bigg{|}U_{j}=a\right)=\Pr\!\left(\big{(}U^{ \prime}_{j}-U_{j}\big{)}>\frac{\big{|}\mu_{a}-\mu_{b}\big{|}}{2}\right), \tag{114}\]  
where \(\frac{\big{|}\mu_{a}-\mu_{b}\big{|}}{2}\) assigns a decision boundary. The tail function \(Q\!\left(\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\right)=\Pr\!\left(\frac{ \big{|}a-b\big{|}}{2}\frac{1}{\eta}<g\right)\), where \(g\in\mathbb{N}(0,1)\), has exponential decay for any \(|a-b|>2\eta\), hence:  
\[\frac{1}{\sqrt{2\pi}\Big{(}\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\Big{)}} \left(1-\frac{1}{\Big{(}\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\Big{)}^{2} }\right)e^{-\frac{\big{|}\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\big{)}^{2} }{2}}<Q\!\left(\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\right)<e^{-\frac{ \big{(}\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\big{)}^{2}}{2}}, \tag{115}\]  
which clearly demonstrates that the error probability of scalar reconciliation exponentially converges to zero. As one can readily obtain from (115), for arbitrary large differences between \(a\) and \(b\), \(Q\!\left(\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\right)\to 0\)[15; 16; 17]. Then, by applying the maximum-likelihood decision theory and the Bayes' rule [15; 16; 17; 18; 19], for a given \(U_{j}\), one obtains error probability via the tail function:  
\[ \begin{array}{ll}\Pr\!\left(U^{\prime}_{j}<\frac{\mu_{a}+\mu_{b}}{2}\Big{|} U_{j}=a\right)&=Q\!\left(\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}\right)\\ &=\Pr\!\left(\frac{\big{|}a-b\big{|}}{2}\frac{1}{\eta}<g\right)\\ &=\Pr\!\left(error\right),\end{array}  \tag{116}\]  
where \(g\in\mathbb{N}(0,1)\) is a standard Gaussian random variable such that \(Q(x)=\Pr(x<g)\), which clearly demonstrates that \(\Pr(error)\) depends only on the distance \(|a-b|\) of \(a\) and \(b\). The exponential decay of \(\Pr(error)\) is depicted in Figure 5.  
The condition \(|a-b|>2\eta\) can be trivially satisfied by the parties in any practical CVQKD scenario; the proposed results complete the proof.