# Finite-Key Analysis of 1-Decoy Method Quantum Key Distribution with Intensity Fluctuation
## 2 The Finite-Key Analysis of 1-Decoy with Intensity Fluctuations
 \tag{16}\]  
In the 2-decoy method, three intensities (i.e., signal states, weak states and vacuum states) are needed to bound the vacuum events. For the 1-decoy method, there are two approaches to obtain the upper bound of the vacuum events. In this paper, we choose the latter, which provides the best secret key rate [15]. Let \(m_{\alpha,n}\) be the number of errors detected at [PERSON]'s side when [PERSON] sent n-photon states. We can obtain the upper bound by considering only the errors relative to the 1-decoy state:  
\[M_{Z,\mu_{i}}^{*}=\sum_{n=0}^{\infty}p_{\mu_{i};n}m_{Z,n}=\sum_{n=0}^{\infty} \frac{p_{\mu_{i}}e^{-\mu_{i}}\mu_{i}^{\,n}}{\tau_{n}n!}m_{Z,n}\geq\frac{p_{\mu_ {i}}e^{-\mu_{i}}m_{Z,0}}{\tau_{0}}\geq\frac{p_{\mu_{i}}e^{-\mu_{i}^{+}}m_{Z,0}} {\tau_{0}^{+}}. \tag{17}\]  
Therefore, we can obtain the upper bound of \(m_{Z,0}\),  
\[m_{Z,0}\leq\frac{\tau_{0}^{+}e^{\mu_{i}^{+}}M_{Z,\mu_{i}}^{*}}{p_{\mu_{i}}}=g( \mu_{i}). \tag{18}\]  
Since there are two intensities in the 1-decoy method, we can obtain a tighter upper bound as  
\[m_{Z,0}\leq min(g(\mu_{1}),g(\mu_{2}))=m_{Z,0}^{L}. \tag{19}\]  
As the detectors respond to the vacuum events \((s_{Z,0})\) randomly, no information can be extracted from these events by [PERSON], and [PERSON] also has only 50% probability of correct detection. Therefore, the expectation value of \(m_{Z,0}\) should be half of the corresponding total events:\[\frac{m_{Z,0}}{s_{Z,0}}=\frac{1}{2}. \tag{20}\]  
In this way, we have the relation:  
\[s_{Z,0}\leq s_{Z,0}^{U}=2(m_{Z,0}+\delta_{A}(N_{Z},\epsilon_{A1})). \tag{21}\]  
Therefore, combined with finite-key analysis, we can obtain the lower bound of \(s_{Z,1}\)  
\[ \begin{split} s_{Z,1}\geq\frac{\tau_{1}^{+}\tau_{1}^{-}\mu_{2}^{ +}}{\mu_{1}^{-}(\mu_{2}^{+}\tau_{1}^{-}-\mu_{1}^{-}\tau_{1}^{+})+(\mu_{2}^{+} )^{2}(\tau_{1}^{+}-\tau_{1}^{-})}(\frac{e^{\mu_{1}^{+}}N_{Z,\mu_{1}}^{*LI}}{p_ {\mu_{1}}}-\frac{e^{\mu_{2}^{-}}N_{Z,\mu_{2}}^{*L}}{p_{\mu_{2}}}\\ -\frac{(\left(\mu_{1}^{-}\right)^{2}-\left(\mu_{2}^{+}\right)^{2} )}{\left(\mu_{2}^{+}\right)^{2}}(\frac{e^{\mu_{2}^{+}}N_{Z,\mu_{2}}^{*L}}{p_ {\mu_{2}}}-\frac{e^{\mu_{2}^{-}}}{\tau_{0}^{-}})).\end{split}  \tag{22}\]  
By multiplying Equation (11) by \(\mu_{2}^{+}\) and Equation (12) by \(-\mu_{1}^{-}\) and adding two inequalities, we can obtain  
\[ \begin{split}\frac{\mu_{2}^{+}e^{\mu_{1}^{+}}N_{Z,\mu_{1}}^{*}} {p_{\mu_{1}}}-\frac{\mu_{1}^{-}e^{\mu_{2}^{-}}N_{Z,\mu_{2}}^{*}}{p_{\mu_{2}}} \geq\frac{(\mu_{2}^{+}-\mu_{1}^{-})s_{Z,0}}{\tau_{0}}+\mu_{2}^{+} \mu_{1}^{-}\sum_{n=2}^{\infty}\frac{\left(\left(\mu_{1}^{-}\right)^{n-1}- \left(\mu_{2}^{+}\right)^{n-1}\right)s_{Z,n}}{\tau_{n}n!}\\ \geq\frac{(\mu_{2}^{+}-\mu_{1}^{-})s_{Z,0}}{\tau_{0}}.\end{split}  \tag{23}\]  
Since \(\mu_{1}^{-}\geq\mu_{2}^{+}\), we can obtain  
\[s_{Z,0}\geq s_{Z,0}^{L}=\frac{\tau_{0}^{-}}{\mu_{2}^{+}-\mu_{1}^{-}}(\frac{ \mu_{2}^{+}e^{\mu_{1}^{+}}N_{Z,\mu_{1}}^{*U}}{p_{\mu_{1}}}-\frac{\mu_{1}^{-}e ^{\mu_{2}^{-}}N_{Z,\mu_{2}}^{*L}}{p_{\mu_{2}}}). \tag{24}\]  
Then, we need to estimate the phase error in the \(Z\) basis by following formula [21; 29]:  
\[\phi_{Z}:=\frac{c_{Z,1}}{s_{Z,1}}\leq\frac{m_{X,1}^{U}}{s_{X,1}^{L }}+f\Bigg{(}\varepsilon_{sec},\frac{m_{X,1}^{U}}{s_{X,1}^{L}},s_{Z,1}^{L},s_{ X,1}^{L}\Bigg{)}, \tag{25}\] \[f(a,b,c,d)=\sqrt{\frac{(c+d)(1-b)b}{cdln2}}log_{2}(\frac{(c+d)21 ^{2}}{cd(1-b)ba^{2}}), \tag{26}\]  
where \(c_{Z,1}\) is the number of phase errors in the single-photon events when in the \(Z\) basis. Here, we estimate the phase error under \(Z\) basis by the bit error under \(X\) basis. For \(s_{X,1}^{L}\), we can use the same method as in \(s_{Z,1}\) for calculation. In addition, \(m_{X,1}\) can be bounded as in the following formulas.