## Appendix H Bound on the Binomial Cumulative Distribution Function  
**Lemma A5** (see ([74] App. A)).: _Let \(0<\varepsilon<1\) and \(k<n\) with \(\frac{k}{n}<\varepsilon\). Then,_  
\[\sum_{j=0}^{k}\binom{n}{j}\varepsilon^{j}(1-\varepsilon)^{n-j}\leq\frac{ \varepsilon(n-k)}{\varepsilon n-k}\cdot 2^{n\left[H(\frac{k}{n})-T_{ \varepsilon}(\frac{k}{n})\right]}. \tag{105}\]  
Proof.: Let us define  
\[k^{\prime} \triangleq n-k,\] \[\varepsilon^{\prime} \triangleq 1-\varepsilon, \tag{106}\]  
i.e., \(k\leftrightarrow k^{\prime}\) and \(\varepsilon\leftrightarrow\varepsilon^{\prime}\) or equivalently  
\[k \leftrightarrow n-k,\] \[\varepsilon \leftrightarrow 1-\varepsilon. \tag{107}\]  
Now, observe that \(\frac{k}{n}>\varepsilon\Rightarrow\frac{k^{\prime}}{n}<\varepsilon^{\prime}\).  
Furthermore, by definition of the binary entropy function and its tangent line, we have  
\[H\bigg{(}\frac{k}{n}\bigg{)}=H\bigg{(}\frac{n-k}{n}\bigg{)}, \tag{108}\]  
and  
\[T_{\varepsilon}\bigg{(}\frac{k}{n}\bigg{)}=T_{1-\varepsilon}\bigg{(}\frac{n-k }{n}\bigg{)}, \tag{109}\]  
where (108) follows by (109) and (109) holds by (100).  
Now, applying the variable exchange of \(j\leftrightarrow n-j\) unto (110), we obtain  
\[\sum_{n-j=k}^{n-j-n}\binom{n}{n-j}\epsilon^{n-j}(1-\epsilon)^{n-(n-j)}\leq 2^{n \left[H(\frac{k}{n})-T_{\epsilon}(\frac{k}{n})\right]}\left[\frac{k(1-\epsilon)} {k(1-\epsilon)-(n-k)\epsilon}\right]. \tag{111}\]  
Observe that, since the index of sum in (110) runs form \(k\) to \(n\), i.e., \(k\leq j\leq n\), in the new system, we have \(k\leq n-j\leq n\), which is equivalent to \(0\leq j\leq n-k\). Further, the Binomial coefficient for \(0\leq j\leq n\) fulfills the subsequent identity:  
\[\binom{n}{n-j}=\binom{n}{j}, \tag{112}\]  
Thereby,  
\[\sum_{j=0}^{n-k}\binom{n}{j}\epsilon^{n-j}(1-\epsilon)^{j}\leq 2^{n \left[H(\frac{k}{n})-T_{\epsilon}(\frac{k}{n})\right]}\left[\frac{k(1- \epsilon)}{k(1-\epsilon)-(n-k)\epsilon}\right]. \tag{113}\]  
Now, applying the exchange of variables given in (108) unto (113), we obtain  
\[\sum_{j=0}^{k}\binom{n}{j}(1-\epsilon)^{n-j}\epsilon^{j} \leq 2^{n\left[H(\frac{n-k}{n})-T_{1-\epsilon}(\frac{n-k}{n}) \right]}\left[\frac{(n-k)\epsilon}{(n-k)\epsilon-k(1-\epsilon)}\right]\] \[=2^{n\left[H(\frac{k}{n})-T_{\epsilon}(\frac{k}{n})\right]} \left[\frac{(n-k)\epsilon}{(n-k)\epsilon-k(1-\epsilon)}\right], \tag{114}\]  
where the equality holds by (107) and (108). Therefore,  
\[\sum_{j=0}^{k}\binom{n}{j}(1-\epsilon)^{n-j}\epsilon^{j}\leq 2^{n \left[H(\frac{k}{n})-T_{\epsilon}(\frac{k}{n})\right]}\left[\frac{(n-k) \epsilon}{(n-k)\epsilon-k(1-\epsilon)}\right]. \tag{115}\]  
Now, we focus on the bracket in (114), which can be simplified as follows:  
\[\frac{(n-k)\epsilon}{(n-k)\epsilon-k(1-\epsilon)}=\frac{\left(\frac{n-k}{n} \right)\epsilon}{\left(\frac{n-k}{n}\right)\epsilon-\frac{k}{n}(1-\epsilon)} =\frac{\epsilon-\frac{k}{n}\epsilon}{\epsilon-\frac{k}{n}}=\frac{\epsilon(n-k )}{\epsilon n-k}, \tag{116}\]  
where the first equality follows by dividing both sides in the left side by factor \(n\). Thereby,  
\[\sum_{j=0}^{k}\binom{n}{j}(1-\epsilon)^{n-j}\epsilon^{j}\leq\frac{\epsilon(n- k)}{\epsilon n-k}\cdot 2^{n\left[H(\frac{k}{n})-T_{\epsilon}(\frac{k}{n}) \right]}. \tag{117}\]  
This completes the proof of Lemma A5.