## 3 Cross-Layer Design for Finite Blocklength Coding
### Achieving the Hard Delay Constraint with Finite Average Power
\begin{split}&\sum_{i=0}^{M-1}\binom{M-1}{i}\frac{(-1)^{i+1}}{i+1 }\phi(i+1)-(M-1)\sum_{i=0}^{M-2}\binom{M-2}{i}\frac{(-1)^{i+1}}{i+2}\phi(i+2) \\ =&-\phi(1)+\sum_{i=1}^{M-1}\binom{M-1}{i}\frac{(-1)^{ i+1}}{i+1}\phi(i+1)+(M-1)\sum_{i=1}^{M-1}\binom{M-2}{i-1}\frac{(-1)^{i+1}}{i+1 }\phi(i+1)\\ =&-\phi(1)+\sum_{i=1}^{M-1}\binom{M-1}{i}\frac{(-1)^ {i+1}}{i+1}\phi(i+1)+i\sum_{i=1}^{M-1}\binom{M-1}{i}\frac{(-1)^{i+1}}{i+1} \phi(i+1)\\ =&-\phi(1)+\sum_{i=1}^{M-1}\binom{M-1}{i}(-1)^{i+1} \phi(i+1)\\ =&\sum_{i=0}^{M-1}\binom{M-1}{i}(-1)^{i+1}\phi(i+1). \end{split}  \tag{26}\]  
Therefore, the lemma is proved. **Theorem 1**. : _The average power across the optimal subchannel of parallel fading channels is expressed as_  
\[ \begin{split}\bar{P}(g_{th})=& PM\ln(g_{th})e^{-g_{ th}}(1-e^{-g_{th}})^{M-1}\\ &+PM\sum_{i=1}^{M-1}\binom{M-1}{i}(-1)^{i+1}\Big{(}\gamma(ig_{th }+g_{th})+e^{-(i+1)g_{th}}\ln(i+1)\Big{)}.\end{split}  \tag{27}\]  
_Remarkably, \(\bar{P}(g_{th})<\infty\) even when \(g_{th}=0\)._  
Proof. : Initially, we employ integration by parts to reshape the integral subject. Following this, the higher-order polynomial is expanded through the implementation of the binomial theorem, thereby facilitating its integration. Subsequently, Lemma 1 is invoked to substitute the associated integral expression. The ensuing expression post-integration is simplified utilizing Lemma 2. Finally, we utilize Lemma 1 to prove the convergence of Theorem 1. Leveraging Equation (19), the average power over the optimal subchannel of parallel fading channels is expressed as \[\int_{S_{th}}^{\infty}\frac{e^{-x}(1-e^{-x})^{M-1}}{x}dx\] \[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\] \[+\int_{g_{th}}^{\infty}\ln(x)e^{-x}\Big{(}(1-e^{-x})^{M-1}-(M-1)e^ {-2x}(1-e^{-x})^{M-2}\Big{)}dx\] \[= \ln(g_{th})e^{-g_{th}}(1-e^{-g_{th}})^{M-1}\] \[+\int_{g_{th}}^{\infty}\ln(x)e^{-x}(1-e^{-x})^{M-1}dx-\int_{g_{th} }^{\infty}(M-1)e^{-2x}(1-e^{-x})^{M-2}dx\] \[= \ln(g_{th})e^{-g_{th}}(1-e^{-g_{th}})^{M-1}\] \[+\int_{g_{th}}^{\infty}\ln(x)e^{-x}\sum_{i=0}^{M-1}{M-1\choose i} (-1)^{i}e^{-ix}dx-\int_{g_{th}}^{\infty}(M-1)e^{-2x}\sum_{i=0}^{M-2}{M-2\choose i }(-1)^{i}e^{-ix}dx\] \[= \ln(g_{th})e^{-g_{th}}(1-e^{-g_{th}})^{M-1}\] \[+\sum_{i=0}^{M-1}{M-1\choose i}(-1)^{i+1}\frac{1}{i+1}\Big{(} \gamma(ig_{th}+g_{th})+e^{-(i+1)g_{th}}\ln(i+1)\Big{)}\] \[-(M-1)\sum_{i=0}^{M-2}{M-1\choose i}(-1)^{i+1}\frac{1}{i+2}\Big{(} \gamma(ig_{th}+2g_{th})+e^{-(i+1)g_{th}}\ln(i+2)\Big{)}. \tag{28}\]  
Next, we employ Lemma 2, where we define  
\[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\]0  
In this way, we have  
\[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\]1 \[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\]2 \[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\]3 \[= \ln(x)e^{-x}(1-e^{-x})^{M-1}|_{gh}^{\infty}\]4  
According to Lemma 1, we prove the convergence of Equation (30). At this point, the theorem is proved. By using Lemmas 1 and 2, the average power consumption is given in Theorem 1, which demonstrates that the average power consumption of a parallel fading channel is finite. Then, using Theorem 1, we show that the system can meet the hard delay constraint with finite power consumption with a Markov chain.