# Polar Code Parameter Recognition Algorithm Based on Dual Space
## 3 Principle and Method of Polar Code Parameter Recognition
### Decision Threshold
\tag{37}\]  
Thus, the expectation of \(Y_{i}\) and \(Y_{i}^{2}\) can be obtained, respectively, as  
\[E(Y_{i})=\int_{0}^{+\infty}w\cdot x\cdot f_{1}(x)\cdot\left(1-F_{1}(x)\right) ^{w-1}dx, \tag{38}\]  
\[E(Y_{i}^{2})=\int_{0}^{+\infty}w\cdot x^{2}\cdot f_{1}(x)\cdot\left(1-F_{1}(x )\right)^{w-1}dx. \tag{39}\]  
Calculating integration by parts,  
\[E(Y_{i})=\left.-x\cdot\left(1-F_{1}(x)\right)^{w}\right|_{0}^{+\infty}+\int_{ 0}^{+\infty}\left(1-F_{1}(x)\right)^{w}dx, \tag{40}\]  
\[E(Y_{i}^{2})=\left.-x^{2}\cdot\left(1-F_{1}(x)\right)^{w}\right|_{0}^{+\infty }+\int_{0}^{+\infty}2x\cdot\left(1-F_{1}(x)\right)^{w}dx. \tag{41}\]  
According to the L'Hospital principle \(\lim\limits_{x\rightarrow+\infty}x(1-F_{1}(x))=0\),  
\(\lim\limits_{x\rightarrow+\infty}x^{2}(1-F_{1}(x))=0\), so (40) and (41) can be rewritten, respectively, as  
\[E(Y_{i})=\int_{0}^{+\infty}\left(1-F_{1}(x)\right)^{w}dx, \tag{42}\]\[E(Y_{i}^{2})=\int_{0}^{+\infty}2x\cdot(1-F_{1}(x))^{w}dx, \tag{43}\]  
where \(F_{1}(x)=\int_{0}^{x}f_{1}(t)dt\), so  
\[F_{1}(x)=0.5(erf(\frac{x+\frac{2}{\sigma^{2}}}{\frac{2\sqrt{2}}{\sigma}})+erf( \frac{x-\frac{2}{\sigma^{2}}}{\frac{2\sqrt{2}}{\sigma^{2}}})), \tag{44}\]  
where \(erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}\). It is easy to know that \(X_{i}^{2}\cdot Y_{i}^{2}=Y_{i}^{2}\) and it has nothing to do with the assumptions, so  
\[E(Y_{i}^{2})=E(X_{i}^{2}\cdot Y_{i}^{2}|H_{0})=E(X_{i}^{2}\cdot Y_{i}^{2}|H_{1 }). \tag{45}\]  
If \(H_{0}\) is true, \(\mathbf{h}_{K}\) is not orthogonal to the codeword, so  
\[E(X_{i}\cdot Y_{i}|H_{0})=0, \tag{46}\]  
\[Var(X_{i}\cdot Y_{i}|H_{0})=E(X_{i}^{2}\cdot Y_{i}^{2}|H_{0})=\int_{0}^{+ \infty}2x\cdot(1-F_{1}(x))^{w}dx. \tag{47}\]  
If \(H_{1}\) is true, and the judgment of \(c_{i,t}\) is correct, let \(R_{1}\) denote the condition of \(c_{i,t}=0\), \(l_{i,t}>0\) and \(R_{2}\) denote the condition of \(c_{i,t}=1\), \(l_{i,t}<0\); then, the PDFs of \(l_{i,t}\) are  
\[f(x|R_{1})=\left\{ \begin{array}{ll}\frac{\sigma\cdot e^{-\sigma^{2}\cdot(x- 2/\sigma^{2})^{2}/8}}{\sqrt{2\pi}(1+erf(1/(\sqrt{2}\sigma)))}&,x>0\\ 0&,x\leq 0\end{array} ,\right. \tag{48}\]  
\[f(x|R_{2})=\left\{ \begin{array}{ll}\frac{\sigma\cdot e^{-\sigma^{2}\cdot(x+ 2/\sigma^{2})^{2}/8}}{\sqrt{2\pi}(1+erf(1/(\sqrt{2}\sigma)))}&,x<0\\ 0&,x\geq 0\end{array} .\right. \tag{49}\]  
Let \(F(x|R_{1})\) and \(F(x|R_{2})\) denote the cumulative functions (CFs) of \(f(x|R_{1})\) and \(f(x|R_{2})\), respectively, so the CFs of \(Y=|l_{i,t}|\) are  
\[F_{Y}(y|R_{1})=\left\{ \begin{array}{ll}F(y|R_{1})-F(-y|R_{1})&,y>0\\ 0&,y\leq 0^{\prime}\end{array} \right. \tag{50}\]  
\[F_{Y}(y|R_{2})=\Pr(Y\leq y|R_{2})=\left\{ \begin{array}{ll}F(y|R_{2})-F(-y|R _{2})&,y>0\\ 0&,y\leq 0^{\prime}\end{array} .\right. \tag{51}\]  
When \(y>0\), \(F(-y|R_{1})=0\), and \(F(y|R_{2})=1\), so \(F_{Y}(y|R_{1})=F(y|R_{1})\) and \(F_{Y}(y|R_{2})=1-F(-y|R_{2})\), and therefore the PDFs of \(Y\) are  
\[f_{Y}(y|R_{1})=\left\{ \begin{array}{ll}f(y|R_{1})&,y>0\\ 0&,y\leq 0^{\prime}\end{array} \right. \tag{52}\]  
\[f_{Y}(y|R_{2})=\left\{ \begin{array}{ll}f(-y|R_{2})&,y>0\\ 0&,y\leq 0^{\prime}\end{array} \right. \tag{53}\]  
It can be obtained from (49), (50), (53), and (54) that, when \(y\geq 0\), \(f_{Y}(y|R_{1}\cup R_{2})\)\(=0\), and, when \(y<0\), that  
\[f_{Y}(y|R_{1}\cup R_{2})=\frac{\sigma\cdot e^{-\sigma^{2}\cdot(y-2/\sigma^{2} )^{2}/8}}{\sqrt{2\pi}(1+erf(1/(\sqrt{2}\sigma)))}. \tag{54}\]If \(H_{1}\) is true, and the judgment of \(c_{i,t}\) is not correct, let \(R_{1}^{\prime}\) denote the condition of \(c_{i,t}=0,l_{i,t}<0\) and \(R_{2}^{\prime}\) denote the condition of \(c_{i,t}=1,l_{i,t}>0\); then, the PDFs of \(l_{i,t}\) are  
\[f(x|R_{1}^{\prime})= \begin{cases}\frac{\sigma\cdot e^{-\sigma^{2}\cdot(x-2/ \sigma^{2})^{2}/8}}{\sqrt{2\pi}(1-erf(1/(\sqrt{2}\sigma)))}&,x>0\\ 0&,x\leq 0\end{cases} , \tag{55}\]  
\[f(x|R_{2}^{\prime})= \begin{cases}\frac{\sigma\cdot e^{-\sigma^{2}\cdot(x+2/ \sigma^{2})^{2}/8}}{\sqrt{2\pi}(1-erf(1/(\sqrt{2}\sigma)))}&,x>0\\ 0&,x\leq 0\end{cases}