# Polar Code Parameter Recognition Algorithm Based on Dual Space
## 3 Principle and Method of Polar Code Parameter Recognition
### Decision Threshold  
Let \(X_{i}=\underset{t=t_{1}}{\overset{t_{w}}{\Pi}}\,sign(l_{i,t})\) and \(Y_{i}=\underset{t=t_{1},\cdots,t_{w}}{\min}(|l_{i,t}|)\). **Theorem 2**. : _Suppose \(X_{1},X_{2},\cdots,X_{l}\) are \(J\) independent random variables, and their distribution functions (DFs) are \(F_{X_{1}}(x),F_{X_{2}}(x),\cdots,F_{X_{J}}(x)\), respectively; the DF for \(Z=\min\bigl{\{}X_{1},X_{2},\cdots,X_{J}\bigr{\}}\) is then_  
\[F_{Z}(z)=1-\underset{i=1}{\overset{I}{\Pi}}(1-F_{X_{i}}(z)). \tag{21}\]  
_The probability density function (PDF) of \(Z=\min\bigl{\{}X_{1},X_{2},\cdots,X_{J}\bigr{\}}\) is_  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]  
Proof. : From the definition of the DF, we know that  
\[F_{Z}(z)=\Pr(Z\leq z)=1-\Pr(Z>z). \tag{23}\]  
That is to say,  
\[F_{Z}(z)=1-\Pr(X_{1}>z)\Pr(X_{2}>z)\cdots\Pr(X_{J}>z) \tag{24}\]  
and, in turn,  
\[F_{Z}(z)=1-(1-\Pr(X_{1}\leq z))(1-\Pr(X_{2}\leq z))\cdots(1-\Pr(X_{J}\leq z)) \tag{25}\]  
and  
\[F_{Z}(z)=1-\underset{i=1}{\overset{J}{\Pi}}(1-F_{X_{i}}(z)). \tag{26}\]  
Taking the derivative of \(F_{Z}(z)\) with respect to \(z\), the PDF of the random variable \(Z\) is obtained as  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {J}{\Pi}}(1-F_{X_{i}}(z)). \tag{27}\]  
\(\square\)  
Inference 1 can be easily obtained from Theorem 2 as follows. **Inference 1. **_When \(X_{1},X_{2},\cdots,X_{J}\) are \(J\) independent and identically distributed random variables, supposing they have the same DF \(F_{X}(x)\) and PDF \(f_{1}(x)\), then the PDF of \(Z=\min\bigl{\{}X_{1},X_{2},\cdots,X_{J}\bigr{\}}\) is_  
\[F_{Z}(z)=1-(1-F_{X}(z))^{J}. \tag{28}\]_The PDF of_ \(Z=\min\{X_{1},X_{2},\cdots,X_{J}\}\) _is_  
\[f_{Z}(z)=J\cdot f_{X}(z)\cdot(1-F_{X}(z))^{J-1}. \tag{29}\]  
Let \(F(x)\) denote \(l_{i,t}\)'s DF, \(f(x)\) denote \(l_{i,t}\)'s PDF, \(F_{1}(x)\) denote \(|l_{i,t}|\)'s DF, and \(f_{1}(x)\) denote \(|l_{i,t}|\)'s PDF, so  
\[F_{1}(x)=\Pr(|l_{i,t}|\leq x)=\Pr(-x\leq l_{i,t}\leq x), \tag{30}\]  
where \(x>0\), so  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]0  
Taking the derivative of x on both sides of (31), we obtain  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]1  
It can be known from (14) that  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]2  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]3  
\(c_{i,j}\) takes the same probability of 0 and 1 without prior information, and we obtain  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]4  
Because \(f(x)=f(-x)\), when \(x>0\),  
\[f_{Z}(z)=\sum\limits_{j=1}^{J}f_{X_{j}}(z)\cdot\underset{i=1,j\
eq j}{\overset {I}{\Pi}}(1-F_{X_{i}}(z)). \tag{22}\]5  
When \(x\leq 0\), \(f_{1}(x)=0\). Let \(f_{2}(x)\) be the PDF of \(Y_{i}\), and then it is known from Inference 1 that  
\[f_{2}(x)=\left\{ \begin{matrix}w\cdot f_{1}(x)\cdot\left(1-F_{1}(x)\right)^{w- 1}&,x>0\\ 0&,x\leq 0\end{matrix} \right.