Polar Coding Notes: A Simple Proof
For any B-DMC $W$, the channels $\{W_N^{(i)}\}$ polarize in the sense that, for any fixed $\delta \in (0, 1)$, as $N$ goes to infinity through powers of two, the fraction of indices $i \in \{1, \dots, N\}$ for which $I(W_N^{(i)}) \in (1 − \delta, 1]$ goes to $I(W)$ and the fraction for which $I(W_N^{(i)}) \in [0, \delta)$ goes to $1−I(W)^{[1]}$.
Mrs. Gerber’s Lemma
Mrs. Gerber’s Lemma provides a lower bound on the entropy of the modulo-$2$ sum of two binary random vectors$^{[2][3]}$.
Let $h^{-1} : [0, 1] \to [0, 1/2]$ be the inverse of the binary entropy function $h(p) = -p\log p - (1-p)\log(1-p)$.
Here we set a binary symmetric channal with crossover probability $p_0$.
The convolution of $a$ and $b$ is denoted by
Convex: The function $f(u) = h(h^{-1}(u)\ast p_0), u \in [0,1]$ is convex in $u$ for every fixed $p_0 \in (0,1/2]^{[2]}$.
Scalar MGL: Let $X$ be a binary random variable and let $U$ be an arbitrary random variable. If $Z \sim Bern(p)$ is independent of $(X, U)$ and $Y = X \oplus Z$, then
Vector MGL: Let $X^n$ be a binary random vector and $U$ be an arbitrary random variable. If $Z^n$ is a vector of independent and identically distributed $Bern(p)$ random variables independent of $(X^n, U)$ and $Y^n = X^n \oplus Z^n$, then
Let $X,Y$ are two binary random variable, from$^{[2]}$
Let
Then from conditional entropy definition
So that
and consider $\beta_0(y)=p(X=1|Y)$ as a RV
Let us consider first
Here $\beta_k = p(X_k=1 | X_1, ... , X_{k-1}), 1 \le k \le n$. Now we have
and
Strict Polarization for Binary-Input Channels
First notice that
and
Since $H(X_1|Y_1) \in (0, 1)$, there exists an $\alpha \in (0, 1/2)$ such that $H(X_1|Y_1) = h(\alpha)$.
Thus, we have that
So what we have concluded is that for every $\delta > 0$, there exists $\kappa(\delta) > 0$ such that if $I(W) \in (\delta, 1 − \delta)$, we have
Proof of Channel Polarization
Given $W$ and $\delta > 0$, define$^{[4][5]}$
Let
We have
$\begin{align}\mu_{n+1} &= {1\over 2^{n+1}} \sum_{s \in \{\pm\}^{n+1}} I(W^s) \\&= {1\over 2^n} \sum_{t \in \{\pm\}^n} {1\over 2} [I(W^{t+})+I(W^{t-})] \\&= {1\over 2^n} \sum_{t \in \{\pm\}^n} I(W^t) \\&= \mu_n = \mu_0 = I(W) \\ \nu_{n+1} &= {1\over 2^{n+1}} \sum_{s \in \{\pm\}^{n+1}} [I(W^s)]^2 \\&= {1\over 2^n} \sum_{t \in \{\pm\}^n} {[I(W^{t+})]^2+[I(W^{t-}]^2 \over 2} \\&= {1\over 2^n} \sum_{t \in \{\pm\}^n} [I(W^t)]^2+[\Delta(W^t)]^2 \tag{${a^2+b^2 \over 2} = ({a+b \over 2})^2+({a-b \over 2})^2$} \\&\ge \nu_n + \theta_n(\delta)\kappa(\delta)^2 \tag{definition of $\theta_n(\delta)$} \end{align}$
The sequence $\nu_n$ is thus bounded and monotone and consequently convergent; in particular $\nu_{n+1}−\nu_n$ converges to zero. As $\theta_n$ is sandwiched by
between two quantities both convergent to zero, we conclude
This means that for large enough $n$, the fraction of mediocre channels (i.e., those with symmetric capacities in $(\delta, 1 − \delta)$) vanishes to zero. But by preservation of mutual information, we also know that if we define
we automatically have
This M.Alsan and E.Telatar's proof is much simpler than the martingale convergence theorem used by Arıkan$^{[1]}$.
Reference:
1. E. Arikan. Channel polarization: A method for constructing capacity-achieving codes for symmetric binary-input memoryless channels. IEEE Trans. on Information Theory, vol.55, no.7, pp.3051–3073, July 2009.
2. A. D. Wyner and J. Ziv. A theorem on the entropy of certain binary sequences and applications (Part I). IEEE Trans.Inform.Theory, vol.19, no.6, pp.769-772, Nov.1973.
3. Abbas El Gamal and Young-Han Kim. Network Information Theory. Cambridge University Press. 2011.
4. M.Alsan and E.Telatar. A simple proof of polarization and polarization for non-stationary memoryless channels. IEEE Trans.Info.Theory, vol.62, no.9,pp.4873-4878. 2016.
5. Vincent Y. F. Tan. EE5139R: Information Theory for Communication Systems:2016/7, Semester 1. https://www.ece.nus.edu.sg/
6. Eren Sasoglu. Polarization and Polar Codes. Foundations and Trends in Communications and Information Theory Vol. 8, No. 4 (2011) 259–381
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