Channel polarization

Binary symmetric channel

Let $p$ be the crossover probability of a binary symmetric channel (BSC).

	00	01	10	11
00	$(1-p)^{2}$	$p(1-p)$	$p(1-p)$	$p^{2}$
01	$p^{2}$	$p(1-p)$	$p(1-p)$	$(1-p)^{2}$
10	$p(1-p)$	$p^{2}$	$(1-p)^{2}$	$p(1-p)$
11	$p(1-p)$	$(1-p)^{2}$	$1-p^{2}$	$p(1-p)$

	00	01	10	11
00	$0.5(1-2p+2p^{2})$	$p(1-p)$	$p(1-p)$	$0.5(1-2p+2p^{2})$
01	$p(1-p)$	$0.5(1-2p+2p^{2})$	$0.5(1-2p+2p^{2})$	$p(1-p)$

It turns out that the channel $W^{-}:U_{1}\rightarrow (Y_{1},Y_{2})$ reduces to another BSC. To see this, consider the case where $(Y_{1},Y_{2})=(0,0)$. To minimize the error probability, we must decide the value of $U_{1}$ that has the greater likelihood. For $0<p<0.5$, $0.5(1-2p+2p^{2})>p(1-p)$ so that the maximum-likelihood (ML) decision is $U_{1}=0$. Using the same argument, we see that the ML decision is $U_{1}=0$ for $(Y_{1},Y_{2})=(1,1)$. More generally, the receiver decision is to set ${\hat {U_{1}}}=Y_{1}\oplus Y_{2}$. Indeed, if the crossover probability is low, it is very likely that $Y_{1}=U_{2}\oplus U_{1}$ and $Y_{2}=U_{2}$. Solving for $U_{2}$ and plugging it back produces the desired result.

	00,11	01,10
00	$1-2p+2p^{2}$	$2p(1-p)$
01	$2p(1-p)$	$1-2p+2p^{2}$

We just saw that despite having a quartenary output alphabet, $W^{-}$ is equivalent to a BSC. To get the effective crossover probability of this synthetic channel, we just determine the probability that the ML decision ${\hat {U_{1}}}$ is not the same as $U_{1}$. This will happen with probability $2p(1-p)$. Intuitively, $W^{-}$ should have less mutual information compared to the original channel $W$ since an independent source $U_{2}$ interferes with $U_{1}$ on top of the effects of the BSC.

Checkpoint: Show that for $0<p<1/2$, $p<2p(1-p)$.

Now, let us consider the other synthetic channel $W^{+}:U_{2}\rightarrow (U_{1},Y_{1},Y_{2})$. This synthetic channel has a greater mutual information compared to the original BSC due to the "stolen" information about $U_{1}$. As with the previous synthetic channel, we can produce the transition probability matrix from $U_{2}$ to $(U_{1},Y_{1},Y_{2})$, which now has an eight-element output alphabet. To facilitate the discussion, the columns of the table below have been grouped according to their entries:

	000	110	001	010	100	111	011	101
0	$0.5(1-p)^{2}$	$0.5(1-p)^{2}$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5p^{2}$	$0.5p^{2}$
1	$0.5p^{2}$	$0.5p^{2}$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5p(1-p)$	$0.5(1-p)^{2}$	$0.5(1-p)^{2}$

In the transition probability matrix, we see that there are four columns where the receiver will not be able to tell whether $U_{2}=0$ or $U_{2}=1$. We can call such a scenario an erasure, and say that the transmitted bit is erased with probability $4\times 0.5p(1-p)=2p(1-p)$. Clearly, we cannot reduce this synthetic channel to a BSC since there are no erasures in a BSC. Regardless, we can still come up the following more manageable reduction:

	000,110	001,010,100,111	011,101
0	$(1-p)^{2}$	$2p(1-p)$	$p^{2}$
1	$p^{2}$	$2p(1-p)$	$(1-p)^{2}$

Checkpoint: Show that the mutual information is preserved after the reduction.

Binary erasure channel

Let $\epsilon$ be the erasure probability of a binary erasure channel (BEC). As with the BSC, we can start with the conditional probability of $(Y_{1},Y_{2})$ given $(U_{1},U_{2})$.

	00	0?	01	?0	??	?1	10	1?	11
00	$(1-\epsilon )^{2}$	$\epsilon (1-\epsilon )$		$\epsilon (1-\epsilon )$	$\epsilon ^{2}$
01					$\epsilon ^{2}$	$\epsilon (1-\epsilon )$		$\epsilon (1-\epsilon )$	$(1-\epsilon )^{2}$
10				$\epsilon (1-\epsilon )$	$\epsilon ^{2}$		$(1-\epsilon )^{2}$	$\epsilon (1-\epsilon )$
11		$\epsilon (1-\epsilon )$	$(1-\epsilon )^{2}$		$\epsilon ^{2}$	$\epsilon (1-\epsilon )$

The synthetic channel $W^{-}:U_{1}\rightarrow (Y_{1},Y_{2})$ can be obtained by marginalizing $U_{2}$,

	00	0?	01	?0	??	?1	10	1?	11
0	$0.5(1-\epsilon )^{2}$	$0.5\epsilon (1-\epsilon )$		$0.5\epsilon (1-\epsilon )$	$\epsilon ^{2}$	$0.5\epsilon (1-\epsilon )$		$0.5\epsilon (1-\epsilon )$	$0.5(1-\epsilon )^{2}$
1		$0.5\epsilon (1-\epsilon )$	$0.5(1-\epsilon )^{2}$	$0.5\epsilon (1-\epsilon )$	$\epsilon ^{2}$	$0.5\epsilon (1-\epsilon )$	$0.5(1-\epsilon )^{2}$	$0.5\epsilon (1-\epsilon )$

A maximum-likelihood receiver will decide that $U_{1}=0$ if is more likely than $U_{1}=0$. If there are ties, we declare an erasure, denoted by $?$. This allows us to reduce the nine-element output alphabet into three groups. The receiver declares that ${\hat {U}}_{1}=0$ if $Y_{1}\oplus Y_{2}=0$ and ${\hat {U}}_{1}=1$ if if $Y_{1}\oplus Y_{2}=1$. This is very similar to the XOR decisions in our discussion of the BSC. The difference now is that several combinations may correspond to an erasure.

	00,11	0?,?0,??,?1,1?	01,11
0	$(1-\epsilon )^{2}$	$2\epsilon -\epsilon ^{2}$
1		$2\epsilon -\epsilon ^{2}$	$(1-\epsilon )^{2}$

The other synthetic channel $W^{+}:U_{2}\rightarrow (U_{1},Y_{1},Y_{2})$ has an output alphabet of size 18. For brevity, the table below shows the reduced transition probability matrix. Upon inspection, we see that the reduced channel is equivalent to another BEC with erasure probability $\epsilon ^{2}$. For all $0<\epsilon <1$, $\epsilon ^{2}<2\epsilon -\epsilon ^{2}$ which guarantees that $W^{+}$ has higher mutual information compared to $W^{-}$.

	000, 00?, 0?0, 1?0, 110, 11?	0??, 1??	0?1, 01?, 011, 10?, 101, 1?1
0	$1-\epsilon ^{2}$	$\epsilon ^{2}$
1		$\epsilon ^{2}$	$1-\epsilon ^{2}$

Finally, observe that the capacity of the original BEC is $C(W)=1-\epsilon$. Since $W^{-}$ and $W^{+}$ are also BECs, then their capacities are given by similar expressions. Adding the capacities of the synthetic channels gives twice the capacity of the original channel,

$(1-\epsilon ^{2})+(1-2\epsilon +\epsilon ^{2})=2-2\epsilon =2(1-\epsilon )$