Exercise 4.1Z: Log Likelihood Ratio at the BEC Model

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BEC channel model

We consider the so-called  "BEC channel model"  (binary erasure channel) with.

• the input variable  $x ∈ \{+1, \, -1\}$,
• the output variable  $y ∈ \{+1, \, -1, \, {\rm E}\}$, and
• the erasure probability  $\lambda$.

Here  $y = {\rm E}$  (erasure) means that the initial value  $y$  could neither be decided as  $+1$  nor as  $-1$ .

Also known are the input probabilities

$${\rm Pr}(x = +1) = 3/4\hspace{0.05cm}, \hspace{0.5cm}{\rm Pr}(x = -1) = 1/4\hspace{0.05cm}.$$

The LLR of the binary random variable  $x$  is given by bipolar approach as follows:

$$L(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = +1)}{{\rm Pr}(x = -1)}\hspace{0.05cm}.$$

Correspondingly, for the conditional LLR in the forward direction for all  $y ∈ \{+1, \, -1, \, {\rm E}\}$:

$$L(y\hspace{0.05cm}|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x = +1)}{{\rm Pr}(y\hspace{0.05cm}|\hspace{0.05cm}x = -1)} \hspace{0.05cm}. $$

Hints::

• This exercise belongs to the chapter  "Soft–in Soft–out Decoder".
• Reference is made in particular to the page  "Reliability Information – Log Likelihood Ratio"  and to the page  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|Binary Erasure Channel].

Questions

1

What is the LLR of the input variable  $x$?

$L(x) \ = \ $

2

What probability  ${\rm Pr}(x = \, -1)$  corresponds to  $L(x) = \, -2$?

${\rm Pr}(x = \, -1) \ = \ $

3

Calculate the conditional  $L$–value  $L(y = {\rm E}\hspace{0.05cm} |\hspace{0.05cm} x)$  in the forward direction.

$L(y = {\rm E} \hspace{0.05cm} |\hspace{0.05cm} x) \ = \ $

4

Which statements are true for the other two conditional LLR?

	$L(y = +1 \hspace{0.05cm} |\hspace{0.05cm} x)$  is positive infinity.
	$L(y = \, -1 \hspace{0.05cm} |\hspace{0.05cm} x)$  is negative infinite in magnitude.
	It holds  $L(y = +1 \hspace{0.05cm} |\hspace{0.05cm} x) = L(y = \, -1 \hspace{0.05cm} |\hspace{0.05cm} x) = 0$.

5

Under what conditions do the results from (3) and (4) hold?

	For  $0 ≤ \lambda ≤ 1$.
	For  $0 < \lambda ≤ 1$.
	For  $0 ≤ \lambda < 1$.
	For  $0 < \lambda < 1$.

Solution

Solution

(1)  With the given symbol probabilities ${\rm Pr}(x = +1) = 3/4$ and ${\rm Pr}(x = -1) = 1/4$, we obtain:

$$L(x)={\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(x = +1)}{{\rm Pr}(x = -1)} ={\rm ln} \hspace{0.15cm} \frac{3/4}{1/4}\hspace{0.15cm}\underline{= 1.099}\hspace{0.05cm}.$$

(2)  According to the definition

$$L(x)={\rm ln} \hspace{0.15cm} \frac{\rm Pr}(x = +1)}{\rm Pr}(x = -1)}$$

yields the following equation of determination for $L(x) = \, -2$:

$$\hspace{0.15cm} \frac{{\rm Pr}(x = +1)}{1-{\rm Pr}(x = +1)} \stackrel{!}{=}{\rm e}^{-2} \approx 0.135 \hspace{0.25cm}\Rightarrow \hspace{0.25cm} 1.135 \cdot {\rm Pr}(x = +1)\stackrel{!}{=}0.135\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(x = +1) = 0.119\hspace{0.05cm},\hspace{0.4cm}{\rm Pr}(x = -1) \hspace{0.15cm}\underline{= 0.881}\hspace{0.05cm}. $$

(3)  For the conditional LLR $L(y = {\rm E} \hspace{0.05cm} |\hspace{0.05cm} x)$ in the forward direction, for the given BEC model:

$$L(y = {\rm E}\hspace{0.05cm}|\hspace{0.05cm}x) = {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y= {\rm E}\hspace{0.05cm}|\hspace{0.05cm}x = +1)}{{\rm Pr}(y= {\rm E}\hspace{0.05cm}|\hspace{0.05cm}x = -1)} = {\rm ln} \hspace{0.15cm} \frac{\lambda}{\lambda}\hspace{0.15cm}\underline{= 0}\hspace{0.05cm}.$$

(4)  Analogous to the sample solution of subtask (3), we obtain for $y = ±1$:

$$L(y = +1\hspace{0.05cm}|\hspace{0.05cm}x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y= +1\hspace{0.05cm}|\hspace{0.05cm}x = +1)}{{\rm Pr}(y= +1\hspace{0.05cm}|\hspace{0.05cm}x = -1)} = {\rm ln} \hspace{0.15cm} \frac{1-\lambda}{0}\hspace{0.15cm}\underline{ \hspace{0.05cm}\Rightarrow \hspace{0.15cm}+\infty }\hspace{0.05cm},$$

$$L(y = -1\hspace{0.05cm}|\hspace{0.05cm}x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}(y= -1\hspace{0.05cm}|\hspace{0.05cm}x = +1)}{{\rm Pr}(y= -1\hspace{0.05cm}|\hspace{0.05cm}x = -1)} = {\rm ln} \hspace{0.15cm} \frac{0}{1-\lambda}\hspace{0.15cm}\underline{ \hspace{0.05cm}\Rightarrow \hspace{0.15cm}-\infty }\hspace{0.05cm}. $$

Accordingly, the proposed solutions 1 and 2 are correct.

(5)  Correct is the last proposed solution:

• For $\lambda = 0$ (ideal channel), $L(y = {\rm E} \hspace{0.05cm} |\hspace{0.05cm} x) = \ln {(0/0)}$   ⇒   indefinite result.
• For $\lambda = 1$ (complete erasure, $y ≡ {\rm E}$) $L(y = +1 \hspace{0.05cm} |\hspace{0.05cm} x)$ and $L(y = \, -1 \hspace{0.05cm} |\hspace{0.05cm} x)$ are indeterminate.