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$y$ is a random variable

\begin{equation} y=ax+n \end{equation}

where $a$ is a scalar, $x \in \{ +1,-1 \}$ and $n \sim \mathcal{N}(0,\sigma_c^2)$.

We define

\begin{equation} L(x)\triangleq \ln \left( \dfrac{Prob(x_i=+1) }{ Prob(x_i=-1)} \right) \end{equation}

and

\begin{equation} L_{CH}=L(x\mid y)= \ln \left( \dfrac{Prob(x_i=+1 \mid y) }{ Prob(x_i=-1\mid y)} \right) = L_c.y+L(x) \end{equation}

Here

\begin{equation} L_c = \dfrac{2a}{\sigma_c^2} \end{equation}

How do we prove that $L_{CH} \sim \mathcal{N}(\pm \sigma_{CH}^2 / 2, \sigma_{CH}^2)$, where $\sigma_{CH}^2=2aL_c$

Ref: Hagenauer, J., "The exit chart - introduction to extrinsic information transfer in iterative processing," in Signal Processing Conference, 2004 12th European , vol., no., pp.1541-1548, 6-10 Sept. 2004

Michael Hardy
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NAASI
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1 Answers1

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As

\begin{equation} y \sim \mathcal{N}(\pm a,\sigma_c^2) \end{equation}

and

\begin{equation} L_c.y \sim \mathcal{N}(\pm 2a^2/ \sigma_c^2,4a^2/ \sigma_c^2) \end{equation}

where $2a^2/ \sigma_c^2=a.L_c=2a.L_c/2=\sigma_{CH}^2/2$ and $4a^2/ \sigma_c^2=2a.L_c=\sigma_{CH}^2$

As $L(x)$ is not a random variable for a given $Prob(x=+1)$ and $Prob(x=-1)$,therefore only $L_c.y$ contributes to the distribution of $L_{CH}$

\begin{equation} L_{CH}\sim \mathcal{N}(\pm \sigma_{CH}^2/2,\sigma_{CH}^2) \end{equation}

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