Exercise 5.6: Error Correlation Duration

Error correlation function of the GE model

The graph shows the error correlation function (ECF) of the Gilbert–Elliott model with the parameters

$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001, \hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.1, \hspace{0.2cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}$$

in logarithmic representation.

This model is discussed in detail in "Exercise 5.6Z". In particular, the error correlation function (ECF) is also calculated in this exercise. With the auxiliary quantities

$$A \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (p_{\rm B}- p_{\rm M}) \cdot (p_{\rm M}- p_{\rm G})\hspace{0.05cm},$$

$$B\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)$$

it can be written for:

$$\varphi_{e}(k) = \left\{ \begin{array}{c} p_{\rm M} \\ p_{\rm M}^2 + A \cdot (1-B)^k \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k > 0 \hspace{0.05cm}.\\ \end{array}$$

This is a burst error channel. To quantitatively describe the statistical bonds, one often uses the correlation term according to the following definition:

$$D_{\rm K} = \frac{1}{\varphi_{e0} - p_{\rm M}^2} \cdot \sum_{k = 1 }^{\infty}\hspace{0.1cm}\big [\varphi_{e}(k) - p_{\rm M}^2 \big]\hspace{0.05cm}.$$

The reference value $\varphi_{e0}$ is obtained by extrapolation of the error correlation function to the point $k = 0$. If, as here, the ECF curve is given analytically, $\varphi_{e0}$ can also be calculated by inserting the value $k = 0$ into the equation which is actually only valid for $k > 0$.

Notes:

The exercise belongs to the chapter "Burst Error Channels".
Reference is made in particular to the section "Error correlation function of the Gilbert-Elliott model".

Questions

$\varphi_e(k = 0) \ = \ $

$\ \cdot 10^{-2}$

$\varphi_{e0} \ = \ $

$\ \cdot 10^{-2}$

	$D_{\rm K} = A \cdot B$,
	$D_{\rm K} = 1/A \, - B$,
	$D_{\rm K} = 1/B \, -1$.

$D_{\rm K} \ = \ $

	$D_{\rm K}$ remains the same if ${\rm Pr}({\rm B\hspace{0.05cm}\|\hspace{0.05cm}G})$ and ${\rm Pr(G\hspace{0.05cm}\|\hspace{0.05cm}B)}$ are interchanged.
	$D_{\rm K}$ depends only on the sum ${\rm Pr(G\hspace{0.05cm}\|\hspace{0.05cm}B) + Pr(B\hspace{0.05cm}\|\hspace{0.05cm}G)}$.
	The red area in the graph is equal to the blue rectangular area.

Solution

(1) The ECF value $\varphi_e(k = 0)$ always indicates the mean error probability $p_{\rm M}$, while the ECF limit for $k → ∞$ is equal to $p_{\rm M}^2$.

From the graph on the information section, one can read $p_{\rm M} \ \underline {= 0.01}$.
In "Exercise 5.6Z", this value is calculated in a different way.

(2) If we insert the parameter $k = 0$ into the lower ECF equation, which is actually only valid for $k > 0$, we obtain the extrapolation value we are looking for.

$$\varphi_{e0} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 + (p_{\rm B}- p_{\rm M}) \cdot (p_{\rm M}- p_{\rm G})\hspace{0.05cm} = 10^{-4} + (0.1- 0.01) \cdot (0.01- 0.001)=10^{-4} + 0.09 \cdot 0.009 \hspace{0.15cm}\underline {\approx 0.091 \cdot 10^{-2}}\hspace{0.05cm}.$$

(3) According to the general definition equation, the following holds for the error correlation period

$$D_{\rm K} = \frac{1}{\varphi_{e0} - p_{\rm M}^2} \cdot \sum_{k = 1 }^{\infty}\hspace{0.1cm} [\varphi_{e}(k) - p_{\rm M}^2]\hspace{0.05cm}.$$

With the expressions

$$A \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (p_{\rm B}- p_{\rm M}) \cdot (p_{\rm M}- p_{\rm G}) = \varphi_{e0} - p_{\rm M}^2\hspace{0.05cm},$$

$$B\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)$$

this equation can be written as follows:

$$D_{\rm K} = {1}/{A} \cdot \sum_{k = 1 }^{\infty}\hspace{0.1cm} A \cdot (1 - B)^k = \sum_{k = 1 }^{\infty}\hspace{0.1cm} (1 - B)^k\hspace{0.05cm}.$$

Using the summation formula of a geometric series, this gives the final result:

$$D_{\rm K} = {1}/{B} - 1 = \frac{1}{{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)} - 1\hspace{0.05cm}.$$

So solution 3 is correct.

(4) With ${\rm Pr(B|G)} = 0.01$ and ${\rm Pr(G|B)} = 0.1$ we get

$$D_{\rm K} = \frac{1}{0.01 + 0.1} - 1 \hspace{0.15cm}\underline {\approx 8.091}\hspace{0.05cm}.$$

(5) Only solution 1 is correct, as shown in the sample solutions to the last subtasks:

Thus the correlation term is fixed, for example:
With ${\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G)} = 0.1$ and $\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}B) = 0.01$ we get the same $D_{\rm K} = 8.091$ as with $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G) = 0.01$ and $\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}B) = 0.1$.
But now the mean error probability $p_{\rm M} \approx 9.1\%$ instead of $1\%$, respectively for $p_{\rm G} = 0.001$ and $p_{\rm B} = 0.1$.
The last statement is also false. This statement would only be true if $\varphi_e(k)$ was plotted linearly and not logarithmically as here.