Exercise 5.3Z: Analysis of the BSC Model

Given error sequence

We consider two different BSC models with the following parameters:

Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
Model $M_2 \text{:} \hspace{0.4cm} p = 0.02$.

The graph shows an error sequence of length $N = 1000$, but it is not known from which of the two models this sequence originates.

The two models are to be analyzed on the basis of

the error distance probabilities

$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$

the error distance distribution

$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$

the error correlation function

$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \ \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$

Notes:

The exercise belongs to the chapter "Binary Symmetric Channel (BSC)".
By counting, we would see that the error sequence of length $N = 1000$ contains exactly $22$ ones.

Questions

	ECF value $\varphi_e(k = 0)$,
	ECF value $\varphi_e(k = 10)$,
	EDD value $V_a(k = 1)$,
	EDD value $V_a(k = 2)$,
	EDD value $V_a(k = 10)$.

	Model $M_1$,
	Model $M_2$.

${\rm E}\big[a\big] \ = \ $

${\rm Pr}(a = 1) \ = \ $

${\rm Pr}(a = 2) \ = \ $

${\rm Pr}(a = {\rm E}\big[a\big]) \ = \ $

$V_a(k = 2) \ = \ $

$V_a(k = 10) \ = \ $

$V_a(k = 11) \ = \ $

Solution

(1) In the BSC model, the mean error probability $p_{\rm M}$ is always equal to the characteristic probability $p$.

For the error correlation function and the error distance distribution are valid

$$\varphi_{e}(k) = \left\{ \begin{array}{c} p \\ p^2 \end{array} \right.\quad \begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm}, \\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array} \hspace{0.4cm}V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$

$p$ can be determined from all the given characteristics, except $V_a(k = 1)$. This EDD value is independent of $p$ equal to $(1–p)^0 = 1$.
Therefore, the solutions 1, 2, 4 and 5 are correct.

(2) The relative error frequency of the given sequence is equal to $h_{\rm F} = 22/1000 \approx 0.022$.

It is quite obvious that the error sequence was generated by the model $M_2$ ⇒ $p_{\rm M} = 0.02$.
Because of the short sequence, $h_{\rm F}$ does not match $p_{\rm M}$ exactly, but at least approximates ⇒ solution 2.

(3) The mean error distance – that is, the expected value of the random variable $a$ – is equal to the inverse of the mean error probability ⇒ ${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}$.

(4) According to the equation ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$ we obtain:

$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {= 0.1}\hspace{0.05cm},$$

$${\rm Pr}(a = 2) = 0.9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$

$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(a = 10)= 0.9^9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.0387}\hspace{0.05cm}.$$

(5) From the relation $V_a(k) = (1–p)^{k–1}$ we obtain

$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k = 1) - V_a(k = 2) = 0.1\hspace{0.05cm},$$

$$V_a(k = 10)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^9 \hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10} \hspace{0.15cm}\underline {=0.3487}.$$

To check in comparison with subtask (4):

$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k = 11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$