Aufgaben:Exercise 5.3Z: Analysis of the BSC Model

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The 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  $\rm (EDD)$

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

• the  error correlation function  $\rm (ECF)$

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

Notes:

• The exercise belongs to the chapter  "Binary Symmetric Channel".

• By counting,  we would see that the error sequence of length  $N = 1000$  contains exactly  $22$  "ones".

Questions

1

Which parameters can be used to infer the mean error probability  $p_{\rm M}$  of the BSC model?

	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  $V_a(k = 10)$.

2

From which model does the given error sequence originate?

	Model $M_1$,
	model $M_2$.

3

What is the mean error distance of model  $M_1$?

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

4

What are the following probabilities for model  $M_1$? 

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

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

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

5

Calculate the following values of the error distance distribution for model  $M_1$: 

$V_a(k = 2) \ = \ $

$V_a(k = 10) \ = \ $

$V_a(k = 11) \ = \ $

Solution

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

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

$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$

• $p$  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):

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