Exercise 1.7Z: BARBARA Generator

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$BARBARA$ Generator

Here we consider a ternary random generator with symbols  $A$,  $B$  and  $R$, which can be described by a homogeneous and stationary first order Markov chain.

• The transition probabilities can be taken from the sketched Markov diagram.
• For the first three subtasks,  $p = 1/4$  should always hold.

Hints:

• The task belongs to the chapter  Markov Chains.

Questions

1

Which of the following statements are true?

	The values of  $p > 0$  and  $q < 1$  are largely arbitrary.
	For the transition probabilities, the following must hold:   $p + q = 1$.
	All symbols have equal ergodic probabilities.
	It holds here:  ${\rm Pr}(A) = 1/2, \; {\rm Pr}(B) = 1/3, \; {\rm Pr}(R) = 1/6$.

2

What are the conditional probabilities  $p_{\rm A}$,  $p_{\rm B}$  and  $p_{\rm C}$ that at times between  $ν+1$  and  $ν+7$  the sequence  $BARBARA$  is output,
if one is in state im Zustand  $A$,  $B$  or  $R$ , respectively,at time  $ν$ ?  Let  $p = 1/4$.

$p_{\rm A} \ = \ $ $\ \cdot 10^{-3}$

$p_{\rm B} \ = \ $ $\ \cdot 10^{-3}$

$p_{\rm C} \ = \ $ $\ \cdot 10^{-3}$

3

What is the overall probability that the generator outputs the sequence  $BARBARA$  ausgibt?
Let  $p = 1/4$ continue to hold.

${\rm Pr}(BARBARA)\ = \ $ $\ \cdot 10^{-3}$

4

How should the parameter  $p_{\rm opt}$  be chosen to make  ${\rm Pr}(BARBARA)$  as large as possible?
What is the resulting probability for  $BARBARA$?

$p_{\rm opt} \ = \ $

$p = p_{\rm opt}\hspace{-0.1cm}: \hspace{0.3cm}{\rm Pr}(BARBARA)\ = \ $ $\ \cdot 10^{-3}$

Musterlösung

Solution

(1)  Correct are the second and third suggested solutions:

• The sum of all outgoing arrows must always be  $1$ . Therefore  $q = 1 - p$ holds.
• Because of the symmetry of the Markov diagram, the ergodic probabilities are all equal:

$${\rm Pr}(A) ={\rm Pr}(B) ={\rm Pr}(R) = 1/3.$$

(2)  If one is in the state  $B$  at the starting time  $\nu=1$  because of  ${\rm Pr}(B\hspace{0.05cm}|\hspace{0.05cm}B) = 0$ the state  $B$  is not possible.

• One fails here already with the initial letter $B$:

$$p_{\rm B} \; \underline{ =0}.$$

• For the calculation of  $p_{\rm A}$  it should be noted:   Starting from  $A$  one goes in the Markov diagram first to  $B$  $($with probability $q)$, then five times clockwise  $($each time with probability $p)$  and finally from  $R$  to  $A$  $($with probability  $q)$.   Meaning:

$$p_{\rm A} = q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 = 3^2 / 4^7 \hspace{0.15cm}\underline {\approx 0.549 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$

• In a similar way, one obtains:

$$p_{\rm R} = q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 = 3 / 4^7 \hspace{0.15cm}\underline {\approx 0.183 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$

(3)  By averaging over the conditional probabilities we obtain:

$${\rm Pr}(BARBARA) = p_{\rm A} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(A) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm B} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(B) \hspace{0.1cm} + \hspace{0.1cm}p_{\rm R} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm Pr}(R).$$

This leads to the result:

$${\rm Pr}(BARBARA) = {1}/{3} \cdot \left( q^2 \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 \hspace{0.1cm} +\hspace{0.1cm}0 \hspace{0.1cm} +\hspace{0.1cm}q \hspace{0.05cm}\cdot \hspace{0.05cm} p^6 \right) = \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \cdot{p+q} = \hspace{-0.15cm} \frac{q \hspace{0.05cm}\cdot \hspace{0.05cm} p^5 }{3} \hspace{0.15cm}\underline { \approx 0.244 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$

(4)  The probability calculated in  (3)  is  $p^5 \cdot (1-p)/3$, where  $q= 1-p$  is considered.

• By setting the differential to zero, we obtain the governing equation:

$$5 \cdot p^4 - 6 \cdot p^5 = 0 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} p_{\rm opt} = 5/6 \hspace{0.15cm}\underline { \approx \rm 0.833}.$$

• This results in a value that is larger than the subtask  (3)  by a factor  $90$  approximately:

$${\rm Pr}(BARBARA) \hspace{0.15cm}\underline { \approx 22 \hspace{0.05cm}\cdot \hspace{0.05cm} 10^{-3}}.$$