Aufgaben:Exercise 2.3: Algebraic Sum of Binary Numbers

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Considered random generator

A random number generator outputs a binary random number   $x_\nu$  at each clock time  $(\nu)$ , which can be  $0$  or  $1$ .

• The value  "1"  occurs with probability  $p = 0.25$ .
• The individual values  $x_\nu$  are statistically independent of each other.

The binary numbers are stored in a shift register with  $I = 6$  memory cells.

At each clock instant,  the contents of this shift register are shifted one place to the right and the algebraic sum  $y_\nu$  of the shift register contents is formed in each case:

$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$

Hints:

• The exercise belongs to the chapter  Binomial Distribution.

• To check your results you can use the interactive applet  Binomial and Poisson distribution .

Questions

1

What values can the sum  $y$  take?  What is the largest possible value?

$y_\max \ = \ $

2

Calculate the probability that  $y$  is greater than  $2$.

${\rm Pr}(y > 2) \ = \ $

3

What is the mean value of the random variable  $y$?

$m_y \ =$

4

Find the rms value of the random variable  $y$.

$\sigma_y \ = \ $

5

Are the random numbers  $y_\nu$  statistically independent?  Justify your result.

	The random numbers are statistically independent.
	The random numbers are statistically dependent.

6

What is the conditional probability that  $y_\nu = \mu$  if  $y_{\nu-1} = \mu$  occured previously?  $(\mu = 0, \ 1, \ \text{...} \ , \ I)$.

${\rm Pr}(y_\nu = \mu \hspace{0.05cm} | \hspace{0.05cm} y_{\nu-1} = \mu ) \ = \ $

Solution

Solution

(1)  Each cell can contain a  $0$  or a  $1$ .  Therefore,  the sum can take all integer values between  $0$  ánd  $6$ :

$$y_{\nu}\in\{0,1,\ \text{...} \ ,6\}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} y_{\rm max} \hspace{0.15cm} \underline{= 6}.$$

(2)  There is a binomial distribution.  Therefore,  with  $p = 0.25$:

$${\rm Pr}(y =0)=(1-p)^{\it I}=0.75^6=0.178,$$

$${\rm Pr}(y=1)=\left({ I \atop {1}}\right)\cdot (1-p)^{I-1}\cdot p= \rm 6\cdot 0.75^5\cdot 0.25=0.356,$$

$${\rm Pr}(y=2)=\left({ I \atop { 2}}\right)\cdot (1-p)^{I-2}\cdot p^{\rm 2}= \rm 15\cdot 0.75^4\cdot 0.25^2=0.297,$$

$$\Rightarrow \hspace{0.3cm}{\rm Pr}(y>2)=1-{\rm Pr}(y=0)-{\rm Pr}( y=1)-{\rm Pr}( y=2)\hspace{0.15cm} \underline{=\rm 0.169}.$$

(3)  According to the general equation,  the mean of the binomial distribution is:

$$m_y= I\cdot p\hspace{0.15cm} \underline{=\rm 1.5}.$$

(4)  Accordingly,  for the rms value of the binomial distribution:

$$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$

(5)  Correct is the  proposed solution 2:

• If   $y_\nu = 0$, then only the values  $0$  and  $1$  can follow at the next time point,  but not  $2$, ... , $6$.
• That is:   The sequence  $ \langle y_\nu \rangle$  has (strong) statistical bindings.

(6)  The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register.  It follows that:

$${\rm Pr} (y_{\nu} = \mu\hspace{0.05cm}| \hspace{0.05cm} y_{\nu-{1}} = \mu) = {\rm Pr}(x_{\nu}= x_{\nu-6}). $$

• Since the symbols  $x_\nu$  are statistically independent of each other,  we can also write for this:

$${\rm Pr}(x_{\nu} = x_{\nu-6}) = {\rm Pr}\big[(x_{\nu}= 1)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6}= 1)\hspace{0.05cm}\cup \hspace{0.05cm}(x_\nu=0)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6} =0)\big]= p^{2}+(1- p)^{2}=\rm 0.25^2 + 0.75^2\hspace{0.15cm} \underline{ = 0.625}. $$