Difference between revisions of "Aufgaben:Exercise 1.4Z: Sum of Ternary Quantities"

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:$$x ∈ {–2, \ 0, +2},$$
 
:$$x ∈ {–2, \ 0, +2},$$
 
 
:$$y ∈ {–1, \ 0, +1}.$$
 
:$$y ∈ {–1, \ 0, +1}.$$
  
These two ternary values each occur with equal probability.  From this, the sum  $s = x + y$  is formed as a new random variable.
+
*These two ternary values each occur with equal probability.   
 
+
*From this,  the sum  $s = x + y$  is formed as a new random variable.
The adjacent scheme shows that the sum  $s$  can take all integer values between  $–3$  and  $+3$ : 
+
*The adjacent scheme shows that the sum  $s$  can take all integer values between  $–3$  and  $+3$ :
 
 
 
:$$ s \in \{-3, -2, -1, \ 0, +1, +2, +3\}.$$
 
:$$ s \in \{-3, -2, -1, \ 0, +1, +2, +3\}.$$
  
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*The topic of this chapter is illustrated with examples in the   (German language)   learning video
 
*The topic of this chapter is illustrated with examples in the   (German language)   learning video
:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]]   $\Rightarrow$   "Statistical dependence and independence".
+
::[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und Unabhängigkeit]]   $\Rightarrow$   "Statistical dependence and independence".
  
  
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${\rm Pr}\big [(x>0) \cap (s>0)\big] \ = \ $ { 0.3333 3% }
 
${\rm Pr}\big [(x>0) \cap (s>0)\big] \ = \ $ { 0.3333 3% }
  
{Calculate the conditional probability that the input variable  $x > 0$ , when  $s > 0$  holds:
+
{Calculate the conditional probability that the input variable  $x > 0$,  when  $s > 0$  holds:
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(x>0\hspace{0.05cm}|\hspace{0.05cm}s>0)\ = \ $ { 0.75 3% }
 
${\rm Pr}(x>0\hspace{0.05cm}|\hspace{0.05cm}s>0)\ = \ $ { 0.75 3% }
  
{Calculate the conditional probability that the sum  $s$  is positive when the input variable is  $x > 0$ :
+
{Calculate the conditional probability that the sum  $s$  is positive,  when the input variable is  $x > 0$ :
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(s>0\hspace{0.05cm}|\hspace{0.05cm}x>0)\ = \ $ { 1 }
 
${\rm Pr}(s>0\hspace{0.05cm}|\hspace{0.05cm}x>0)\ = \ $ { 1 }
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[[File:P_ID99__Sto_Z_1_4_a.png|right|frame|Ternary variables in the Venn diagram]]
 
[[File:P_ID99__Sto_Z_1_4_a.png|right|frame|Ternary variables in the Venn diagram]]
 
In the adjacent graph
 
In the adjacent graph
*the three fields belonging to the event  $„x > 0“$  are outlined in purple,  
+
*the three fields belonging to the event  $\big[x > 0\big]$  are outlined in purple,  
*the fields for  $„s > 0“$  are highlighted in yellow.
+
*the fields for  $\big[ s > 0\big]$  are highlighted in yellow.
  
  
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'''(3)'''  Using the results of subtasks  '''(1)'''  and  '''(2)''' , it follows:
+
'''(3)'''  Using the results of subtasks  '''(1)'''  and  '''(2)''',  it follows:
 
:$$\rm Pr \big[(\it x > \rm 0) \hspace{0.05cm}| \hspace{0.05cm} (\it s > \rm 0)\big]  =  \frac{{\rm Pr} [(\it x > \rm 0) \cap (\it s > \rm 0)]}{{\rm Pr}(\it s > \rm 0)}= \frac{3/9}{4/9}\hspace{0.15cm}\underline {= 0.75}.$$
 
:$$\rm Pr \big[(\it x > \rm 0) \hspace{0.05cm}| \hspace{0.05cm} (\it s > \rm 0)\big]  =  \frac{{\rm Pr} [(\it x > \rm 0) \cap (\it s > \rm 0)]}{{\rm Pr}(\it s > \rm 0)}= \frac{3/9}{4/9}\hspace{0.15cm}\underline {= 0.75}.$$
  

Latest revision as of 15:41, 30 November 2021

Sum of two ternary variables  $x$  and  $y$

Let be given the ternary random variables

$$x ∈ {–2, \ 0, +2},$$
$$y ∈ {–1, \ 0, +1}.$$
  • These two ternary values each occur with equal probability. 
  • From this,  the sum  $s = x + y$  is formed as a new random variable.
  • The adjacent scheme shows that the sum  $s$  can take all integer values between  $–3$  and  $+3$ :
$$ s \in \{-3, -2, -1, \ 0, +1, +2, +3\}.$$




Hints:

  • The topic of this chapter is illustrated with examples in the  (German language)  learning video
Statistische Abhängigkeit und Unabhängigkeit   $\Rightarrow$   "Statistical dependence and independence".


Questions

1

Calculate the probability that the sum  $s$  is positive:

${\rm Pr}(s>0) \ = \ $

2

Calculate the probability that both the input  $x$  and the sum  $s$  are positive:

${\rm Pr}\big [(x>0) \cap (s>0)\big] \ = \ $

3

Calculate the conditional probability that the input variable  $x > 0$,  when  $s > 0$  holds:

${\rm Pr}(x>0\hspace{0.05cm}|\hspace{0.05cm}s>0)\ = \ $

4

Calculate the conditional probability that the sum  $s$  is positive,  when the input variable is  $x > 0$ :

${\rm Pr}(s>0\hspace{0.05cm}|\hspace{0.05cm}x>0)\ = \ $


Solution

Ternary variables in the Venn diagram

In the adjacent graph

  • the three fields belonging to the event  $\big[x > 0\big]$  are outlined in purple,
  • the fields for  $\big[ s > 0\big]$  are highlighted in yellow.


All sought probabilities can be determined here with the help of the classical definition.

(1)  This event is marked by the fields with yellow background:

$$\rm Pr (\it s > \rm 0) = \rm 4/9 \hspace{0.15cm}\underline { \approx \rm 0.444}.$$


(2)  The following facts hold here:

$$\rm Pr \big[(\it x > \rm 0) \cap (\it s>\rm 0) \big ] = \rm Pr(\it x > \rm 0) =\rm 3/9\hspace{0.15cm}\underline { \approx \rm 0.333}. $$


(3)  Using the results of subtasks  (1)  and  (2),  it follows:

$$\rm Pr \big[(\it x > \rm 0) \hspace{0.05cm}| \hspace{0.05cm} (\it s > \rm 0)\big] = \frac{{\rm Pr} [(\it x > \rm 0) \cap (\it s > \rm 0)]}{{\rm Pr}(\it s > \rm 0)}= \frac{3/9}{4/9}\hspace{0.15cm}\underline {= 0.75}.$$


(4)  Analogous to subtask  (3)  now holds:

$$\rm Pr(\it s > \rm 0 \hspace{0.05cm} | \hspace{0.05cm} \it x > \rm 0)=\frac{Pr \big[(\it x > \rm 0) \cap (\it s > \rm 0) \big]}{Pr(\it x >\rm 0)}=\rm \frac{3/9}{3/9}\hspace{0.15cm}\underline {= 1}.$$