Theory of Stochastic Signals/Statistical Dependence and Independence - Revision history

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← Older revision		Revision as of 14:15, 13 December 2022
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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".

@@ Line 166: / Line 166: @@
 *If we meet a female student, we can use the inference probability
 :$${\rm Pr}(L \hspace{-0.05cm}\mid  \hspace{-0.05cm}F) = \frac{ {\rm Pr}(F \hspace{-0.05cm}\mid  \hspace{-0.05cm}L)\cdot  {\rm Pr}(L) }{ {\rm Pr}(F \hspace{-0.05cm}\mid  \hspace{-0.05cm}L) \cdot  {\rm Pr}(L) +{\rm Pr}(F \hspace{-0.05cm}\mid  \hspace{-0.05cm}T) \cdot  {\rm Pr}(T)}=\rm \frac{0.6\cdot 0.7}{0.6\cdot 0.7 + 0.1\cdot 0.3}=\frac{14}{15}\approx 93.3 \%$$
-:predict that she will study at LMU.&nbsp; A quite realistic result&nbsp; (at least in the past).}}
+:to predict that she will study at LMU.&nbsp; A quite realistic result&nbsp; (at least in the past).}}

@@ Line 149: / Line 149: @@
 {{BlaueBox|TEXT=
-$\text{Definition:}$&nbsp; From this equation, using&nbsp;  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability|$\text{Bayes' theorem:}$]] &nbsp;  &rArr; &nbsp; &raquo;'''Inference probability'''&laquo;:
+$\text{Definition:}$&nbsp; From this equation, using&nbsp;  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_probability|$\text{Bayes' theorem:}$]] &nbsp;  &rArr; &nbsp; &raquo;'''Inference probability'''&laquo;:
 :$${\rm Pr}(A_i \hspace{0.05cm} \vert \hspace{0.05cm} B) = \frac{ {\rm Pr}( B \mid A_i)\cdot {\rm Pr}(A_i )}{ {\rm Pr}(B)} = \frac{ {\rm Pr}(B \hspace{0.05cm} \vert \hspace{0.05cm} A_i)\cdot {\rm Pr}(A_i )}{\sum_{k={\rm1} }^{I}{\rm Pr}(B \hspace{0.05cm} \vert \hspace{0.05cm} A_k)\cdot{\rm Pr}(A_k) }.$$}}

@@ Line 87: / Line 87: @@
-and refer again to the event of&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#General_Definition_of_Statistical_Dependence|$\text{Example 1}$]]:&nbsp;
+and refer again to the event of&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#General_definition_of_statistical_dependence|$\text{Example 1}$]]:&nbsp;
 *$A_3$:&nbsp; The sum of the two cubes is&nbsp; $S = 7$ &nbsp; &rArr; &nbsp; ${\rm Pr}(A_3) = 1/6$.
@@ Line 98: / Line 98: @@
 *This last result can also be derived using&nbsp; &raquo;'''Bayes'&nbsp; theorem'''&laquo;, &nbsp; for example:
 :$${\rm Pr}(A_4 \hspace{0.05cm} \vert\hspace{0.05cm} A_1) =  \frac{ {\rm Pr}(A_1 \hspace{0.05cm} \vert\hspace{0.05cm} A_4)\cdot {\rm Pr} ( A_4)} {  {\rm Pr}(A_1)}  = \frac{2/5 \cdot 5/36}{1/2}  = 1/9.$$
-*In contrast,&nbsp; the following conditional probabilities hold for&nbsp; $A_1$&nbsp; and the statistically independent event&nbsp; $A_3$,&nbsp; see&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abhängigkeit_und_Unabhängigkeit#General_Definition_of_Statistical_Dependence|$\text{Example 1}$]]:
+*In contrast,&nbsp; the following conditional probabilities hold for&nbsp; $A_1$&nbsp; and the statistically independent event&nbsp; $A_3$,&nbsp; see&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#General_definition_of_statistical_dependence|$\text{Example 1}$]]:
 :$${\rm Pr}(A_{\rm 1} \hspace{0.05cm}\vert \hspace{0.05cm} A_{\rm 3}) = {\rm Pr}(A_{\rm 1}) = \rm 1/2\hspace{0.5cm}{\rm bzw.}\hspace{0.5cm}{\rm Pr}(A_{\rm 3} \hspace{0.05cm} \vert \hspace{0.05cm} A_{\rm 1}) = {\rm Pr}(A_{\rm 3}) = 1/6.$$}}

@@ Line 6: / Line 6: @@
 }}
-==General Definition of Statistical Dependence==
+==General definition of statistical dependence==
 <br>
 So far we have not paid much attention to&nbsp; "statistical dependence"&nbsp;  between events,&nbsp; even though we have already used it as in the case of two disjoint sets: &nbsp; If an element belongs to&nbsp; $A$,&nbsp; it cannot with certainty also be contained in the disjoint set&nbsp; $B$.
@@ Line 52: / Line 52: @@
 '''This example shows that disjunctivity is a particularly pronounced form of statistical dependence'''. }}
-==Conditional Probability==
+==Conditional probability==
 <br>
 If there are statistical bindings between the two events&nbsp; $A$&nbsp; and&nbsp; $B$,&nbsp; the&nbsp; (unconditional)&nbsp; probabilities&nbsp; ${\rm Pr}(A)$&nbsp; and&nbsp; ${\rm Pr}(B)$&nbsp; do not describe the situation unambiguously in the statistical sense.&nbsp; So-called&nbsp; "conditional probabilities"&nbsp; are then required.

@@ Line 10: / Line 10: @@
 So far we have not paid much attention to&nbsp; "statistical dependence"&nbsp;  between events,&nbsp; even though we have already used it as in the case of two disjoint sets: &nbsp; If an element belongs to&nbsp; $A$,&nbsp; it cannot with certainty also be contained in the disjoint set&nbsp; $B$.
-The strongest form of dependence at all is such a&nbsp; '''deterministic dependence'''&nbsp; between two sets or two events.&nbsp; Less pronounced is the statistical dependence.&nbsp; Let us start with its complement:
+The strongest form of dependence at all is such a&nbsp; &raquo;'''deterministic dependence'''&laquo;&nbsp; between two sets or two events.&nbsp; Less pronounced is the statistical dependence.&nbsp; Let us start with its complement:
 {{BlaueBox|TEXT=
 $\text{Definitions:}$&nbsp;
-*Two events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are called&nbsp; '''statistically independent''',&nbsp; if the probability of the intersection&nbsp; $A ∩ B$&nbsp;  is equal to the product of the individual probabilities:
+*Two events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are called&nbsp; &raquo;'''statistically independent'''&laquo;,&nbsp; if the probability of the intersection&nbsp; $A ∩ B$&nbsp;  is equal to the product of the individual probabilities:
 :$${\rm Pr}(A \cap B) = {\rm Pr}(A)\cdot {\rm Pr}(B).$$
-*If this condition is not satisfied,&nbsp; then the events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are '''statistically dependent''':
+*If this condition is not satisfied,&nbsp; then the events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are &raquo;'''statistically dependent'''&laquo;:
 :$${\rm Pr}(A \cap B) \ne {\rm Pr}(A)\cdot {\rm Pr}(B).$$}}
@@ Line 59: / Line 59: @@
 $\text{Definitions:}$&nbsp;
-*The&nbsp; '''conditional probability''' of&nbsp; $A$&nbsp; under condition&nbsp; $B$&nbsp; can be calculated as follows:
+*The&nbsp; &raquo;'''conditional probability'''&laquo; of&nbsp; $A$&nbsp; under condition&nbsp; $B$&nbsp; can be calculated as follows:
 :$${\rm Pr}(A\hspace{0.05cm} \vert \hspace{0.05cm} B) = \frac{ {\rm Pr}(A \cap B)}{ {\rm Pr}(B)}.$$
@@ Line 65: / Line 65: @@
 :$${\rm Pr}(B\hspace{0.05cm} \vert \hspace{0.05cm}A) = \frac{ {\rm Pr}(A \cap B)}{ {\rm Pr}(A)}.$$
-*Combining these two equations,&nbsp; we get&nbsp; [https://en.wikipedia.org/wiki/Thomas_Bayes Bayes']&nbsp; Theorem:
+*Combining these two equations,&nbsp; we get&nbsp; [https://en.wikipedia.org/wiki/Thomas_Bayes $\text{Bayes'}$]&nbsp; theorem:
 :$${\rm Pr}(B \hspace{0.05cm} \vert \hspace{0.05cm} A) = \frac{ {\rm Pr}(A\hspace{0.05cm} \vert \hspace{0.05cm} B)\cdot {\rm Pr}(B)}{ {\rm Pr}(A)}.$$}}
@@ Line 71: / Line 71: @@
 Below are some properties of conditional probabilities:
 *Also a conditional probability lies always  between&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; including these two limits: &nbsp; $0 \le {\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) \le 1$.
-*With constant  condition&nbsp; $B$,&nbsp; all calculation rules given in the chapter&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics|Set Theory Basics]]&nbsp;  for the unconditional probabilities&nbsp; ${\rm Pr}(A)$&nbsp; and&nbsp; ${\rm Pr}(B)$&nbsp; still apply.
+*With constant  condition&nbsp; $B$,&nbsp; all calculation rules given in the chapter&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics|"Set Theory Basics"]]&nbsp;  for the unconditional probabilities&nbsp; ${\rm Pr}(A)$&nbsp; and&nbsp; ${\rm Pr}(B)$&nbsp; still apply.
 *If the existing events&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; are disjoint,&nbsp; then&nbsp; ${\rm Pr}(A\hspace{0.05cm} | \hspace{0.05cm} B) = {\rm Pr}(B\hspace{0.05cm} | \hspace{0.05cm}A)= 0$&nbsp; $($ event&nbsp; $A$&nbsp; "exists"&nbsp; if&nbsp; ${\rm Pr}(A) > 0)$.
 *If&nbsp; $B$&nbsp; is a proper or improper subset of&nbsp; $A$,&nbsp; then&nbsp; ${\rm Pr}(A \hspace{0.05cm} | \hspace{0.05cm} B) =1$. &nbsp;
@@ Line 87: / Line 87: @@
-and refer again to the event of&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#General_Definition_of_Statistical_Dependence|Example 1]]:&nbsp;
+and refer again to the event of&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#General_Definition_of_Statistical_Dependence|$\text{Example 1}$]]:&nbsp;
 *$A_3$:&nbsp; The sum of the two cubes is&nbsp; $S = 7$ &nbsp; &rArr; &nbsp; ${\rm Pr}(A_3) = 1/6$.
@@ Line 96: / Line 96: @@
 *Since the events&nbsp; $A_1$&nbsp; and&nbsp; $A_4$&nbsp; are statistically dependent,&nbsp; the conditional probability&nbsp; ${\rm Pr}(A_1 \hspace{0.05cm}\vert \hspace{0.05cm} A_4) = 2/5$&nbsp;  (two of the five squares outlined in green are highlighted in red)&nbsp; is not equal to the absolute probability&nbsp; ${\rm Pr}(A_1) = 1/2$&nbsp; (half of all squares are highlighted in red).
 *Similarly,&nbsp; the conditional probability&nbsp; ${\rm Pr}(A_4 \hspace{0.05cm} \vert \hspace{0.05cm} A_1) = 2/18 = 1/9$&nbsp;  (two of the&nbsp; $18$&nbsp; fields with a red background are outlined in green)&nbsp; is unequal to the absolute probability&nbsp; ${\rm Pr}(A_4) = 5/36$&nbsp; (a total of five of the&nbsp; $36$&nbsp; fields are outlined in green).
-*This last result can also be derived using&nbsp; '''Bayes'&nbsp; Theorem''', &nbsp; for example:
+*This last result can also be derived using&nbsp; &raquo;'''Bayes'&nbsp; theorem'''&laquo;, &nbsp; for example:
 :$${\rm Pr}(A_4 \hspace{0.05cm} \vert\hspace{0.05cm} A_1) =  \frac{ {\rm Pr}(A_1 \hspace{0.05cm} \vert\hspace{0.05cm} A_4)\cdot {\rm Pr} ( A_4)} {  {\rm Pr}(A_1)}  = \frac{2/5 \cdot 5/36}{1/2}  = 1/9.$$
-*In contrast,&nbsp; the following conditional probabilities hold for&nbsp; $A_1$&nbsp; and the statistically independent event&nbsp; $A_3$,&nbsp; see&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abhängigkeit_und_Unabhängigkeit#General_Definition_of_Statistical_Dependence| Example 1]]:
+*In contrast,&nbsp; the following conditional probabilities hold for&nbsp; $A_1$&nbsp; and the statistically independent event&nbsp; $A_3$,&nbsp; see&nbsp; [[Theory_of_Stochastic_Signals/Statistische_Abhängigkeit_und_Unabhängigkeit#General_Definition_of_Statistical_Dependence|$\text{Example 1}$]]:
 :$${\rm Pr}(A_{\rm 1} \hspace{0.05cm}\vert \hspace{0.05cm} A_{\rm 3}) = {\rm Pr}(A_{\rm 1}) = \rm 1/2\hspace{0.5cm}{\rm bzw.}\hspace{0.5cm}{\rm Pr}(A_{\rm 3} \hspace{0.05cm} \vert \hspace{0.05cm} A_{\rm 1}) = {\rm Pr}(A_{\rm 3}) = 1/6.$$}}
@@ Line 104: / Line 104: @@
 ==General multiplication theorem==
 <br>
-We consider several events denoted as&nbsp;  $A_i$&nbsp; with&nbsp; $1 ≤ i ≤ I$.&nbsp; However,&nbsp; these events&nbsp; $A_i$&nbsp; now no longer represent a&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Complete_system|"complete system"]]&nbsp;, viz,
+We consider several events denoted as&nbsp;  $A_i$&nbsp; with&nbsp; $1 ≤ i ≤ I$.&nbsp; However,&nbsp; these events&nbsp; $A_i$&nbsp; now no longer represent a&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Complete_system|$\text{complete system}$]]&nbsp;, viz,
 *they are not pairwise disjoint to each other,&nbsp; and
 *there may also be statistical bindings between the individual events.
@@ Line 111: / Line 111: @@
 {{BlaueBox|TEXT=
 $\text{Definition:}$&nbsp;
-For the so-called&nbsp; '''joint probability''', i.e. for the probability of the intersection of all&nbsp; $I$&nbsp; events&nbsp; $A_i$,&nbsp; holds in this case:
+For the so-called&nbsp; &raquo;'''joint probability'''&laquo;, i.e. for the probability of the intersection of all&nbsp; $I$&nbsp; events&nbsp; $A_i$,&nbsp; holds in this case:
 :$${\rm Pr}(A_{\rm 1} \cap \hspace{0.02cm}\text{ ...}\hspace{0.1cm} \cap A_{I})  =
   {\rm Pr}(A_{I})\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm Pr}(A_{I \rm -1} \hspace{0.05cm}\vert  \hspace{0.05cm} A_I) \hspace{0.05cm}\cdot \hspace{0.05cm}{\rm Pr}(A_{I \rm -2} \hspace{0.05cm}\vert\hspace{0.05cm} A_{I - \rm 1}\cap A_I)\hspace{0.05cm} \cdot  \hspace{0.02cm}\text{ ...}  \hspace{0.1cm}  \cdot\hspace{0.05cm} {\rm Pr}(A_{\rm 1} \hspace{0.05cm}\vert  \hspace{0.05cm}A_{\rm 2} \cap \hspace{0.02cm}\text{ ...}  \hspace{0.1cm}\cap A_{ I}).$$
@@ Line 134: / Line 134: @@
 ==Inference probability==
 <br>
-Given again events&nbsp; $A_i$&nbsp; with&nbsp; $1 ≤ i ≤ I$&nbsp; that form a&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Complete_system|"complete system"]].&nbsp; That is:
+Given again events&nbsp; $A_i$&nbsp; with&nbsp; $1 ≤ i ≤ I$&nbsp; that form a&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Complete_system|$\text{complete system}$]].&nbsp; That is:
 *All events are pairwise disjoint&nbsp; $(A_i ∩ A_j = ϕ$&nbsp; for all&nbsp; $i ≠ j$&nbsp;).
 *The union gives the universal set:
@@ Line 149: / Line 149: @@
 {{BlaueBox|TEXT=
-$\text{Definition:}$&nbsp; From this equation, using&nbsp;  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability|Bayes' theorem:]] &nbsp;  &rArr; &nbsp; '''Inference probability''':
+$\text{Definition:}$&nbsp; From this equation, using&nbsp;  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability|$\text{Bayes' theorem:}$]] &nbsp;  &rArr; &nbsp; &raquo;'''Inference probability'''&laquo;:
 :$${\rm Pr}(A_i \hspace{0.05cm} \vert \hspace{0.05cm} B) = \frac{ {\rm Pr}( B \mid A_i)\cdot {\rm Pr}(A_i )}{ {\rm Pr}(B)} = \frac{ {\rm Pr}(B \hspace{0.05cm} \vert \hspace{0.05cm} A_i)\cdot {\rm Pr}(A_i )}{\sum_{k={\rm1} }^{I}{\rm Pr}(B \hspace{0.05cm} \vert \hspace{0.05cm} A_k)\cdot{\rm Pr}(A_k) }.$$}}
@@ Line 170: / Line 170: @@
 The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video
-:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|Statistische Abhängigkeit und  Unabhängigkeit]] &nbsp; $\Rightarrow$ &nbsp; "Statistical Dependence and Independence".
+:[[Statistische_Abhängigkeit_und_Unabhängigkeit_(Lernvideo)|"Statistische Abhängigkeit und Unabhängigkeit"]] &nbsp; $\Rightarrow$ &nbsp; "Statistical Dependence and Independence".
 ==Exercises for the chapter==

@@ Line 82: / Line 82: @@
 [[File:EN_Sto_T_1_3_S2.png |frame| Example of statistically dependent events|right]]
-Here we consider ties between the two events
+Here we consider bindings between the two events
 *$A_1$:&nbsp; The outcome of the red cube is&nbsp; $R < 4$&nbsp; (red background) &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1) = 1/2$,
 *$A_4$:&nbsp; The sum of the two cubes is&nbsp; $S = 8$&nbsp; (green outline)  &nbsp; &rArr; &nbsp; ${\rm Pr}(A_4) = 5/36$,

@@ Line 46: / Line 46: @@
 *The two events&nbsp; $A_1$&nbsp; and&nbsp; $A_2$&nbsp; are statistically independent because the probability&nbsp; ${\rm Pr}(A_1 ∩ A_2) = 1/6$&nbsp; of the intersection is equal to the product of the two individual probabilities&nbsp; ${\rm Pr}(A_1) = 1/2$&nbsp; and&nbsp; ${\rm Pr}(A_2) = 1/3$&nbsp;.&nbsp; Given the problem definition,&nbsp; any other result would also have been very surprising.
 *But also the events&nbsp; $A_1$&nbsp; and&nbsp; $A_3$&nbsp; are statistically independent because of&nbsp; ${\rm Pr}(A_1) = 1/2$,&nbsp; ${\rm Pr}(A_3) = 1/6$&nbsp; and&nbsp; ${\rm Pr}(A_1 ∩ A_3) = 1/12$.&nbsp; The probability of intersection&nbsp; $(1/12)$&nbsp; arises because three of the&nbsp; $36$&nbsp; squares are both highlighted in red and outlined in green.
-*In contrast,&nbsp; there are statistical ties between&nbsp; $A_1$&nbsp; and&nbsp; $A_4$&nbsp; because the probability of intersection &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1 ∩ A_4) = 1/18 = 4/72$&nbsp; is not equal to the product&nbsp; ${\rm Pr}(A_1) \cdot {\rm Pr}(A_4)= 1/2 \cdot 5/36 = 5/72$&nbsp;.
+*In contrast,&nbsp; there are statistical bindings between&nbsp; $A_1$&nbsp; and&nbsp; $A_4$&nbsp; because the probability of intersection &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1 ∩ A_4) = 1/18 = 4/72$&nbsp; is not equal to the product&nbsp; ${\rm Pr}(A_1) \cdot {\rm Pr}(A_4)= 1/2 \cdot 5/36 = 5/72$&nbsp;.
 *The two events&nbsp; $A_1$&nbsp; and&nbsp; $A_5$&nbsp; are even disjoint &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1 ∩ A_5) = 0$: &nbsp; none of the boxes with red background is labeled&nbsp; $S=10$&nbsp;.
@@ Line 54: / Line 54: @@
 ==Conditional Probability==
 <br>
-If there are statistical ties between the two events&nbsp; $A$&nbsp; and&nbsp; $B$,&nbsp; the&nbsp; (unconditional)&nbsp; probabilities&nbsp; ${\rm Pr}(A)$&nbsp; and&nbsp; ${\rm Pr}(B)$&nbsp; do not describe the situation unambiguously in the statistical sense.&nbsp; So-called&nbsp; "conditional probabilities"&nbsp; are then required.
+If there are statistical bindings between the two events&nbsp; $A$&nbsp; and&nbsp; $B$,&nbsp; the&nbsp; (unconditional)&nbsp; probabilities&nbsp; ${\rm Pr}(A)$&nbsp; and&nbsp; ${\rm Pr}(B)$&nbsp; do not describe the situation unambiguously in the statistical sense.&nbsp; So-called&nbsp; "conditional probabilities"&nbsp; are then required.
 {{BlaueBox|TEXT=
@@ Line 92: / Line 92: @@
 Regarding this graph,&nbsp; note:
-*There are statistical ties between  the both events&nbsp; $A_1$&nbsp; and&nbsp; $A_4$,&nbsp; since the probability of intersection  &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1 ∩ A_4) = 2/36 = 4/72$&nbsp; is not equal to the product&nbsp; ${\rm Pr}(A_1) \cdot {\rm Pr}(A_4)= 1/2 \cdot 5/36 = 5/72$&nbsp;.
+*There are statistical bindings between  the both events&nbsp; $A_1$&nbsp; and&nbsp; $A_4$,&nbsp; since the probability of intersection  &nbsp; &rArr; &nbsp; ${\rm Pr}(A_1 ∩ A_4) = 2/36 = 4/72$&nbsp; is not equal to the product&nbsp; ${\rm Pr}(A_1) \cdot {\rm Pr}(A_4)= 1/2 \cdot 5/36 = 5/72$&nbsp;.
 *The conditional probability&nbsp; ${\rm Pr}(A_1 \hspace{0.05cm} \vert \hspace{0.05cm} A_4) = 2/5$&nbsp; can be calculated from the quotient of the joint probability&nbsp; ${\rm Pr}(A_1 ∩ A_4) = 2/36$&nbsp; and the probability &nbsp; ${\rm Pr}(A_4) = 5/36$&nbsp;.
 *Since the events&nbsp; $A_1$&nbsp; and&nbsp; $A_4$&nbsp; are statistically dependent,&nbsp; the conditional probability&nbsp; ${\rm Pr}(A_1 \hspace{0.05cm}\vert \hspace{0.05cm} A_4) = 2/5$&nbsp;  (two of the five squares outlined in green are highlighted in red)&nbsp; is not equal to the absolute probability&nbsp; ${\rm Pr}(A_1) = 1/2$&nbsp; (half of all squares are highlighted in red).
@@ Line 106: / Line 106: @@
 We consider several events denoted as&nbsp;  $A_i$&nbsp; with&nbsp; $1 ≤ i ≤ I$.&nbsp; However,&nbsp; these events&nbsp; $A_i$&nbsp; now no longer represent a&nbsp; [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Complete_system|"complete system"]]&nbsp;, viz,
 *they are not pairwise disjoint to each other,&nbsp; and
-*there may also be statistical ties between the individual events.
+*there may also be statistical bindings between the individual events.