Conditional dependence

<h2 id="example">Example</h2>
<p>In essence <a href="/facts/Probability/fgYvMhwb">probability</a> is influenced by a person's information about the possible occurrence of an event. For example, let the event 
  
    
      
        A
      
    
    {\displaystyle A}
  
 be 'I have a new phone'; event 
  
    
      
        B
      
    
    {\displaystyle B}
  
 be 'I have a new watch'; and event 
  
    
      
        C
      
    
    {\displaystyle C}
  
 be 'I am happy'; and suppose that having either a new phone or a new watch increases the probability of my being happy. Let us assume that the event 
  
    
      
        C
      
    
    {\displaystyle C}
  
 has occurred – meaning 'I am happy'. Now if another person sees my new watch, he/she will reason that my likelihood of being happy was increased by my new watch, so there is less need to attribute my happiness to a new phone.
</p><p>To make the example more numerically specific, suppose that there are four possible states 
  
    
      
        Ω
        =
        
          {
          
            
              s
              
                1
              
            
            ,
            
              s
              
                2
              
            
            ,
            
              s
              
                3
              
            
            ,
            
              s
              
                4
              
            
          
          }
        
        ,
      
    
    {\displaystyle \Omega =\left\{s_{1},s_{2},s_{3},s_{4}\right\},}
  
 given in the middle four columns of the following table, in which the occurrence of event 
  
    
      
        A
      
    
    {\displaystyle A}
  
 is signified by a 
  
    
      
        1
      
    
    {\displaystyle 1}
  
 in row 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and its non-occurrence is signified by a 
  
    
      
        0
        ,
      
    
    {\displaystyle 0,}
  
 and likewise for 
  
    
      
        B
      
    
    {\displaystyle B}
  
 and 
  
    
      
        C
        .
      
    
    {\displaystyle C.}
  
 That is, 
  
    
      
        A
        =
        
          {
          
            
              s
              
                2
              
            
            ,
            
              s
              
                4
              
            
          
          }
        
        ,
        B
        =
        
          {
          
            
              s
              
                3
              
            
            ,
            
              s
              
                4
              
            
          
          }
        
        ,
      
    
    {\displaystyle A=\left\{s_{2},s_{4}\right\},B=\left\{s_{3},s_{4}\right\},}
  
 and 
  
    
      
        C
        =
        
          {
          
            
              s
              
                2
              
            
            ,
            
              s
              
                3
              
            
            ,
            
              s
              
                4
              
            
          
          }
        
        .
      
    
    {\displaystyle C=\left\{s_{2},s_{3},s_{4}\right\}.}
  
 The probability of 
  
    
      
        
          s
          
            i
          
        
      
    
    {\displaystyle s_{i}}
  
 is 
  
    
      
        1
        
          /
        
        4
      
    
    {\displaystyle 1/4}
  
 for every 
  
    
      
        i
        .
      
    
    {\displaystyle i.}
  
 
</p>
<table><tbody><tr><th>Event</th><th>                    P        ⁡        (                  s                      1                          )        =        1                  /                4              {\displaystyle \operatorname {P} (s_{1})=1/4}  </th><th>                    P        ⁡        (                  s                      2                          )        =        1                  /                4              {\displaystyle \operatorname {P} (s_{2})=1/4}  </th><th>                    P        ⁡        (                  s                      3                          )        =        1                  /                4              {\displaystyle \operatorname {P} (s_{3})=1/4}  </th><th>                    P        ⁡        (                  s                      4                          )        =        1                  /                4              {\displaystyle \operatorname {P} (s_{4})=1/4}  </th><th>Probability of event</th></tr><tr><td>                    A              {\displaystyle A}  </td><td>0</td><td>1</td><td>0</td><td>1</td><th>                                                        1              2                                            {\displaystyle {\tfrac {1}{2}}}  </th></tr><tr><td>                    B              {\displaystyle B}  </td><td>0</td><td>0</td><td>1</td><td>1</td><th>                                                        1              2                                            {\displaystyle {\tfrac {1}{2}}}  </th></tr><tr><td>                    C              {\displaystyle C}  </td><td>0</td><td>1</td><td>1</td><td>1</td><th>                                                        3              4                                            {\displaystyle {\tfrac {3}{4}}}  </th></tr></tbody></table>
<p>and so
</p>
<table><tbody><tr><th>Event</th><th>                              s                      1                                {\displaystyle s_{1}}  </th><th>                              s                      2                                {\displaystyle s_{2}}  </th><th>                              s                      3                                {\displaystyle s_{3}}  </th><th>                              s                      4                                {\displaystyle s_{4}}  </th><th>Probability of event</th></tr><tr><td>                    A        ∩        B              {\displaystyle A\cap B}  </td><td>0</td><td>0</td><td>0</td><td>1</td><th>                                                        1              4                                            {\displaystyle {\tfrac {1}{4}}}  </th></tr><tr><td>                    A        ∩        C              {\displaystyle A\cap C}  </td><td>0</td><td>1</td><td>0</td><td>1</td><th>                                                        1              2                                            {\displaystyle {\tfrac {1}{2}}}  </th></tr><tr><td>                    B        ∩        C              {\displaystyle B\cap C}  </td><td>0</td><td>0</td><td>1</td><td>1</td><th>                                                        1              2                                            {\displaystyle {\tfrac {1}{2}}}  </th></tr><tr><td>                    A        ∩        B        ∩        C              {\displaystyle A\cap B\cap C}  </td><td>0</td><td>0</td><td>0</td><td>1</td><th>                                                        1              4                                            {\displaystyle {\tfrac {1}{4}}}  </th></tr></tbody></table>
<p>In this example, 
  
    
      
        C
      
    
    {\displaystyle C}
  
 occurs <a href="/facts/If_and_only_if/bYSxGJ66">if and only if</a> at least one of 
  
    
      
        A
        ,
        B
      
    
    {\displaystyle A,B}
  
 occurs. Unconditionally (that is, without reference to 
  
    
      
        C
      
    
    {\displaystyle C}
  
), 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are <a href="/facts/Independence_(probability_theory)/NUzQtnUL">independent</a> of each other because 
  
    
      
        P
        ⁡
        (
        A
        )
      
    
    {\displaystyle \operatorname {P} (A)}
  
—the sum of the probabilities associated with a 
  
    
      
        1
      
    
    {\displaystyle 1}
  
 in row 
  
    
      
        A
      
    
    {\displaystyle A}
  
—is 
  
    
      
        
          
            
              1
              2
            
          
        
        ,
      
    
    {\displaystyle {\tfrac {1}{2}},}
  
 while

P
        ⁡
        (
        A
        ∣
        B
        )
        =
        P
        ⁡
        (
        A
        
           and 
        
        B
        )
        
          /
        
        P
        ⁡
        (
        B
        )
        =
        
          
            
              
                1
                
                  /
                
                4
              
              
                1
                
                  /
                
                2
              
            
          
        
        =
        
          
            
              1
              2
            
          
        
        =
        P
        ⁡
        (
        A
        )
        .
      
    
    {\displaystyle \operatorname {P} (A\mid B)=\operatorname {P} (A{\text{ and }}B)/\operatorname {P} (B)={\tfrac {1/4}{1/2}}={\tfrac {1}{2}}=\operatorname {P} (A).}

But conditional on 
  
    
      
        C
      
    
    {\displaystyle C}
  
 having occurred (the last three columns in the table), we have

P
        ⁡
        (
        A
        ∣
        C
        )
        =
        P
        ⁡
        (
        A
        
           and 
        
        C
        )
        
          /
        
        P
        ⁡
        (
        C
        )
        =
        
          
            
              
                1
                
                  /
                
                2
              
              
                3
                
                  /
                
                4
              
            
          
        
        =
        
          
            
              2
              3
            
          
        
      
    
    {\displaystyle \operatorname {P} (A\mid C)=\operatorname {P} (A{\text{ and }}C)/\operatorname {P} (C)={\tfrac {1/2}{3/4}}={\tfrac {2}{3}}}

while

P
        ⁡
        (
        A
        ∣
        C
        
           and 
        
        B
        )
        =
        P
        ⁡
        (
        A
        
           and 
        
        C
        
           and 
        
        B
        )
        
          /
        
        P
        ⁡
        (
        C
        
           and 
        
        B
        )
        =
        
          
            
              
                1
                
                  /
                
                4
              
              
                1
                
                  /
                
                2
              
            
          
        
        =
        
          
            
              1
              2
            
          
        
        <
        P
        ⁡
        (
        A
        ∣
        C
        )
        .
      
    
    {\displaystyle \operatorname {P} (A\mid C{\text{ and }}B)=\operatorname {P} (A{\text{ and }}C{\text{ and }}B)/\operatorname {P} (C{\text{ and }}B)={\tfrac {1/4}{1/2}}={\tfrac {1}{2}}<\operatorname {P} (A\mid C).}

Since in the presence of 
  
    
      
        C
      
    
    {\displaystyle C}
  
 the probability of 
  
    
      
        A
      
    
    {\displaystyle A}
  
 is affected by the presence or absence of 
  
    
      
        B
        ,
        A
      
    
    {\displaystyle B,A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are mutually dependent conditional on 
  
    
      
        C
        .
      
    
    {\displaystyle C.}

</p>
<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Conditional_independence/udUXFaMw">Conditional independence</a> – Probability theory concept</li>
<li><a href="/facts/De_Finetti%2527s_theorem/dLpasxgn">de Finetti's theorem</a> – Conditional independence of exchangeable observations</li>
<li><a href="/facts/Conditional_expectation/3GxG1Q3x">Conditional expectation</a> – Expected value of a random variable given that certain conditions are known to occur</li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>Introduction to Artificial Intelligence by Sebastian Thrun and Peter Norvig, 2011 "Unit 3: Conditional Dependence"[permanent dead link] <a href="https://www.ai-class.com/course/video/videolecture/33" target="_blank">https://www.ai-class.com/course/video/videolecture/33</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
<li id="fn:2"><p>Introduction to learning Bayesian Networks from Data by Dirk Husmeier [1][permanent dead link] "Introduction to Learning Bayesian Networks from Data -Dirk Husmeier" <a href="http://www.bioss.sari.ac.uk/staff/dirk/papers/sbb_bnets.pdf" target="_blank">http://www.bioss.sari.ac.uk/staff/dirk/papers/sbb_bnets.pdf</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li>
<li id="fn:3"><p>Conditional Independence in Statistical theory "Conditional Independence in Statistical Theory", A. P. Dawid" Archived 2013-12-27 at the Wayback Machine <a href="http://edlab-www.cs.umass.edu/cs589/2010-lectures/conditional%20independence%20in%20statistical%20theory.pdf" target="_blank">http://edlab-www.cs.umass.edu/cs589/2010-lectures/conditional%20independence%20in%20statistical%20theory.pdf</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li>
<li id="fn:4"><p>Probabilistic independence on Britannica "Probability->Applications of conditional probability->independence (equation 7) " <a href="http://www.britannica.com/EBchecked/topic/477530/probability-theory/32768/Applications-of-conditional-probability#toc32769" target="_blank">http://www.britannica.com/EBchecked/topic/477530/probability-theory/32768/Applications-of-conditional-probability#toc32769</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li>
<li id="fn:5"><p>Introduction to Artificial Intelligence by Sebastian Thrun and Peter Norvig, 2011 "Unit 3: Explaining Away"[permanent dead link] <a href="https://www.ai-class.com/course/video/quizquestion/60" target="_blank">https://www.ai-class.com/course/video/quizquestion/60</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li>
<li id="fn:6"><p>Bouckaert, Remco R. (1994). "11. Conditional dependence in probabilistic networks". In Cheeseman, P.; Oldford, R. W. (eds.). Selecting Models from Data, Artificial Intelligence and Statistics IV. Lecture Notes in Statistics. Vol. 89. Springer-Verlag. pp. 101–111, especially 104. ISBN 978-0-387-94281-0. <a href="978-0-387-94281-0" target="_blank">978-0-387-94281-0</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li>
<li id="fn:7"><p>Conditional Independence in Statistical theory "Conditional Independence in Statistical Theory", A. P. Dawid Archived 2013-12-27 at the Wayback Machine <a href="http://edlab-www.cs.umass.edu/cs589/2010-lectures/conditional%20independence%20in%20statistical%20theory.pdf" target="_blank">http://edlab-www.cs.umass.edu/cs589/2010-lectures/conditional%20independence%20in%20statistical%20theory.pdf</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li>
</ol>

Conditional dependence open-in-new

Conditional dependence