From independence to conditional independence












1












$begingroup$


Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.



Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.






probability probability-theory random-variables independence







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$endgroup$








  • 2




    $begingroup$
    No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
    $endgroup$
    – lulu
    Apr 27 '17 at 15:03












  • $begingroup$
    Is in your example $X$ independent of $Z$?
    $endgroup$
    – STF
    Apr 27 '17 at 15:06






  • 1




    $begingroup$
    Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
    $endgroup$
    – lulu
    Apr 27 '17 at 15:07










  • $begingroup$
    OK, what additional condition I would need to go from independence to conditional independence?
    $endgroup$
    – STF
    Apr 27 '17 at 15:08










  • $begingroup$
    Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
    $endgroup$
    – lulu
    Apr 27 '17 at 15:09
















1












$begingroup$


Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.



Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.






probability probability-theory random-variables independence







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
    $endgroup$
    – lulu
    Apr 27 '17 at 15:03












  • $begingroup$
    Is in your example $X$ independent of $Z$?
    $endgroup$
    – STF
    Apr 27 '17 at 15:06






  • 1




    $begingroup$
    Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
    $endgroup$
    – lulu
    Apr 27 '17 at 15:07










  • $begingroup$
    OK, what additional condition I would need to go from independence to conditional independence?
    $endgroup$
    – STF
    Apr 27 '17 at 15:08










  • $begingroup$
    Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
    $endgroup$
    – lulu
    Apr 27 '17 at 15:09














1












1








1


1



$begingroup$


Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.



Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.






probability probability-theory random-variables independence







share|cite|improve this question









$endgroup$




Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.



Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.






probability probability-theory random-variables independence




probability probability-theory random-variables independence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 27 '17 at 14:55









STFSTF

681420




681420








  • 2




    $begingroup$
    No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
    $endgroup$
    – lulu
    Apr 27 '17 at 15:03












  • $begingroup$
    Is in your example $X$ independent of $Z$?
    $endgroup$
    – STF
    Apr 27 '17 at 15:06






  • 1




    $begingroup$
    Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
    $endgroup$
    – lulu
    Apr 27 '17 at 15:07










  • $begingroup$
    OK, what additional condition I would need to go from independence to conditional independence?
    $endgroup$
    – STF
    Apr 27 '17 at 15:08










  • $begingroup$
    Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
    $endgroup$
    – lulu
    Apr 27 '17 at 15:09














  • 2




    $begingroup$
    No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
    $endgroup$
    – lulu
    Apr 27 '17 at 15:03












  • $begingroup$
    Is in your example $X$ independent of $Z$?
    $endgroup$
    – STF
    Apr 27 '17 at 15:06






  • 1




    $begingroup$
    Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
    $endgroup$
    – lulu
    Apr 27 '17 at 15:07










  • $begingroup$
    OK, what additional condition I would need to go from independence to conditional independence?
    $endgroup$
    – STF
    Apr 27 '17 at 15:08










  • $begingroup$
    Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
    $endgroup$
    – lulu
    Apr 27 '17 at 15:09








2




2




$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03






$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03














$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06




$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06




1




1




$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07




$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07












$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08




$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08












$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09




$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09










2 Answers
2






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1












$begingroup$

It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.



    And further we can obtain:



    $$begin{align}
    begin{array}{rc}
    Xperp Y | Z
    implies&
    p(X cap Y|Z) = p(X|Z)p(Y|Z) \
    implies&
    frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
    implies&
    p(X|Y,Z) = p(X|Z)
    end{array}
    end{align}$$





    And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:



    $$begin{align}
    p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
    & = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
    & = frac{p(Y)p(Z|X,Y)}{p(Z)}
    end{align}$$



    So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.



    I hope the above reduction and reasoning hold.






    share|cite|improve this answer











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      2 Answers
      2






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      2 Answers
      2






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      1












      $begingroup$

      It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.






          share|cite|improve this answer









          $endgroup$



          It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 12:36









          Kavi Rama MurthyKavi Rama Murthy

          56.5k42159




          56.5k42159























              0












              $begingroup$

              $Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.



              And further we can obtain:



              $$begin{align}
              begin{array}{rc}
              Xperp Y | Z
              implies&
              p(X cap Y|Z) = p(X|Z)p(Y|Z) \
              implies&
              frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
              implies&
              p(X|Y,Z) = p(X|Z)
              end{array}
              end{align}$$





              And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:



              $$begin{align}
              p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
              & = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
              & = frac{p(Y)p(Z|X,Y)}{p(Z)}
              end{align}$$



              So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.



              I hope the above reduction and reasoning hold.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.



                And further we can obtain:



                $$begin{align}
                begin{array}{rc}
                Xperp Y | Z
                implies&
                p(X cap Y|Z) = p(X|Z)p(Y|Z) \
                implies&
                frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
                implies&
                p(X|Y,Z) = p(X|Z)
                end{array}
                end{align}$$





                And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:



                $$begin{align}
                p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
                & = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
                & = frac{p(Y)p(Z|X,Y)}{p(Z)}
                end{align}$$



                So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.



                I hope the above reduction and reasoning hold.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.



                  And further we can obtain:



                  $$begin{align}
                  begin{array}{rc}
                  Xperp Y | Z
                  implies&
                  p(X cap Y|Z) = p(X|Z)p(Y|Z) \
                  implies&
                  frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
                  implies&
                  p(X|Y,Z) = p(X|Z)
                  end{array}
                  end{align}$$





                  And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:



                  $$begin{align}
                  p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
                  & = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
                  & = frac{p(Y)p(Z|X,Y)}{p(Z)}
                  end{align}$$



                  So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.



                  I hope the above reduction and reasoning hold.






                  share|cite|improve this answer











                  $endgroup$



                  $Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.



                  And further we can obtain:



                  $$begin{align}
                  begin{array}{rc}
                  Xperp Y | Z
                  implies&
                  p(X cap Y|Z) = p(X|Z)p(Y|Z) \
                  implies&
                  frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
                  implies&
                  p(X|Y,Z) = p(X|Z)
                  end{array}
                  end{align}$$





                  And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:



                  $$begin{align}
                  p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
                  & = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
                  & = frac{p(Y)p(Z|X,Y)}{p(Z)}
                  end{align}$$



                  So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.



                  I hope the above reduction and reasoning hold.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 13 at 8:02

























                  answered Jan 13 at 3:16









                  lernerlerner

                  312216




                  312216






























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