Does independence between random variables imply independence between related events?












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Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?



If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?






random-variable independence







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  • $begingroup$
    See stats.stackexchange.com/questions/94872/… inter alia.
    $endgroup$
    – whuber♦
    Nov 22 '18 at 14:55
















1












$begingroup$


Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?



If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?






random-variable independence







share|cite|improve this question











$endgroup$












  • $begingroup$
    See stats.stackexchange.com/questions/94872/… inter alia.
    $endgroup$
    – whuber♦
    Nov 22 '18 at 14:55














1












1








1





$begingroup$


Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?



If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?






random-variable independence







share|cite|improve this question











$endgroup$




Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?



If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?






random-variable independence




random-variable independence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 11:53







mickkk

















asked Nov 22 '18 at 11:39









mickkkmickkk

379314




379314












  • $begingroup$
    See stats.stackexchange.com/questions/94872/… inter alia.
    $endgroup$
    – whuber♦
    Nov 22 '18 at 14:55


















  • $begingroup$
    See stats.stackexchange.com/questions/94872/… inter alia.
    $endgroup$
    – whuber♦
    Nov 22 '18 at 14:55
















$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55




$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55










2 Answers
2






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oldest

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$begingroup$

I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.



Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
    $$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
    But, $$F_{X,Y}(a,b)
    stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
    while $$F_{X}(a)
    stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
    stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
    and so $(*)$ is saying that
    $$P(Acap B) = P(A)P(B),$$
    that is, $A$ and $B$ are independent events.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.



      Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.



        Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.



          Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.






          share|cite|improve this answer









          $endgroup$



          I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.



          Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 15:01









          user158565user158565

          5,3641518




          5,3641518

























              1












              $begingroup$

              If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
              $$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
              But, $$F_{X,Y}(a,b)
              stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
              while $$F_{X}(a)
              stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
              stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
              and so $(*)$ is saying that
              $$P(Acap B) = P(A)P(B),$$
              that is, $A$ and $B$ are independent events.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
                $$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
                But, $$F_{X,Y}(a,b)
                stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
                while $$F_{X}(a)
                stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
                stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
                and so $(*)$ is saying that
                $$P(Acap B) = P(A)P(B),$$
                that is, $A$ and $B$ are independent events.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
                  $$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
                  But, $$F_{X,Y}(a,b)
                  stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
                  while $$F_{X}(a)
                  stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
                  stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
                  and so $(*)$ is saying that
                  $$P(Acap B) = P(A)P(B),$$
                  that is, $A$ and $B$ are independent events.






                  share|cite|improve this answer









                  $endgroup$



                  If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
                  $$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
                  But, $$F_{X,Y}(a,b)
                  stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
                  while $$F_{X}(a)
                  stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
                  stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
                  and so $(*)$ is saying that
                  $$P(Acap B) = P(A)P(B),$$
                  that is, $A$ and $B$ are independent events.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 '18 at 15:03









                  Dilip SarwateDilip Sarwate

                  30k252147




                  30k252147






























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