Does independence between random variables imply independence between related events?












1














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?










share|cite|improve this question
























  • See stats.stackexchange.com/questions/94872/… inter alia.
    – whuber
    Nov 22 at 14:55
















1














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?










share|cite|improve this question
























  • See stats.stackexchange.com/questions/94872/… inter alia.
    – whuber
    Nov 22 at 14:55














1












1








1







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?










share|cite|improve this question















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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 11:53

























asked Nov 22 at 11:39









mickkk

374314




374314












  • See stats.stackexchange.com/questions/94872/… inter alia.
    – whuber
    Nov 22 at 14:55


















  • See stats.stackexchange.com/questions/94872/… inter alia.
    – whuber
    Nov 22 at 14:55
















See stats.stackexchange.com/questions/94872/… inter alia.
– whuber
Nov 22 at 14:55




See stats.stackexchange.com/questions/94872/… inter alia.
– whuber
Nov 22 at 14:55










2 Answers
2






active

oldest

votes


















2














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





























    1














    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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "65"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f378255%2fdoes-independence-between-random-variables-imply-independence-between-related-ev%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      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


























        2














        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
























          2












          2








          2






          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












          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 at 15:01









          user158565

          5,2551318




          5,2551318

























              1














              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


























                1














                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
























                  1












                  1








                  1






                  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












                  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 at 15:03









                  Dilip Sarwate

                  29.8k252147




                  29.8k252147






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Cross Validated!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f378255%2fdoes-independence-between-random-variables-imply-independence-between-related-ev%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      AnyDesk - Fatal Program Failure

                      How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                      QoS: MAC-Priority for clients behind a repeater