Mutual independence in probability












0












$begingroup$


Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.



Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.






probability







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.



    Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.






    probability







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.



      Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.






      probability







      share|cite|improve this question











      $endgroup$




      Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.



      Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.






      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 18:52









      Mee Seong Im

      2,8151617




      2,8151617










      asked Jan 15 at 18:04









      NoteBookNoteBook

      1167




      1167






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."




          Do you know how to compute $P(A) = P(X_1 = X_2)$?




          $P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.




          Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.




          Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?




          $A cap B cap C = {X_1 = X_2 = X_3}$




          Then can you compute the probability of $A cap B cap C$?




          $P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            spotted my mistake, thank you
            $endgroup$
            – NoteBook
            Jan 15 at 18:23





















          0












          $begingroup$

          Assume $X_1$, $X_2$, and $X_3$ are mutually independent.



          Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
          $$
          A = {X_1=X_2}, qquad
          B = { X_1=X_3}, qquad
          C = {X_2=X_3 }.
          $$
          Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.



          Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
          begin{align*}
          P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
          &qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
          &= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
          &= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
          &= frac{1}{2},
          end{align*}
          where the second equality holds since $X_1$ and $X_2$ are independent.



          Similarly, $P(B)=P(C)=frac{1}{2}$.



          So
          begin{align*}
          P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
          \
          &qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
          &= P(X_1=1)P(X_2=1)P(X_3=1) \
          &qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
          &= frac{1}{2^3} + frac{1}{2^3} \
          &= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
          end{align*}



          On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.



          Thus the events $A$, $B$, and $C$ are not mutually independent.






          share|cite|improve this answer











          $endgroup$














            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: "69"
            };
            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: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            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
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074732%2fmutual-independence-in-probability%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









            1












            $begingroup$

            Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."




            Do you know how to compute $P(A) = P(X_1 = X_2)$?




            $P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.




            Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.




            Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?




            $A cap B cap C = {X_1 = X_2 = X_3}$




            Then can you compute the probability of $A cap B cap C$?




            $P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              spotted my mistake, thank you
              $endgroup$
              – NoteBook
              Jan 15 at 18:23


















            1












            $begingroup$

            Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."




            Do you know how to compute $P(A) = P(X_1 = X_2)$?




            $P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.




            Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.




            Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?




            $A cap B cap C = {X_1 = X_2 = X_3}$




            Then can you compute the probability of $A cap B cap C$?




            $P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              spotted my mistake, thank you
              $endgroup$
              – NoteBook
              Jan 15 at 18:23
















            1












            1








            1





            $begingroup$

            Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."




            Do you know how to compute $P(A) = P(X_1 = X_2)$?




            $P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.




            Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.




            Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?




            $A cap B cap C = {X_1 = X_2 = X_3}$




            Then can you compute the probability of $A cap B cap C$?




            $P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$







            share|cite|improve this answer











            $endgroup$



            Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."




            Do you know how to compute $P(A) = P(X_1 = X_2)$?




            $P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.




            Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.




            Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?




            $A cap B cap C = {X_1 = X_2 = X_3}$




            Then can you compute the probability of $A cap B cap C$?




            $P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 15 at 18:13

























            answered Jan 15 at 18:08









            angryavianangryavian

            42.5k23481




            42.5k23481












            • $begingroup$
              spotted my mistake, thank you
              $endgroup$
              – NoteBook
              Jan 15 at 18:23




















            • $begingroup$
              spotted my mistake, thank you
              $endgroup$
              – NoteBook
              Jan 15 at 18:23


















            $begingroup$
            spotted my mistake, thank you
            $endgroup$
            – NoteBook
            Jan 15 at 18:23






            $begingroup$
            spotted my mistake, thank you
            $endgroup$
            – NoteBook
            Jan 15 at 18:23













            0












            $begingroup$

            Assume $X_1$, $X_2$, and $X_3$ are mutually independent.



            Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
            $$
            A = {X_1=X_2}, qquad
            B = { X_1=X_3}, qquad
            C = {X_2=X_3 }.
            $$
            Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.



            Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
            begin{align*}
            P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
            &qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
            &= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
            &= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
            &= frac{1}{2},
            end{align*}
            where the second equality holds since $X_1$ and $X_2$ are independent.



            Similarly, $P(B)=P(C)=frac{1}{2}$.



            So
            begin{align*}
            P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
            \
            &qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
            &= P(X_1=1)P(X_2=1)P(X_3=1) \
            &qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
            &= frac{1}{2^3} + frac{1}{2^3} \
            &= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
            end{align*}



            On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.



            Thus the events $A$, $B$, and $C$ are not mutually independent.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Assume $X_1$, $X_2$, and $X_3$ are mutually independent.



              Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
              $$
              A = {X_1=X_2}, qquad
              B = { X_1=X_3}, qquad
              C = {X_2=X_3 }.
              $$
              Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.



              Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
              begin{align*}
              P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
              &qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
              &= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
              &= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
              &= frac{1}{2},
              end{align*}
              where the second equality holds since $X_1$ and $X_2$ are independent.



              Similarly, $P(B)=P(C)=frac{1}{2}$.



              So
              begin{align*}
              P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
              \
              &qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
              &= P(X_1=1)P(X_2=1)P(X_3=1) \
              &qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
              &= frac{1}{2^3} + frac{1}{2^3} \
              &= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
              end{align*}



              On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.



              Thus the events $A$, $B$, and $C$ are not mutually independent.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Assume $X_1$, $X_2$, and $X_3$ are mutually independent.



                Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
                $$
                A = {X_1=X_2}, qquad
                B = { X_1=X_3}, qquad
                C = {X_2=X_3 }.
                $$
                Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.



                Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
                begin{align*}
                P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
                &qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
                &= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
                &= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
                &= frac{1}{2},
                end{align*}
                where the second equality holds since $X_1$ and $X_2$ are independent.



                Similarly, $P(B)=P(C)=frac{1}{2}$.



                So
                begin{align*}
                P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
                \
                &qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
                &= P(X_1=1)P(X_2=1)P(X_3=1) \
                &qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
                &= frac{1}{2^3} + frac{1}{2^3} \
                &= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
                end{align*}



                On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.



                Thus the events $A$, $B$, and $C$ are not mutually independent.






                share|cite|improve this answer











                $endgroup$



                Assume $X_1$, $X_2$, and $X_3$ are mutually independent.



                Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
                $$
                A = {X_1=X_2}, qquad
                B = { X_1=X_3}, qquad
                C = {X_2=X_3 }.
                $$
                Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.



                Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
                begin{align*}
                P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
                &qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
                &= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
                &= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
                &= frac{1}{2},
                end{align*}
                where the second equality holds since $X_1$ and $X_2$ are independent.



                Similarly, $P(B)=P(C)=frac{1}{2}$.



                So
                begin{align*}
                P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
                \
                &qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
                &= P(X_1=1)P(X_2=1)P(X_3=1) \
                &qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
                &= frac{1}{2^3} + frac{1}{2^3} \
                &= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
                end{align*}



                On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.



                Thus the events $A$, $B$, and $C$ are not mutually independent.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 15 at 19:04

























                answered Jan 15 at 18:50









                Mee Seong ImMee Seong Im

                2,8151617




                2,8151617






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • 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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074732%2fmutual-independence-in-probability%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

                    Human spaceflight

                    Image
                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                    Read more

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    Image
                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                    Read more

                    張江高科駅

                    Image
                    張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
                    Read more