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

                    File:DeusFollowingSea.jpg

                    Image
                    File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                    Read more