Independence of 4 random variables











up vote
0
down vote

favorite












$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










share|cite|improve this question


















  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07

















up vote
0
down vote

favorite












$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










share|cite|improve this question


















  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07















up vote
0
down vote

favorite









up vote
0
down vote

favorite











$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










share|cite|improve this question













$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!







probability probability-theory independence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 15 at 12:05









DrXYZ

1




1








  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07
















  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07










2




2




You probably mean events and not random variables. In that case your thinking is correct.
– Kavi Rama Murthy
Nov 15 at 12:07






You probably mean events and not random variables. In that case your thinking is correct.
– Kavi Rama Murthy
Nov 15 at 12:07












1 Answer
1






active

oldest

votes

















up vote
0
down vote













As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






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: "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',
    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%2f2999613%2findependence-of-4-random-variables%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



    Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



    Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



    But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



    eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



    If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



    So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



      Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



      Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



      But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



      eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



      If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



      So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



        Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



        Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



        But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



        eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



        If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



        So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






        share|cite|improve this answer












        As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



        Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



        Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



        But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



        eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



        If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



        So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 15 at 14:04









        Arnaud Mégret

        351414




        351414






























            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.





            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%2fmath.stackexchange.com%2fquestions%2f2999613%2findependence-of-4-random-variables%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

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            ComboBox Display Member on multiple fields

            Is it possible to collect Nectar points via Trainline?