${V_n}$ and ${W_n}$ are two sequences of random variables, show that$V_n-W_noverset{p}to0$.












1












$begingroup$


Let ${V_n}$ and ${W_n}$ be two sequences of random variables satisfying the following conditions.



(i) For all $delta>0$, there exists $lambda$ s.t. $P(|W_n|>lambda)<delta$.



(ii) For all $k$ and all $varepsilon>0$
$$lim_{ntoinfty}P(V_nle k,W_nge k+varepsilon)=0$$
$$lim_{ntoinfty}P(V_nge k+varepsilon ,W_nle k)=0.$$



Then $V_n-W_noverset{p}to0$,as $ntoinfty$.



How to understand and prove this proposition?






probability probability-theory measure-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:02










  • $begingroup$
    The observation is very helpful. But how to use condition (i). Is it a redundant condition?
    $endgroup$
    – J.Mike
    Dec 27 '18 at 5:14








  • 1




    $begingroup$
    If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:45








  • 1




    $begingroup$
    Unless you say how $k$ is chosen the arguments in above comments are not valid.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 5:49










  • $begingroup$
    Does condition (i) holds for some/all $n$?
    $endgroup$
    – user251257
    Dec 27 '18 at 12:20
















1












$begingroup$


Let ${V_n}$ and ${W_n}$ be two sequences of random variables satisfying the following conditions.



(i) For all $delta>0$, there exists $lambda$ s.t. $P(|W_n|>lambda)<delta$.



(ii) For all $k$ and all $varepsilon>0$
$$lim_{ntoinfty}P(V_nle k,W_nge k+varepsilon)=0$$
$$lim_{ntoinfty}P(V_nge k+varepsilon ,W_nle k)=0.$$



Then $V_n-W_noverset{p}to0$,as $ntoinfty$.



How to understand and prove this proposition?






probability probability-theory measure-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:02










  • $begingroup$
    The observation is very helpful. But how to use condition (i). Is it a redundant condition?
    $endgroup$
    – J.Mike
    Dec 27 '18 at 5:14








  • 1




    $begingroup$
    If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:45








  • 1




    $begingroup$
    Unless you say how $k$ is chosen the arguments in above comments are not valid.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 5:49










  • $begingroup$
    Does condition (i) holds for some/all $n$?
    $endgroup$
    – user251257
    Dec 27 '18 at 12:20














1












1








1


0



$begingroup$


Let ${V_n}$ and ${W_n}$ be two sequences of random variables satisfying the following conditions.



(i) For all $delta>0$, there exists $lambda$ s.t. $P(|W_n|>lambda)<delta$.



(ii) For all $k$ and all $varepsilon>0$
$$lim_{ntoinfty}P(V_nle k,W_nge k+varepsilon)=0$$
$$lim_{ntoinfty}P(V_nge k+varepsilon ,W_nle k)=0.$$



Then $V_n-W_noverset{p}to0$,as $ntoinfty$.



How to understand and prove this proposition?






probability probability-theory measure-theory probability-distributions random-variables







share|cite|improve this question









$endgroup$




Let ${V_n}$ and ${W_n}$ be two sequences of random variables satisfying the following conditions.



(i) For all $delta>0$, there exists $lambda$ s.t. $P(|W_n|>lambda)<delta$.



(ii) For all $k$ and all $varepsilon>0$
$$lim_{ntoinfty}P(V_nle k,W_nge k+varepsilon)=0$$
$$lim_{ntoinfty}P(V_nge k+varepsilon ,W_nle k)=0.$$



Then $V_n-W_noverset{p}to0$,as $ntoinfty$.



How to understand and prove this proposition?






probability probability-theory measure-theory probability-distributions random-variables




probability probability-theory measure-theory probability-distributions random-variables






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 4:49









J.MikeJ.Mike

336110




336110








  • 1




    $begingroup$
    Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:02










  • $begingroup$
    The observation is very helpful. But how to use condition (i). Is it a redundant condition?
    $endgroup$
    – J.Mike
    Dec 27 '18 at 5:14








  • 1




    $begingroup$
    If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:45








  • 1




    $begingroup$
    Unless you say how $k$ is chosen the arguments in above comments are not valid.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 5:49










  • $begingroup$
    Does condition (i) holds for some/all $n$?
    $endgroup$
    – user251257
    Dec 27 '18 at 12:20














  • 1




    $begingroup$
    Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:02










  • $begingroup$
    The observation is very helpful. But how to use condition (i). Is it a redundant condition?
    $endgroup$
    – J.Mike
    Dec 27 '18 at 5:14








  • 1




    $begingroup$
    If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
    $endgroup$
    – Jonas
    Dec 27 '18 at 5:45








  • 1




    $begingroup$
    Unless you say how $k$ is chosen the arguments in above comments are not valid.
    $endgroup$
    – Kavi Rama Murthy
    Dec 27 '18 at 5:49










  • $begingroup$
    Does condition (i) holds for some/all $n$?
    $endgroup$
    – user251257
    Dec 27 '18 at 12:20








1




1




$begingroup$
Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
$endgroup$
– Jonas
Dec 27 '18 at 5:02




$begingroup$
Observe that if $V_n leq k$ and $W_n geq k + varepsilon$, it follows that $W_n - V_n geq varepsilon$, so that the conditions imply that $ lim_{n to infty} P(W_n - V_n geq varepsilon) = 0$, and similarly $ lim_{n to infty} P(V_n - W_n geq varepsilon) = 0$.
$endgroup$
– Jonas
Dec 27 '18 at 5:02












$begingroup$
The observation is very helpful. But how to use condition (i). Is it a redundant condition?
$endgroup$
– J.Mike
Dec 27 '18 at 5:14






$begingroup$
The observation is very helpful. But how to use condition (i). Is it a redundant condition?
$endgroup$
– J.Mike
Dec 27 '18 at 5:14






1




1




$begingroup$
If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
$endgroup$
– Jonas
Dec 27 '18 at 5:45






$begingroup$
If $V_n = W_n = infty$, then $V_n - W_n = infty - infty$ is not defined, and similarly if $V_n = W_n = - infty$. However, condition (i) implies that $lim_{lambda to infty} P(|W_n| > lambda) = 0$, so that in particular, $ P(|W_n| = infty) = 0 $. We can use this to argue rigorously that the set of values where $V_n - W_n$ is not defined is negligible, and hence does not affect the desired conclusion.
$endgroup$
– Jonas
Dec 27 '18 at 5:45






1




1




$begingroup$
Unless you say how $k$ is chosen the arguments in above comments are not valid.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 5:49




$begingroup$
Unless you say how $k$ is chosen the arguments in above comments are not valid.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 5:49












$begingroup$
Does condition (i) holds for some/all $n$?
$endgroup$
– user251257
Dec 27 '18 at 12:20




$begingroup$
Does condition (i) holds for some/all $n$?
$endgroup$
– user251257
Dec 27 '18 at 12:20










1 Answer
1






active

oldest

votes


















1












$begingroup$

Thanks for all the comments. I find a proof in the book Order Statistics by David and Nagaraja (2003) on page 286.



Fix $varepsilon>0, delta>0$, by condtion (i) we can choose integers $m$ and $n_0$ s.t.
$$P(|W_n|>mvarepsilon)<delta,quad for~~~ nge n_0.$$
Hence
begin{align*}
P(|V_n&-W_n|>2varepsilon)<delta+P(|W_n|le mvarepsilon,|V_n-W_n|>2varepsilon)\
&=delta+sum_{j=-m}^{m-1}P(jvarepsilonle W_nle (j+1)varepsilon,|V_n-W_n|>2varepsilon)\
&ledelta+sum_{j=-m}^{m-1}Big[PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n>(j+2)varepsilonBig)\
&+PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n<(j-1)varepsilonBig)Big]
end{align*}
which tends to $0$ as $ntoinfty$ by condtion (ii).






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%2f3053600%2fv-n-and-w-n-are-two-sequences-of-random-variables-show-thatv-n-w-n%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









    1












    $begingroup$

    Thanks for all the comments. I find a proof in the book Order Statistics by David and Nagaraja (2003) on page 286.



    Fix $varepsilon>0, delta>0$, by condtion (i) we can choose integers $m$ and $n_0$ s.t.
    $$P(|W_n|>mvarepsilon)<delta,quad for~~~ nge n_0.$$
    Hence
    begin{align*}
    P(|V_n&-W_n|>2varepsilon)<delta+P(|W_n|le mvarepsilon,|V_n-W_n|>2varepsilon)\
    &=delta+sum_{j=-m}^{m-1}P(jvarepsilonle W_nle (j+1)varepsilon,|V_n-W_n|>2varepsilon)\
    &ledelta+sum_{j=-m}^{m-1}Big[PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n>(j+2)varepsilonBig)\
    &+PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n<(j-1)varepsilonBig)Big]
    end{align*}
    which tends to $0$ as $ntoinfty$ by condtion (ii).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Thanks for all the comments. I find a proof in the book Order Statistics by David and Nagaraja (2003) on page 286.



      Fix $varepsilon>0, delta>0$, by condtion (i) we can choose integers $m$ and $n_0$ s.t.
      $$P(|W_n|>mvarepsilon)<delta,quad for~~~ nge n_0.$$
      Hence
      begin{align*}
      P(|V_n&-W_n|>2varepsilon)<delta+P(|W_n|le mvarepsilon,|V_n-W_n|>2varepsilon)\
      &=delta+sum_{j=-m}^{m-1}P(jvarepsilonle W_nle (j+1)varepsilon,|V_n-W_n|>2varepsilon)\
      &ledelta+sum_{j=-m}^{m-1}Big[PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n>(j+2)varepsilonBig)\
      &+PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n<(j-1)varepsilonBig)Big]
      end{align*}
      which tends to $0$ as $ntoinfty$ by condtion (ii).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Thanks for all the comments. I find a proof in the book Order Statistics by David and Nagaraja (2003) on page 286.



        Fix $varepsilon>0, delta>0$, by condtion (i) we can choose integers $m$ and $n_0$ s.t.
        $$P(|W_n|>mvarepsilon)<delta,quad for~~~ nge n_0.$$
        Hence
        begin{align*}
        P(|V_n&-W_n|>2varepsilon)<delta+P(|W_n|le mvarepsilon,|V_n-W_n|>2varepsilon)\
        &=delta+sum_{j=-m}^{m-1}P(jvarepsilonle W_nle (j+1)varepsilon,|V_n-W_n|>2varepsilon)\
        &ledelta+sum_{j=-m}^{m-1}Big[PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n>(j+2)varepsilonBig)\
        &+PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n<(j-1)varepsilonBig)Big]
        end{align*}
        which tends to $0$ as $ntoinfty$ by condtion (ii).






        share|cite|improve this answer









        $endgroup$



        Thanks for all the comments. I find a proof in the book Order Statistics by David and Nagaraja (2003) on page 286.



        Fix $varepsilon>0, delta>0$, by condtion (i) we can choose integers $m$ and $n_0$ s.t.
        $$P(|W_n|>mvarepsilon)<delta,quad for~~~ nge n_0.$$
        Hence
        begin{align*}
        P(|V_n&-W_n|>2varepsilon)<delta+P(|W_n|le mvarepsilon,|V_n-W_n|>2varepsilon)\
        &=delta+sum_{j=-m}^{m-1}P(jvarepsilonle W_nle (j+1)varepsilon,|V_n-W_n|>2varepsilon)\
        &ledelta+sum_{j=-m}^{m-1}Big[PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n>(j+2)varepsilonBig)\
        &+PBig(jvarepsilonle W_nle (j+1)varepsilon,V_n<(j-1)varepsilonBig)Big]
        end{align*}
        which tends to $0$ as $ntoinfty$ by condtion (ii).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 18:59









        J.MikeJ.Mike

        336110




        336110






























            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%2f3053600%2fv-n-and-w-n-are-two-sequences-of-random-variables-show-thatv-n-w-n%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

            Index of /

            This page is only for reference, If you need detailed information, please check here
            Read more

            Tribalistas

            Image
              Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
            Read more

            Listed building

            Image
            Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
            Read more