Probability X1 ≥ X2In convergence in probability or a.s. convergence w.r.t which measure is the...

Linearity Assumption

Can you prevent a man in the middle from reading the message?

Can you say "leftside right"?

How bad is a Computer Science course that doesn't teach Design Patterns?

What could cause an entire planet of humans to become aphasic?

What is formjacking?

What's the reason that we have different quantities of days each month?

Coworker is trying to get me to sign his petition to run for office. How to decline politely?

Third wheel character

Players preemptively rolling, even though their rolls are useless or are checking the wrong skills

User input happy birthday program

In the Lost in Space intro why was Dr. Smith actor listed as a special guest star?

Crack the bank account's password!

How to deal with an underperforming subordinate?

How unreachable are Jupiter's moons from Mars with the technology developed for going to Mars?

Why is it that Bernie Sanders is always called a "socialist"?

Connecting to SMTP server from AWS Lambda

Limits of Rolle theorem

Is Screenshot Time-tracking Common?

How can I handle players killing my NPC outside of combat?

If I tried and failed to start my own business, how do I apply for a job without job experience?

Buying a "Used" Router

Why is Acetic acid (pKa = 4.76) stronger than carbonic acid (pKa = 6.36)?

Is there a way to pause a running process on Linux systems and resume later?



Probability X1 ≥ X2


In convergence in probability or a.s. convergence w.r.t which measure is the probability?Comparison of waiting times to geometric distributionShow that for a Geometric distribution, the probability generating function is given by $frac{ps}{1-qs}$, $q=1-p$Different ways of generating Geometric distributionsHow to calculate the sum or difference of two probability generating functions?How to check if functions of i.i.d random variables are dependent or independent?Calculating probability of floor'ed log-normal distributionConcepts of Probability all messed upGeometric distribution with random, varying success probabilityWhat can we say about $N_{i}$ where $N=N_{1}+cdots+N_{m}$, $Nthicksim Geom(frac{1-p}{p})$ and conditional distribution of $N_{j}$ is binomial













2












$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    4 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    4 hours ago
















2












$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    4 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    4 hours ago














2












2








2





$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$




Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution




random-variable geometric-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Sra

















asked 5 hours ago









SraSra

464




464








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    4 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    4 hours ago














  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    4 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    4 hours ago








1




1




$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago




$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago




1




1




$begingroup$
Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
4 hours ago




$begingroup$
Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
4 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

It can't be $50%$ because $P(X_1=X_2)>0$



One approach:



Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago



















0












$begingroup$

Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



begin{align}
Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
end{align}



This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






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: "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%2f394042%2fprobability-x1-%25e2%2589%25a5-x2%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









    5












    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago
















    5












    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago














    5












    5








    5





    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$



    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 4 hours ago

























    answered 4 hours ago









    Glen_b♦Glen_b

    212k22406754




    212k22406754












    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago


















    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago
















    $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago




    $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago













    0












    $begingroup$

    Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



    begin{align}
    Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
    end{align}



    This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



      begin{align}
      Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
      end{align}



      This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



        begin{align}
        Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
        end{align}



        This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






        share|cite|improve this answer









        $endgroup$



        Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



        begin{align}
        Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
        end{align}



        This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 16 mins ago









        Paulo C. Marques F.Paulo C. Marques F.

        16.9k35397




        16.9k35397






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f394042%2fprobability-x1-%25e2%2589%25a5-x2%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

            Discografia di Klaus Schulze Indice Album in studio | Album dal vivo | Singoli | Antologie | Colonne...

            Image
            Discografie di artisti tedeschi Voce principale: Klaus Schulze . Discografia di Klaus Schulze Klaus Schulze e Lisa Gerrard durante un concerto Album in studio 56 ↙ Album dal vivo 15 ↙ Raccolte 37 ↙ Singoli 4 ↙ Album video 1 ↙ Colonne sonore 4 ↙ Indice 1 Album in studio 1.1 Solisti 1.1.1 Come Richard Wahnfried 1.2 Collaborazioni 1.2.1 Con Rainer Bloss 1.2.2 Con Rainer Bloss ed Ernst Fuchs 1.2.3 Con Lisa Gerrard 1.2.4 Con Andreas Grosser 1.2.5 Con Pete Namlook 1.2.6 Con Pete Namlook e Bill Laswell 1.2.7 Con Michael Shrieve 1.2.8 Con Solar Moon 1.2.9 Con artisti vari 2 Album dal vivo 3 Singoli 4 Antologie 5 Colonne sonore 6 Videografia 7 Collegamenti musica Album in studio | Solisti | 1972 – Irrlicht 1973 – Cyborg 1974 – Blackdance 1975 – Timewind 1975 – Picture Music 1976 – Moondawn 1977 – Mirage 1978 – "X" 1979 – Dune...
            Read more

            Armoriale delle famiglie italiane (Car) Indice Armi | Bibliografia | Menu di navigazioneBlasone...

            Image
            Armoriali delle famiglie italiane Blasone Cesenate Famiglie nobili di SiciliaFamiglie nobili di SiciliaTropea MagazineTropea MagazineNobili napoletani Nobili napoletani[1]Nobili napoletaniNobili napoletaniTropea Magazine Nobili napoletaniNobili napoletani Famiglie nobili di SiciliaBlasonario CremoneseFamiglie nobili di Sicilia Archivio Storico CapitolinoRaccolta stemmi TrippiniFamiglie nobili di SiciliaNobili napoletaniArmas de los cavalleros de Veneçia Tropea Magazine Famiglie nobili di Sicilia Nobili napoletaniFamiglie nobili di SiciliaRaccolta stemmi Topoguz Famiglie nobili di SiciliaBlasone CesenateRaccolta stemmi TrippiniArchivio Storico Capitolino Famiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Famiglie nobili di SiciliaFamiglie nobili di SiciliaBlasone Cesenate Raccolta stemmi TrippiniFamiglie nobili di SiciliaFamiglie nobili di SiciliaFamiglie nobili di Sicilia Blasonario CremoneseBlasonario Cr...
            Read more

            Lupi Siderali Indice Storia | Organizzazione | La Tredicesima Compagnia | Aspetto | Membri Importanti...

            Capitoli degli Space Marine Space MarineImperium della Terrawargame tridimensionaleWarhammer 40.000PrimarcalupiuroImperatoreRune di DissimulazioneAngeli OscuriEresia di HorusStirpe dei MilleOcchio del TerroreCaosTzeentchHorusCodex AstartesSeconda FondazioneFratelli del LupoOcchio del TerroreAlbero della VitareliquieGuardia ImperialeLogan Grimnartruppe d'assaltoImperium dell'UomoveteraniStirpe dei MilleCaosOcchio del TerroreAngeli SanguinariEresia di Horus Lupi Siderali Nome originale Space Wolves Primarca Leman Russ Grido di battaglia Per Russ e per il Padre di Tutti (un feroce ululato) Colori Grigio Altri capitoli I Lupi Siderali sono una delle venti Legioni della Prima Fondazione degli Space Marine che servono l'Imperium della Terra nel wargame tridimensionale Warhammer 40.000. Sono formidabili nel corpo a corpo, i Confratelli desiderano solo la gloria sul campo di battaglia. Indice 1 Storia 1.1 Leman Russ 1.2 La ...
            Read more