Probability - point inside disk












0












$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07


















0












$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07
















0












0








0





$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







share|cite|improve this question











$endgroup$





$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution




probability normal-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 16:12









gt6989b

35k22557




35k22557










asked Dec 10 '18 at 16:05









AlphaDelphiAlphaDelphi

11




11








  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07
















  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07










4




4




$begingroup$
Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
$endgroup$
– Federico
Dec 10 '18 at 16:07






$begingroup$
Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
$endgroup$
– Federico
Dec 10 '18 at 16:07












2 Answers
2






active

oldest

votes


















1












$begingroup$

HINT



For any event $A$, you have
$$
mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
$$
and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
    $$
    begin{split}
    mathbb P(B_R) &=
    int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
    &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
    &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
    &= 1-e^{R^2/2}.
    end{split}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
      $endgroup$
      – gt6989b
      Dec 10 '18 at 16:18










    • $begingroup$
      @gt6989b Well, I don't see that as a good reason to -1...
      $endgroup$
      – Federico
      Dec 10 '18 at 16:18











    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%2f3034102%2fprobability-point-inside-disk%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












    $begingroup$

    HINT



    For any event $A$, you have
    $$
    mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
    $$
    and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



    Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      HINT



      For any event $A$, you have
      $$
      mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
      $$
      and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



      Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        HINT



        For any event $A$, you have
        $$
        mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
        $$
        and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



        Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






        share|cite|improve this answer









        $endgroup$



        HINT



        For any event $A$, you have
        $$
        mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
        $$
        and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



        Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 16:11









        gt6989bgt6989b

        35k22557




        35k22557























            -1












            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18
















            -1












            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18














            -1












            -1








            -1





            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$



            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 16:17









            FedericoFederico

            5,144514




            5,144514












            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18


















            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18
















            $begingroup$
            -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
            $endgroup$
            – gt6989b
            Dec 10 '18 at 16:18




            $begingroup$
            -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
            $endgroup$
            – gt6989b
            Dec 10 '18 at 16:18












            $begingroup$
            @gt6989b Well, I don't see that as a good reason to -1...
            $endgroup$
            – Federico
            Dec 10 '18 at 16:18




            $begingroup$
            @gt6989b Well, I don't see that as a good reason to -1...
            $endgroup$
            – Federico
            Dec 10 '18 at 16:18


















            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%2f3034102%2fprobability-point-inside-disk%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)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more