Multiple Random Variable for a uniform distribution











up vote
0
down vote

favorite












A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.



My attempt:
$$int_{-1}^1int_{-1}^{2-x} frac{1}{4},dy,dx$$
Did I set up the double integral right?










share|cite|improve this question






















  • Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
    – angryavian
    yesterday












  • @angryavian, yeah! But I want to use double integral in solving it. How do I do that?
    – Lady
    yesterday










  • @herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
    – Lady
    yesterday






  • 1




    $|X+Y|$ is at most 2!, so the probability =1!.
    – herb steinberg
    yesterday






  • 1




    Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
    – herb steinberg
    yesterday















up vote
0
down vote

favorite












A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.



My attempt:
$$int_{-1}^1int_{-1}^{2-x} frac{1}{4},dy,dx$$
Did I set up the double integral right?










share|cite|improve this question






















  • Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
    – angryavian
    yesterday












  • @angryavian, yeah! But I want to use double integral in solving it. How do I do that?
    – Lady
    yesterday










  • @herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
    – Lady
    yesterday






  • 1




    $|X+Y|$ is at most 2!, so the probability =1!.
    – herb steinberg
    yesterday






  • 1




    Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
    – herb steinberg
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.



My attempt:
$$int_{-1}^1int_{-1}^{2-x} frac{1}{4},dy,dx$$
Did I set up the double integral right?










share|cite|improve this question













A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.



My attempt:
$$int_{-1}^1int_{-1}^{2-x} frac{1}{4},dy,dx$$
Did I set up the double integral right?







statistics statistical-inference






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Lady

978




978












  • Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
    – angryavian
    yesterday












  • @angryavian, yeah! But I want to use double integral in solving it. How do I do that?
    – Lady
    yesterday










  • @herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
    – Lady
    yesterday






  • 1




    $|X+Y|$ is at most 2!, so the probability =1!.
    – herb steinberg
    yesterday






  • 1




    Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
    – herb steinberg
    yesterday


















  • Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
    – angryavian
    yesterday












  • @angryavian, yeah! But I want to use double integral in solving it. How do I do that?
    – Lady
    yesterday










  • @herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
    – Lady
    yesterday






  • 1




    $|X+Y|$ is at most 2!, so the probability =1!.
    – herb steinberg
    yesterday






  • 1




    Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
    – herb steinberg
    yesterday
















Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
– angryavian
yesterday






Instead of setting up an integral, it may be easier to find the area of the region ${(x,y) : |x+y| < 2}$ that lies inside the square, and divide by $4$.
– angryavian
yesterday














@angryavian, yeah! But I want to use double integral in solving it. How do I do that?
– Lady
yesterday




@angryavian, yeah! But I want to use double integral in solving it. How do I do that?
– Lady
yesterday












@herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
– Lady
yesterday




@herbsteinberg, I did that. But doesn't the above fall within the square? Hence the limit of the integral.
– Lady
yesterday




1




1




$|X+Y|$ is at most 2!, so the probability =1!.
– herb steinberg
yesterday




$|X+Y|$ is at most 2!, so the probability =1!.
– herb steinberg
yesterday




1




1




Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
– herb steinberg
yesterday




Are you sure that you have stated the problem correctly? As is, it seems pointless. It would make more sense if the question involved $|X+Y|lt 1$.
– herb steinberg
yesterday










2 Answers
2






active

oldest

votes

















up vote
0
down vote













Comment: @herbsteinberg is correct that the problem as stated
is pointless, because the answer is obviously $1.$ The integral would be
$int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$



The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
Because the joint distribution of $(X,Y)$ on the square is uniform it
seems clear that $P(|X+Y|<1) = 3/4.$



If you want to use integral calculus, it is probably best to break
the integral into two parts, perhaps to the left and right of the vertical green
line.



enter image description here



Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
[A larger number of points, such as m = 10^6, would give a much closer
approximation, but an ugly plot.]



set.seed(1112);  m = 50000
x = runif(m, -1,1); y = runif(m, -1,1)
plot(x,y, pch=".")
cond = (abs(x+y)<1)
points(x[cond],y[cond], pch=".", col="blue")
abline(h=0, col="green2", lwd=2); abline(v=0, col="green2", lwd=2)
mean(cond)
[1] 0.7519 # aprx P(|X + Y| < 1) = 3/4





share|cite|improve this answer






























    up vote
    0
    down vote













    Note thet:
    $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
    P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$

    See the graph:
    enter image description here



    If the question is to find $P(|X+Y|<1)$, then:
    $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
    int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$



    See the graph:
    enter image description here






    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%2f2996118%2fmultiple-random-variable-for-a-uniform-distribution%23new-answer', 'question_page');
      }
      );

      Post as a guest
































      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote













      Comment: @herbsteinberg is correct that the problem as stated
      is pointless, because the answer is obviously $1.$ The integral would be
      $int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$



      The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
      Because the joint distribution of $(X,Y)$ on the square is uniform it
      seems clear that $P(|X+Y|<1) = 3/4.$



      If you want to use integral calculus, it is probably best to break
      the integral into two parts, perhaps to the left and right of the vertical green
      line.



      enter image description here



      Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
      a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
      [A larger number of points, such as m = 10^6, would give a much closer
      approximation, but an ugly plot.]



      set.seed(1112);  m = 50000
      x = runif(m, -1,1); y = runif(m, -1,1)
      plot(x,y, pch=".")
      cond = (abs(x+y)<1)
      points(x[cond],y[cond], pch=".", col="blue")
      abline(h=0, col="green2", lwd=2); abline(v=0, col="green2", lwd=2)
      mean(cond)
      [1] 0.7519 # aprx P(|X + Y| < 1) = 3/4





      share|cite|improve this answer



























        up vote
        0
        down vote













        Comment: @herbsteinberg is correct that the problem as stated
        is pointless, because the answer is obviously $1.$ The integral would be
        $int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$



        The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
        Because the joint distribution of $(X,Y)$ on the square is uniform it
        seems clear that $P(|X+Y|<1) = 3/4.$



        If you want to use integral calculus, it is probably best to break
        the integral into two parts, perhaps to the left and right of the vertical green
        line.



        enter image description here



        Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
        a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
        [A larger number of points, such as m = 10^6, would give a much closer
        approximation, but an ugly plot.]



        set.seed(1112);  m = 50000
        x = runif(m, -1,1); y = runif(m, -1,1)
        plot(x,y, pch=".")
        cond = (abs(x+y)<1)
        points(x[cond],y[cond], pch=".", col="blue")
        abline(h=0, col="green2", lwd=2); abline(v=0, col="green2", lwd=2)
        mean(cond)
        [1] 0.7519 # aprx P(|X + Y| < 1) = 3/4





        share|cite|improve this answer

























          up vote
          0
          down vote










          up vote
          0
          down vote









          Comment: @herbsteinberg is correct that the problem as stated
          is pointless, because the answer is obviously $1.$ The integral would be
          $int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$



          The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
          Because the joint distribution of $(X,Y)$ on the square is uniform it
          seems clear that $P(|X+Y|<1) = 3/4.$



          If you want to use integral calculus, it is probably best to break
          the integral into two parts, perhaps to the left and right of the vertical green
          line.



          enter image description here



          Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
          a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
          [A larger number of points, such as m = 10^6, would give a much closer
          approximation, but an ugly plot.]



          set.seed(1112);  m = 50000
          x = runif(m, -1,1); y = runif(m, -1,1)
          plot(x,y, pch=".")
          cond = (abs(x+y)<1)
          points(x[cond],y[cond], pch=".", col="blue")
          abline(h=0, col="green2", lwd=2); abline(v=0, col="green2", lwd=2)
          mean(cond)
          [1] 0.7519 # aprx P(|X + Y| < 1) = 3/4





          share|cite|improve this answer














          Comment: @herbsteinberg is correct that the problem as stated
          is pointless, because the answer is obviously $1.$ The integral would be
          $int_{-1}^1 int_{-1}^1 frac 1 4, dy,dx.$



          The plot below shows the region of the square corresponding to $P(|X+Y|<1).$
          Because the joint distribution of $(X,Y)$ on the square is uniform it
          seems clear that $P(|X+Y|<1) = 3/4.$



          If you want to use integral calculus, it is probably best to break
          the integral into two parts, perhaps to the left and right of the vertical green
          line.



          enter image description here



          Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides
          a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places.
          [A larger number of points, such as m = 10^6, would give a much closer
          approximation, but an ugly plot.]



          set.seed(1112);  m = 50000
          x = runif(m, -1,1); y = runif(m, -1,1)
          plot(x,y, pch=".")
          cond = (abs(x+y)<1)
          points(x[cond],y[cond], pch=".", col="blue")
          abline(h=0, col="green2", lwd=2); abline(v=0, col="green2", lwd=2)
          mean(cond)
          [1] 0.7519 # aprx P(|X + Y| < 1) = 3/4






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 17 hours ago

























          answered 17 hours ago









          BruceET

          34.6k71440




          34.6k71440






















              up vote
              0
              down vote













              Note thet:
              $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
              P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$

              See the graph:
              enter image description here



              If the question is to find $P(|X+Y|<1)$, then:
              $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
              int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$



              See the graph:
              enter image description here






              share|cite|improve this answer

























                up vote
                0
                down vote













                Note thet:
                $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
                P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$

                See the graph:
                enter image description here



                If the question is to find $P(|X+Y|<1)$, then:
                $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
                int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$



                See the graph:
                enter image description here






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Note thet:
                  $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
                  P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$

                  See the graph:
                  enter image description here



                  If the question is to find $P(|X+Y|<1)$, then:
                  $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
                  int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$



                  See the graph:
                  enter image description here






                  share|cite|improve this answer












                  Note thet:
                  $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 text{or} Y<-X+2)=\
                  P(Y>-X-2ge -1 text{or} Y<-X+2le 1)=int_{-1}^1 int_{-1}^1 frac14dydx=1.$$

                  See the graph:
                  enter image description here



                  If the question is to find $P(|X+Y|<1)$, then:
                  $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y text{or} Y<-X+1)=\
                  int_{-1}^0 int_{-x-1}^1 frac14dydx+int_0^1 int_{-1}^{-x+1}dydx=frac38+frac38=frac34.$$



                  See the graph:
                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 16 hours ago









                  farruhota

                  17.1k2736




                  17.1k2736






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2996118%2fmultiple-random-variable-for-a-uniform-distribution%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest




















































































                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Aardman Animations

                      Are they similar matrix