Joint PDF P[X+Y<=0.5]












1












$begingroup$


I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



I need to find the $P(x+yleq 0.5)$.



For the double integration, I have the following bounds:




  • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


  • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



Here is the initial set up:



$int_0^.5 int_0^{.5-x}(x +y) dydx$.



My steps of integration:



1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






probability







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



    Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



    I need to find the $P(x+yleq 0.5)$.



    For the double integration, I have the following bounds:




    • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


    • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



    Here is the initial set up:



    $int_0^.5 int_0^{.5-x}(x +y) dydx$.



    My steps of integration:



    1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



    2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



    3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



    I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






    probability







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability







      share|cite|improve this question











      $endgroup$




      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 10:42







      MitterHai

















      asked Dec 5 '18 at 10:19









      MitterHaiMitterHai

      84




      84






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43



















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40











          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%2f3026902%2fjoint-pdf-pxy-0-5%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









          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43














          0












          0








          0





          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$



          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 11:22









          farruhotafarruhota

          20.2k2738




          20.2k2738












          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43


















          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42




          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42












          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43




          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43











          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40














          1












          1








          1





          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$



          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 10:22









          Kavi Rama MurthyKavi Rama Murthy

          57.7k42160




          57.7k42160












          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40


















          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38




          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38












          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22




          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22












          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40




          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40


















          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%2f3026902%2fjoint-pdf-pxy-0-5%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

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
          Read more

          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
          Read more

          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

          Image
          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
          Read more