Integration of PDF's to find valid constant











up vote
0
down vote

favorite












I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






Let $X$ and $Y$ have joint PDF



$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






My approach



Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}



Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






Correct solution



begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}



This gives us $c = 4$.






I have two specific questions:




  1. Why did the correct solution switch the order of integration?


  2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






Any feedback is appreciated. Thank you.






EDIT



The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






    Let $X$ and $Y$ have joint PDF



    $$f_{X, Y}(x, y) = cxy$$
    for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






    It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






    My approach



    Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



    begin{align}
    int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
    & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
    & = frac{c(1-x^4)}{8}
    end{align}



    Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






    Correct solution



    begin{align}
    int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
    & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
    & = frac{c}{4} \
    end{align}



    This gives us $c = 4$.






    I have two specific questions:




    1. Why did the correct solution switch the order of integration?


    2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






    Any feedback is appreciated. Thank you.






    EDIT



    The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






      Let $X$ and $Y$ have joint PDF



      $$f_{X, Y}(x, y) = cxy$$
      for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






      It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






      My approach



      Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



      begin{align}
      int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
      & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
      & = frac{c(1-x^4)}{8}
      end{align}



      Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






      Correct solution



      begin{align}
      int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
      & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
      & = frac{c}{4} \
      end{align}



      This gives us $c = 4$.






      I have two specific questions:




      1. Why did the correct solution switch the order of integration?


      2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






      Any feedback is appreciated. Thank you.






      EDIT



      The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.










      share|cite|improve this question















      I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:






      Let $X$ and $Y$ have joint PDF



      $$f_{X, Y}(x, y) = cxy$$
      for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.






      It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.






      My approach



      Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,



      begin{align}
      int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
      & = int_{y=x}^{y=1}frac{c}{2}y^3dy \
      & = frac{c(1-x^4)}{8}
      end{align}



      Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.






      Correct solution



      begin{align}
      int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
      & = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
      & = frac{c}{4} \
      end{align}



      This gives us $c = 4$.






      I have two specific questions:




      1. Why did the correct solution switch the order of integration?


      2. The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?






      Any feedback is appreciated. Thank you.






      EDIT



      The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.







      probability integration density-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 15:06

























      asked Nov 18 at 12:31









      Sean

      23510




      23510






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            Nov 18 at 13:05










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            Nov 18 at 13:06






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            Nov 18 at 13:14








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            Nov 18 at 13:19






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            Nov 18 at 13:32











          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%2f3003474%2fintegration-of-pdfs-to-find-valid-constant%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








          up vote
          1
          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            Nov 18 at 13:05










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            Nov 18 at 13:06






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            Nov 18 at 13:14








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            Nov 18 at 13:19






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            Nov 18 at 13:32















          up vote
          1
          down vote



          accepted










          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer





















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            Nov 18 at 13:05










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            Nov 18 at 13:06






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            Nov 18 at 13:14








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            Nov 18 at 13:19






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            Nov 18 at 13:32













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?






          share|cite|improve this answer












          Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
          In your solution you are fixing x, but what even is the value of x you are fixing it to?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 12:48









          Sorin Tirc

          77210




          77210












          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            Nov 18 at 13:05










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            Nov 18 at 13:06






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            Nov 18 at 13:14








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            Nov 18 at 13:19






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            Nov 18 at 13:32


















          • That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
            – Sean
            Nov 18 at 13:05










          • Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
            – Sean
            Nov 18 at 13:06






          • 1




            You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
            – Sorin Tirc
            Nov 18 at 13:14








          • 1




            What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
            – Sorin Tirc
            Nov 18 at 13:19






          • 1




            Yes, absolutely ;)
            – Sorin Tirc
            Nov 18 at 13:32
















          That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
          – Sean
          Nov 18 at 13:05




          That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
          – Sean
          Nov 18 at 13:05












          Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
          – Sean
          Nov 18 at 13:06




          Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
          – Sean
          Nov 18 at 13:06




          1




          1




          You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
          – Sorin Tirc
          Nov 18 at 13:14






          You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
          – Sorin Tirc
          Nov 18 at 13:14






          1




          1




          What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
          – Sorin Tirc
          Nov 18 at 13:19




          What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
          – Sorin Tirc
          Nov 18 at 13:19




          1




          1




          Yes, absolutely ;)
          – Sorin Tirc
          Nov 18 at 13:32




          Yes, absolutely ;)
          – Sorin Tirc
          Nov 18 at 13:32


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3003474%2fintegration-of-pdfs-to-find-valid-constant%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

          QoS: MAC-Priority for clients behind a repeater

          Ивакино (Тотемский район)

          Can't locate Autom4te/ChannelDefs.pm in @INC (when it definitely is there)