Double integrals - how are the boundaries chosen?











up vote
0
down vote

favorite












I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



    After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



    $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



    Can someone please explain the last equality
    $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



      After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



      $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



      Can someone please explain the last equality
      $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.










      share|cite|improve this question













      I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.



      After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get



      $P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$



      Can someone please explain the last equality
      $int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.







      integration convolution density-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 19 at 10:15









      Tesla

      916426




      916426






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer





















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24











          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%2f3004748%2fdouble-integrals-how-are-the-boundaries-chosen%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










          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer





















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24















          up vote
          1
          down vote



          accepted










          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer





















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).






          share|cite|improve this answer












          For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 10:18









          Kavi Rama Murthy

          45.8k31853




          45.8k31853












          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24


















          • Got it, thank you very much! And why am I allowed to take the functions out of the integral?
            – Tesla
            Nov 19 at 10:22






          • 1




            When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
            – Kavi Rama Murthy
            Nov 19 at 10:24
















          Got it, thank you very much! And why am I allowed to take the functions out of the integral?
          – Tesla
          Nov 19 at 10:22




          Got it, thank you very much! And why am I allowed to take the functions out of the integral?
          – Tesla
          Nov 19 at 10:22




          1




          1




          When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
          – Kavi Rama Murthy
          Nov 19 at 10:24




          When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
          – Kavi Rama Murthy
          Nov 19 at 10:24


















          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%2f3004748%2fdouble-integrals-how-are-the-boundaries-chosen%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

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

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