Transformation method of two random variables












0












$begingroup$


My problem:



Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



(a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



(b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
With the transformation method I obtain for $U$ the following density function
begin{equation} f_U(u)=2e^{-2u} end{equation}



My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






probability-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    My problem:



    Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



    (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



    (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



    My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
    begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
    With the transformation method I obtain for $U$ the following density function
    begin{equation} f_U(u)=2e^{-2u} end{equation}



    My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






    probability-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      My problem:



      Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



      (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



      (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



      My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
      begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
      With the transformation method I obtain for $U$ the following density function
      begin{equation} f_U(u)=2e^{-2u} end{equation}



      My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






      probability-theory







      share|cite|improve this question









      $endgroup$




      My problem:



      Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0leq y_1leq y_2 < infty$ and $0$ otherwise be the joint density function.



      (a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.



      (b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.



      My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as
      begin{equation} U=Y_2/2 hspace 0.5cm and hspace 0.5cm V=Y_1+1 end{equation}
      With the transformation method I obtain for $U$ the following density function
      begin{equation} f_U(u)=2e^{-2u} end{equation}



      My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...






      probability-theory




      probability-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 20:55









      Joey AdamsJoey Adams

      457




      457






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39













          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%2f3067929%2ftransformation-method-of-two-random-variables%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









          1












          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39


















          1












          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39
















          1












          1








          1





          $begingroup$

          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.






          share|cite|improve this answer











          $endgroup$



          Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=int f(y_1,y_2)dy_2 =int_{y_1}^{infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<infty$; $f_{Y_2}(y_2)=int f(y_1,y_2)dy_1 =int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 0:35

























          answered Jan 9 at 23:48









          Kavi Rama MurthyKavi Rama Murthy

          54.4k32055




          54.4k32055












          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39




















          • $begingroup$
            So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
            $endgroup$
            – Joey Adams
            Jan 10 at 0:05






          • 1




            $begingroup$
            @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
            $endgroup$
            – Kavi Rama Murthy
            Jan 10 at 0:36










          • $begingroup$
            Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
            $endgroup$
            – Joey Adams
            Jan 10 at 0:39


















          $begingroup$
          So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
          $endgroup$
          – Joey Adams
          Jan 10 at 0:05




          $begingroup$
          So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function.
          $endgroup$
          – Joey Adams
          Jan 10 at 0:05




          1




          1




          $begingroup$
          @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
          $endgroup$
          – Kavi Rama Murthy
          Jan 10 at 0:36




          $begingroup$
          @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$
          $endgroup$
          – Kavi Rama Murthy
          Jan 10 at 0:36












          $begingroup$
          Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
          $endgroup$
          – Joey Adams
          Jan 10 at 0:39






          $begingroup$
          Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$
          $endgroup$
          – Joey Adams
          Jan 10 at 0:39




















          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%2f3067929%2ftransformation-method-of-two-random-variables%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

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          Dobbiaco

          Image
          Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
          Read more