pdf of transformed variable












1












$begingroup$


Given pdf $f_X(x)=frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = int_{-2}^xfrac{x+2}{18}dx=frac{x^2}{36}+frac{x}{9}+frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?



begin{align*}
F_Y(y)&=P(Y leq y) \
&=P(frac{12}{|X|}leq y) \
&= P(|X| geq frac{12}{y}) \
&= 1 - P(|X| leq frac{12}{y})\
&=1- P(-frac{12}{y} leq X leq frac{12}{y}) \
&= 1 - F_X(frac{12}{y}) + F_X(-frac{12}{y})\
&= 1 - frac{8}{3y}
end{align*}
Then I just differentiate to get the pdf:
begin{align*}
f_Y(y)&=frac{d}{dy}F_Y(y) \
&= frac{8}{3y^2}
end{align*}



But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.






probability probability-theory probability-distributions density-function







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Given pdf $f_X(x)=frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = int_{-2}^xfrac{x+2}{18}dx=frac{x^2}{36}+frac{x}{9}+frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?



    begin{align*}
    F_Y(y)&=P(Y leq y) \
    &=P(frac{12}{|X|}leq y) \
    &= P(|X| geq frac{12}{y}) \
    &= 1 - P(|X| leq frac{12}{y})\
    &=1- P(-frac{12}{y} leq X leq frac{12}{y}) \
    &= 1 - F_X(frac{12}{y}) + F_X(-frac{12}{y})\
    &= 1 - frac{8}{3y}
    end{align*}
    Then I just differentiate to get the pdf:
    begin{align*}
    f_Y(y)&=frac{d}{dy}F_Y(y) \
    &= frac{8}{3y^2}
    end{align*}



    But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.






    probability probability-theory probability-distributions density-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Given pdf $f_X(x)=frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = int_{-2}^xfrac{x+2}{18}dx=frac{x^2}{36}+frac{x}{9}+frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?



      begin{align*}
      F_Y(y)&=P(Y leq y) \
      &=P(frac{12}{|X|}leq y) \
      &= P(|X| geq frac{12}{y}) \
      &= 1 - P(|X| leq frac{12}{y})\
      &=1- P(-frac{12}{y} leq X leq frac{12}{y}) \
      &= 1 - F_X(frac{12}{y}) + F_X(-frac{12}{y})\
      &= 1 - frac{8}{3y}
      end{align*}
      Then I just differentiate to get the pdf:
      begin{align*}
      f_Y(y)&=frac{d}{dy}F_Y(y) \
      &= frac{8}{3y^2}
      end{align*}



      But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.






      probability probability-theory probability-distributions density-function







      share|cite|improve this question









      $endgroup$




      Given pdf $f_X(x)=frac{x+2}{18}$ where $-2 < x < 4$, I wanted to find another r.v. $Y = frac{12}{|X|}$. I think the support of $Y$ would be $3 < y < infty$ but I wasn't super sure. I found the cdf of $X$, which was $F_X(x) = int_{-2}^xfrac{x+2}{18}dx=frac{x^2}{36}+frac{x}{9}+frac{1}{9}$, $-2 < x < 4$. I tried using the cdf of $X$ to find the cdf of $Y$ and consequently find the pdf of $Y$ but have been struggling to do so. Could anyone help me with the derivation?



      begin{align*}
      F_Y(y)&=P(Y leq y) \
      &=P(frac{12}{|X|}leq y) \
      &= P(|X| geq frac{12}{y}) \
      &= 1 - P(|X| leq frac{12}{y})\
      &=1- P(-frac{12}{y} leq X leq frac{12}{y}) \
      &= 1 - F_X(frac{12}{y}) + F_X(-frac{12}{y})\
      &= 1 - frac{8}{3y}
      end{align*}
      Then I just differentiate to get the pdf:
      begin{align*}
      f_Y(y)&=frac{d}{dy}F_Y(y) \
      &= frac{8}{3y^2}
      end{align*}



      But this pdf doesn't integrate to $1$ so I'm not sure what's wrong.






      probability probability-theory probability-distributions density-function




      probability probability-theory probability-distributions density-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 25 at 1:15









      Michael JacksonMichael Jackson

      82




      82






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Take care with the absolute signage and the supports for the functions.



          Note that $12/lvert Xrvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;infty)$, and the domain $[2;4)$ onto $(3;6]$.



          More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Yleq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).



          $$begin{split}mathsf P(Yleq y) &=mathsf P(lvert Xrvertleq 12/y)mathbf 1_{6<y}+mathsf P(lvert Xrvertleq 12/y)mathbf 1_{3<yleq 6}\ &= mathsf P(-12/yleq Xleq 12/y)mathbf 1_{6<y}+mathsf P(0leq Xleq 12/y)mathbf 1_{3<yleq 6}\ &= -F_X(-12/y)mathbf 1_{6<y}-F_X(0)mathbf 1_{3<yleq 6}+F_X(12/y)mathbf 1_{3<y}end{split}$$



          So



          $$begin{split}f_Y(y)&=begin{vmatrix}dfrac{partial (-12/y)}{partial y}end{vmatrix}f_X(-12/y)mathbf 1_{yin(6;infty)}+begin{vmatrix}dfrac{partial (12/y)}{partial y}end{vmatrix}f_X(12/y)mathbf 1_{yin(3;infty)}\ &=dfrac{(-24+4y)}{3y^3}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<y}\ &= dfrac{8}{3y^2}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<yleq 6}end{split}$$






          share|cite|improve this answer











          $endgroup$













            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%2f3086587%2fpdf-of-transformed-variable%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









            0












            $begingroup$

            Take care with the absolute signage and the supports for the functions.



            Note that $12/lvert Xrvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;infty)$, and the domain $[2;4)$ onto $(3;6]$.



            More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Yleq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).



            $$begin{split}mathsf P(Yleq y) &=mathsf P(lvert Xrvertleq 12/y)mathbf 1_{6<y}+mathsf P(lvert Xrvertleq 12/y)mathbf 1_{3<yleq 6}\ &= mathsf P(-12/yleq Xleq 12/y)mathbf 1_{6<y}+mathsf P(0leq Xleq 12/y)mathbf 1_{3<yleq 6}\ &= -F_X(-12/y)mathbf 1_{6<y}-F_X(0)mathbf 1_{3<yleq 6}+F_X(12/y)mathbf 1_{3<y}end{split}$$



            So



            $$begin{split}f_Y(y)&=begin{vmatrix}dfrac{partial (-12/y)}{partial y}end{vmatrix}f_X(-12/y)mathbf 1_{yin(6;infty)}+begin{vmatrix}dfrac{partial (12/y)}{partial y}end{vmatrix}f_X(12/y)mathbf 1_{yin(3;infty)}\ &=dfrac{(-24+4y)}{3y^3}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<y}\ &= dfrac{8}{3y^2}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<yleq 6}end{split}$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Take care with the absolute signage and the supports for the functions.



              Note that $12/lvert Xrvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;infty)$, and the domain $[2;4)$ onto $(3;6]$.



              More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Yleq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).



              $$begin{split}mathsf P(Yleq y) &=mathsf P(lvert Xrvertleq 12/y)mathbf 1_{6<y}+mathsf P(lvert Xrvertleq 12/y)mathbf 1_{3<yleq 6}\ &= mathsf P(-12/yleq Xleq 12/y)mathbf 1_{6<y}+mathsf P(0leq Xleq 12/y)mathbf 1_{3<yleq 6}\ &= -F_X(-12/y)mathbf 1_{6<y}-F_X(0)mathbf 1_{3<yleq 6}+F_X(12/y)mathbf 1_{3<y}end{split}$$



              So



              $$begin{split}f_Y(y)&=begin{vmatrix}dfrac{partial (-12/y)}{partial y}end{vmatrix}f_X(-12/y)mathbf 1_{yin(6;infty)}+begin{vmatrix}dfrac{partial (12/y)}{partial y}end{vmatrix}f_X(12/y)mathbf 1_{yin(3;infty)}\ &=dfrac{(-24+4y)}{3y^3}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<y}\ &= dfrac{8}{3y^2}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<yleq 6}end{split}$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Take care with the absolute signage and the supports for the functions.



                Note that $12/lvert Xrvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;infty)$, and the domain $[2;4)$ onto $(3;6]$.



                More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Yleq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).



                $$begin{split}mathsf P(Yleq y) &=mathsf P(lvert Xrvertleq 12/y)mathbf 1_{6<y}+mathsf P(lvert Xrvertleq 12/y)mathbf 1_{3<yleq 6}\ &= mathsf P(-12/yleq Xleq 12/y)mathbf 1_{6<y}+mathsf P(0leq Xleq 12/y)mathbf 1_{3<yleq 6}\ &= -F_X(-12/y)mathbf 1_{6<y}-F_X(0)mathbf 1_{3<yleq 6}+F_X(12/y)mathbf 1_{3<y}end{split}$$



                So



                $$begin{split}f_Y(y)&=begin{vmatrix}dfrac{partial (-12/y)}{partial y}end{vmatrix}f_X(-12/y)mathbf 1_{yin(6;infty)}+begin{vmatrix}dfrac{partial (12/y)}{partial y}end{vmatrix}f_X(12/y)mathbf 1_{yin(3;infty)}\ &=dfrac{(-24+4y)}{3y^3}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<y}\ &= dfrac{8}{3y^2}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<yleq 6}end{split}$$






                share|cite|improve this answer











                $endgroup$



                Take care with the absolute signage and the supports for the functions.



                Note that $12/lvert Xrvert$ folds both the domains $(-2;0)$ and $(0;2)$ onto $(6;infty)$, and the domain $[2;4)$ onto $(3;6]$.



                More clearly, only when $Y>6$ it is mapped to by a positive and negative value for $X$. When $3<Yleq 6$ then $Y$ is mapped to by only a positive value of $X$ (on $[2;4)$).



                $$begin{split}mathsf P(Yleq y) &=mathsf P(lvert Xrvertleq 12/y)mathbf 1_{6<y}+mathsf P(lvert Xrvertleq 12/y)mathbf 1_{3<yleq 6}\ &= mathsf P(-12/yleq Xleq 12/y)mathbf 1_{6<y}+mathsf P(0leq Xleq 12/y)mathbf 1_{3<yleq 6}\ &= -F_X(-12/y)mathbf 1_{6<y}-F_X(0)mathbf 1_{3<yleq 6}+F_X(12/y)mathbf 1_{3<y}end{split}$$



                So



                $$begin{split}f_Y(y)&=begin{vmatrix}dfrac{partial (-12/y)}{partial y}end{vmatrix}f_X(-12/y)mathbf 1_{yin(6;infty)}+begin{vmatrix}dfrac{partial (12/y)}{partial y}end{vmatrix}f_X(12/y)mathbf 1_{yin(3;infty)}\ &=dfrac{(-24+4y)}{3y^3}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<y}\ &= dfrac{8}{3y^2}mathbf 1_{6<y}+dfrac{(24+4y)}{3y^3}mathbf 1_{3<yleq 6}end{split}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 25 at 3:21

























                answered Jan 25 at 3:04









                Graham KempGraham Kemp

                86.8k43579




                86.8k43579






























                    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%2f3086587%2fpdf-of-transformed-variable%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

                    MongoDB - Not Authorized To Execute Command

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                    Read more

                    How to fix TextFormField cause rebuild widget in Flutter

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                    Read more

                    Npm cannot find a required file even through it is in the searched directory

                    Image
                    up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
                    Read more