Why can we treat MGF in this way

Multi tool use
Multi tool use












1












$begingroup$


For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



We write



$$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



That is , we use the standard normal density to compute the moment generating function.



Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?










share|cite|edit









$endgroup$

















    1












    $begingroup$


    For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



    We write



    $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



    That is , we use the standard normal density to compute the moment generating function.



    Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?










    share|cite|edit









    $endgroup$















      1












      1








      1





      $begingroup$


      For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



      We write



      $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



      That is , we use the standard normal density to compute the moment generating function.



      Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?










      share|cite|edit









      $endgroup$




      For the standard proof that if $Z sim N(0,1)$ than $Z^{2} sim chi^{2}_{1}$



      We write



      $$M_{Z^{2}}(t)=int_{mathbb{R}}exp(tz^{2})frac{1}{sqrt{2pi}}exp(frac{-z^{2}}{2}) dz$$



      That is , we use the standard normal density to compute the moment generating function.



      Why do we not need to use the density of $Z^{2}$ and instead are able to use the density of just $Z$, I assume because it is a continuous one to one function?







      mgf






      share|cite|edit













      share|cite|edit











      share|cite|edit




      share|cite|edit










      asked 5 hours ago









      QualityQuality

      24919




      24919






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          In principle, you don't know the density of $Z$, so you can't use it.



          What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






          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: "65"
            };
            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: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            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
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f391679%2fwhy-can-we-treat-mgf-in-this-way%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









            2












            $begingroup$

            In principle, you don't know the density of $Z$, so you can't use it.



            What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              In principle, you don't know the density of $Z$, so you can't use it.



              What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                In principle, you don't know the density of $Z$, so you can't use it.



                What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.






                share|cite|improve this answer









                $endgroup$



                In principle, you don't know the density of $Z$, so you can't use it.



                What's being done when you use the density of $X$ is an application of the law of the unconscious statistician.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Lucas FariasLucas Farias

                469417




                469417






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Cross Validated!


                    • 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%2fstats.stackexchange.com%2fquestions%2f391679%2fwhy-can-we-treat-mgf-in-this-way%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







                    rGOw5Q3Qt4xCrGrCOxzlP,HdqS6Mk33f7 hy0dTJFquuEicOjEvAcl N,t594rNpF8LI e1gLpft1sQV4bgZMMrjDw 6CFHLmX
                    cXBZjfXURn3Z,hafRSMBw314yono6tr0SQusg2eI153qbDt3XviD0w

                    Popular posts from this blog

                    Running commands remotely through an RDP connection

                    apt-mirror Malformed checksum error

                    Matlab 2018b additional configuration steps message while installing?