expectation of absolute value of cauchy distribution & general questions about expectation of absolute...












1














The Cauchy distribution has PDF:



$$f_X(x) = frac{1}{pi(1 + x^2)}$$



Its expectation does not exist.



If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:



$$E(|X|) = int_{-infty}^{infty} |x| cdot frac{1}{pi(1 + x^2)} dx = int_{-infty}^{0} -x cdot frac{1}{pi(1 + x^2)} dx + int_{0}^{infty} x cdot frac{1}{pi(1 + x^2)} dx$$



If I'm not mistaken, neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?



Furthermore, is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist), or is it specific to just this distribution?



And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?










share|cite|improve this question





























    1














    The Cauchy distribution has PDF:



    $$f_X(x) = frac{1}{pi(1 + x^2)}$$



    Its expectation does not exist.



    If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:



    $$E(|X|) = int_{-infty}^{infty} |x| cdot frac{1}{pi(1 + x^2)} dx = int_{-infty}^{0} -x cdot frac{1}{pi(1 + x^2)} dx + int_{0}^{infty} x cdot frac{1}{pi(1 + x^2)} dx$$



    If I'm not mistaken, neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?



    Furthermore, is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist), or is it specific to just this distribution?



    And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?










    share|cite|improve this question



























      1












      1








      1







      The Cauchy distribution has PDF:



      $$f_X(x) = frac{1}{pi(1 + x^2)}$$



      Its expectation does not exist.



      If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:



      $$E(|X|) = int_{-infty}^{infty} |x| cdot frac{1}{pi(1 + x^2)} dx = int_{-infty}^{0} -x cdot frac{1}{pi(1 + x^2)} dx + int_{0}^{infty} x cdot frac{1}{pi(1 + x^2)} dx$$



      If I'm not mistaken, neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?



      Furthermore, is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist), or is it specific to just this distribution?



      And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?










      share|cite|improve this question















      The Cauchy distribution has PDF:



      $$f_X(x) = frac{1}{pi(1 + x^2)}$$



      Its expectation does not exist.



      If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:



      $$E(|X|) = int_{-infty}^{infty} |x| cdot frac{1}{pi(1 + x^2)} dx = int_{-infty}^{0} -x cdot frac{1}{pi(1 + x^2)} dx + int_{0}^{infty} x cdot frac{1}{pi(1 + x^2)} dx$$



      If I'm not mistaken, neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?



      Furthermore, is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist), or is it specific to just this distribution?



      And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?







      probability probability-theory statistics probability-distributions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 20:31

























      asked Nov 27 '18 at 20:26









      0k33

      12010




      12010






















          1 Answer
          1






          active

          oldest

          votes


















          1















          If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:




          Yes.




          Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?




          Yes. (Or, to be more precise, it's $infty$.)




          Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?



          And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?




          For any random variable, having $mathbb E[|X|] < infty$ and having $mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
          $$mathbb E[X] = int_{-infty}^{infty} x cdot f(x) , textrm{d} x$$
          where $f(x)$ is the associated density function. We can always rewrite this as
          $$mathbb E[X] = int_{-infty}^0 x cdot f(x) , textrm d x + int_0^{infty} x cdot f(x) , textrm d x.$$
          Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
          $$mathbb E[|X|] = int_{-infty}^0 |x| cdot f(x) , textrm d x + int_0^{infty} |x| cdot f(x) , textrm d x$$
          must exist as well. The logic works just as well in reverse to show that $mathbb E[|X|] < infty implies mathbb E[X] < infty$.






          share|cite|improve this answer





















            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%2f3016262%2fexpectation-of-absolute-value-of-cauchy-distribution-general-questions-about-e%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















            If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:




            Yes.




            Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?




            Yes. (Or, to be more precise, it's $infty$.)




            Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?



            And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?




            For any random variable, having $mathbb E[|X|] < infty$ and having $mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
            $$mathbb E[X] = int_{-infty}^{infty} x cdot f(x) , textrm{d} x$$
            where $f(x)$ is the associated density function. We can always rewrite this as
            $$mathbb E[X] = int_{-infty}^0 x cdot f(x) , textrm d x + int_0^{infty} x cdot f(x) , textrm d x.$$
            Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
            $$mathbb E[|X|] = int_{-infty}^0 |x| cdot f(x) , textrm d x + int_0^{infty} |x| cdot f(x) , textrm d x$$
            must exist as well. The logic works just as well in reverse to show that $mathbb E[|X|] < infty implies mathbb E[X] < infty$.






            share|cite|improve this answer


























              1















              If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:




              Yes.




              Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?




              Yes. (Or, to be more precise, it's $infty$.)




              Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?



              And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?




              For any random variable, having $mathbb E[|X|] < infty$ and having $mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
              $$mathbb E[X] = int_{-infty}^{infty} x cdot f(x) , textrm{d} x$$
              where $f(x)$ is the associated density function. We can always rewrite this as
              $$mathbb E[X] = int_{-infty}^0 x cdot f(x) , textrm d x + int_0^{infty} x cdot f(x) , textrm d x.$$
              Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
              $$mathbb E[|X|] = int_{-infty}^0 |x| cdot f(x) , textrm d x + int_0^{infty} |x| cdot f(x) , textrm d x$$
              must exist as well. The logic works just as well in reverse to show that $mathbb E[|X|] < infty implies mathbb E[X] < infty$.






              share|cite|improve this answer
























                1












                1








                1







                If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:




                Yes.




                Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?




                Yes. (Or, to be more precise, it's $infty$.)




                Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?



                And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?




                For any random variable, having $mathbb E[|X|] < infty$ and having $mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
                $$mathbb E[X] = int_{-infty}^{infty} x cdot f(x) , textrm{d} x$$
                where $f(x)$ is the associated density function. We can always rewrite this as
                $$mathbb E[X] = int_{-infty}^0 x cdot f(x) , textrm d x + int_0^{infty} x cdot f(x) , textrm d x.$$
                Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
                $$mathbb E[|X|] = int_{-infty}^0 |x| cdot f(x) , textrm d x + int_0^{infty} |x| cdot f(x) , textrm d x$$
                must exist as well. The logic works just as well in reverse to show that $mathbb E[|X|] < infty implies mathbb E[X] < infty$.






                share|cite|improve this answer













                If we wanted to compute the expectation of the absolute value of this distribution, would it be correct to do the following:




                Yes.




                Neither of the integrals in the above sum of integrals converges, so does this mean that the expectation of the absolute value also does not exist?




                Yes. (Or, to be more precise, it's $infty$.)




                Is this a general result (that if the expectation of the original distribution does not exist, the expectation of the absolute value of the distribution also does not exist)?



                And what about the converse: if the expectation of the absolute value of the distribution does not exist, what can we conclude (if anything) about the expectation of the original distribution?




                For any random variable, having $mathbb E[|X|] < infty$ and having $mathbb E[X]$ exist as a finite number are indeed equivalent. Proof for continuous variables (discrete variables are similar): Note that
                $$mathbb E[X] = int_{-infty}^{infty} x cdot f(x) , textrm{d} x$$
                where $f(x)$ is the associated density function. We can always rewrite this as
                $$mathbb E[X] = int_{-infty}^0 x cdot f(x) , textrm d x + int_0^{infty} x cdot f(x) , textrm d x.$$
                Note that the left integrand is purely negative and the right integrand is purely positive. The rightmost integral therefore either converges to a finite positive number, or it diverges to $infty$; similarly, the leftmost integral either converges to a finite negative number, or it diverges to $-infty$. The only case in which the total expectation exists (meaning yields a finite number) is if both the integrals converge to finite numbers, in which case it is not hard to see why
                $$mathbb E[|X|] = int_{-infty}^0 |x| cdot f(x) , textrm d x + int_0^{infty} |x| cdot f(x) , textrm d x$$
                must exist as well. The logic works just as well in reverse to show that $mathbb E[|X|] < infty implies mathbb E[X] < infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 '18 at 20:42









                Aaron Montgomery

                4,712523




                4,712523






























                    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%2f3016262%2fexpectation-of-absolute-value-of-cauchy-distribution-general-questions-about-e%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

                    Quarter-circle Tiles

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Mont Emei