Likelihood Function for the Uniform Density $(theta, theta+1)$












1












$begingroup$


Let the random variable X have a uniform density given by



$f(x;theta)$~$R(theta,theta+1)$



What is the maximum likelihood function according to the samples $X_1,ldots,X_n$?



The question is much like Likelihood Function for the Uniform Density. But there seems no satisfying answer. So far I get $X_{max}-1lethetale X_{min}$ and my guess is $hattheta=frac{X_{max}+X_{min}-1}{2}$. Am I right? I need a theoretical explaination.






statistics probability-theory statistical-inference







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let the random variable X have a uniform density given by



    $f(x;theta)$~$R(theta,theta+1)$



    What is the maximum likelihood function according to the samples $X_1,ldots,X_n$?



    The question is much like Likelihood Function for the Uniform Density. But there seems no satisfying answer. So far I get $X_{max}-1lethetale X_{min}$ and my guess is $hattheta=frac{X_{max}+X_{min}-1}{2}$. Am I right? I need a theoretical explaination.






    statistics probability-theory statistical-inference







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      Let the random variable X have a uniform density given by



      $f(x;theta)$~$R(theta,theta+1)$



      What is the maximum likelihood function according to the samples $X_1,ldots,X_n$?



      The question is much like Likelihood Function for the Uniform Density. But there seems no satisfying answer. So far I get $X_{max}-1lethetale X_{min}$ and my guess is $hattheta=frac{X_{max}+X_{min}-1}{2}$. Am I right? I need a theoretical explaination.






      statistics probability-theory statistical-inference







      share|cite|improve this question











      $endgroup$




      Let the random variable X have a uniform density given by



      $f(x;theta)$~$R(theta,theta+1)$



      What is the maximum likelihood function according to the samples $X_1,ldots,X_n$?



      The question is much like Likelihood Function for the Uniform Density. But there seems no satisfying answer. So far I get $X_{max}-1lethetale X_{min}$ and my guess is $hattheta=frac{X_{max}+X_{min}-1}{2}$. Am I right? I need a theoretical explaination.






      statistics probability-theory statistical-inference




      statistics probability-theory statistical-inference






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 13 '17 at 12:21









      Community♦

      1




      1










      asked Mar 22 '14 at 12:48









      FrazerFrazer

      1697




      1697






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Hint: When you say that "So far I get $X_{max}-1lethetale X_{min}$" you are actually saying that the likelihood $L(theta ; x_1 cdots x_n)$ (regarded as a function of $theta$) is zero outside that range. Now, what is the likelihood if $theta$ is inside that range?



          Update: you know that the density of each uniform sample is $f(x;theta) = 1$ if $theta <xle theta+1$, $0$ otherwise.



          Then the likelihood $L(theta;x_1 cdots x_n) =prod f(x_i;theta) $ is $1$ if $theta <x_ile theta+1$ $forall i$, $0$ otherwise. In terms of $theta$ as variable, this is equivalent to say that $L(theta;x_1 cdots x_n)=1$ in $X_{max}-1lethetale X_{min}$. You need to find the maximum of $L(theta;x_1 cdots x_n)$ as a function of $theta$. But $L(theta;x_1 cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks, "any value" is really beyond my thought...
            $endgroup$
            – Frazer
            Mar 22 '14 at 15:20










          • $begingroup$
            @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
            $endgroup$
            – Kumara
            Dec 17 '18 at 11:49






          • 1




            $begingroup$
            Undefined. Anyway, that cannot happen under the assumed model
            $endgroup$
            – leonbloy
            Dec 17 '18 at 14:52



















          1












          $begingroup$

          Here a two thoughts




          • If any $X_i$ lies outside of $[theta,theta+1]$, then the likelyhood function is zero

          • If all the $X_i$ lie inside $[theta_1,theta_1+1]$ and also inside $[theta_2,theta_2+1]$, then the likelyhoods are the same, because $f(x,theta_1) = f(x,theta_2)$ for those $x$ which lie in the intersection of the two intervals






          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%2f722101%2flikelihood-function-for-the-uniform-density-theta-theta1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Hint: When you say that "So far I get $X_{max}-1lethetale X_{min}$" you are actually saying that the likelihood $L(theta ; x_1 cdots x_n)$ (regarded as a function of $theta$) is zero outside that range. Now, what is the likelihood if $theta$ is inside that range?



            Update: you know that the density of each uniform sample is $f(x;theta) = 1$ if $theta <xle theta+1$, $0$ otherwise.



            Then the likelihood $L(theta;x_1 cdots x_n) =prod f(x_i;theta) $ is $1$ if $theta <x_ile theta+1$ $forall i$, $0$ otherwise. In terms of $theta$ as variable, this is equivalent to say that $L(theta;x_1 cdots x_n)=1$ in $X_{max}-1lethetale X_{min}$. You need to find the maximum of $L(theta;x_1 cdots x_n)$ as a function of $theta$. But $L(theta;x_1 cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, "any value" is really beyond my thought...
              $endgroup$
              – Frazer
              Mar 22 '14 at 15:20










            • $begingroup$
              @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
              $endgroup$
              – Kumara
              Dec 17 '18 at 11:49






            • 1




              $begingroup$
              Undefined. Anyway, that cannot happen under the assumed model
              $endgroup$
              – leonbloy
              Dec 17 '18 at 14:52
















            2












            $begingroup$

            Hint: When you say that "So far I get $X_{max}-1lethetale X_{min}$" you are actually saying that the likelihood $L(theta ; x_1 cdots x_n)$ (regarded as a function of $theta$) is zero outside that range. Now, what is the likelihood if $theta$ is inside that range?



            Update: you know that the density of each uniform sample is $f(x;theta) = 1$ if $theta <xle theta+1$, $0$ otherwise.



            Then the likelihood $L(theta;x_1 cdots x_n) =prod f(x_i;theta) $ is $1$ if $theta <x_ile theta+1$ $forall i$, $0$ otherwise. In terms of $theta$ as variable, this is equivalent to say that $L(theta;x_1 cdots x_n)=1$ in $X_{max}-1lethetale X_{min}$. You need to find the maximum of $L(theta;x_1 cdots x_n)$ as a function of $theta$. But $L(theta;x_1 cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, "any value" is really beyond my thought...
              $endgroup$
              – Frazer
              Mar 22 '14 at 15:20










            • $begingroup$
              @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
              $endgroup$
              – Kumara
              Dec 17 '18 at 11:49






            • 1




              $begingroup$
              Undefined. Anyway, that cannot happen under the assumed model
              $endgroup$
              – leonbloy
              Dec 17 '18 at 14:52














            2












            2








            2





            $begingroup$

            Hint: When you say that "So far I get $X_{max}-1lethetale X_{min}$" you are actually saying that the likelihood $L(theta ; x_1 cdots x_n)$ (regarded as a function of $theta$) is zero outside that range. Now, what is the likelihood if $theta$ is inside that range?



            Update: you know that the density of each uniform sample is $f(x;theta) = 1$ if $theta <xle theta+1$, $0$ otherwise.



            Then the likelihood $L(theta;x_1 cdots x_n) =prod f(x_i;theta) $ is $1$ if $theta <x_ile theta+1$ $forall i$, $0$ otherwise. In terms of $theta$ as variable, this is equivalent to say that $L(theta;x_1 cdots x_n)=1$ in $X_{max}-1lethetale X_{min}$. You need to find the maximum of $L(theta;x_1 cdots x_n)$ as a function of $theta$. But $L(theta;x_1 cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.






            share|cite|improve this answer











            $endgroup$



            Hint: When you say that "So far I get $X_{max}-1lethetale X_{min}$" you are actually saying that the likelihood $L(theta ; x_1 cdots x_n)$ (regarded as a function of $theta$) is zero outside that range. Now, what is the likelihood if $theta$ is inside that range?



            Update: you know that the density of each uniform sample is $f(x;theta) = 1$ if $theta <xle theta+1$, $0$ otherwise.



            Then the likelihood $L(theta;x_1 cdots x_n) =prod f(x_i;theta) $ is $1$ if $theta <x_ile theta+1$ $forall i$, $0$ otherwise. In terms of $theta$ as variable, this is equivalent to say that $L(theta;x_1 cdots x_n)=1$ in $X_{max}-1lethetale X_{min}$. You need to find the maximum of $L(theta;x_1 cdots x_n)$ as a function of $theta$. But $L(theta;x_1 cdots x_n)$ is constant (in that range) (I recommend you to draw it if you don't quite get it). Hence, $theta_{ML}$ is not well defined, we can pick any value in that range. In particular, $(X_{min}+X_{max})/2$ is as valid as any other.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 24 '18 at 12:38

























            answered Mar 22 '14 at 13:15









            leonbloyleonbloy

            41.5k647108




            41.5k647108












            • $begingroup$
              Thanks, "any value" is really beyond my thought...
              $endgroup$
              – Frazer
              Mar 22 '14 at 15:20










            • $begingroup$
              @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
              $endgroup$
              – Kumara
              Dec 17 '18 at 11:49






            • 1




              $begingroup$
              Undefined. Anyway, that cannot happen under the assumed model
              $endgroup$
              – leonbloy
              Dec 17 '18 at 14:52


















            • $begingroup$
              Thanks, "any value" is really beyond my thought...
              $endgroup$
              – Frazer
              Mar 22 '14 at 15:20










            • $begingroup$
              @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
              $endgroup$
              – Kumara
              Dec 17 '18 at 11:49






            • 1




              $begingroup$
              Undefined. Anyway, that cannot happen under the assumed model
              $endgroup$
              – leonbloy
              Dec 17 '18 at 14:52
















            $begingroup$
            Thanks, "any value" is really beyond my thought...
            $endgroup$
            – Frazer
            Mar 22 '14 at 15:20




            $begingroup$
            Thanks, "any value" is really beyond my thought...
            $endgroup$
            – Frazer
            Mar 22 '14 at 15:20












            $begingroup$
            @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
            $endgroup$
            – Kumara
            Dec 17 '18 at 11:49




            $begingroup$
            @leonbloy: I have a question. What if for $X_1,dots,X_n$ are such that $X_{min}<X_{max}-1$ ? In this case $L(theta) = 0$ for all $theta$. So, what will mle in this case?
            $endgroup$
            – Kumara
            Dec 17 '18 at 11:49




            1




            1




            $begingroup$
            Undefined. Anyway, that cannot happen under the assumed model
            $endgroup$
            – leonbloy
            Dec 17 '18 at 14:52




            $begingroup$
            Undefined. Anyway, that cannot happen under the assumed model
            $endgroup$
            – leonbloy
            Dec 17 '18 at 14:52











            1












            $begingroup$

            Here a two thoughts




            • If any $X_i$ lies outside of $[theta,theta+1]$, then the likelyhood function is zero

            • If all the $X_i$ lie inside $[theta_1,theta_1+1]$ and also inside $[theta_2,theta_2+1]$, then the likelyhoods are the same, because $f(x,theta_1) = f(x,theta_2)$ for those $x$ which lie in the intersection of the two intervals






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Here a two thoughts




              • If any $X_i$ lies outside of $[theta,theta+1]$, then the likelyhood function is zero

              • If all the $X_i$ lie inside $[theta_1,theta_1+1]$ and also inside $[theta_2,theta_2+1]$, then the likelyhoods are the same, because $f(x,theta_1) = f(x,theta_2)$ for those $x$ which lie in the intersection of the two intervals






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Here a two thoughts




                • If any $X_i$ lies outside of $[theta,theta+1]$, then the likelyhood function is zero

                • If all the $X_i$ lie inside $[theta_1,theta_1+1]$ and also inside $[theta_2,theta_2+1]$, then the likelyhoods are the same, because $f(x,theta_1) = f(x,theta_2)$ for those $x$ which lie in the intersection of the two intervals






                share|cite|improve this answer









                $endgroup$



                Here a two thoughts




                • If any $X_i$ lies outside of $[theta,theta+1]$, then the likelyhood function is zero

                • If all the $X_i$ lie inside $[theta_1,theta_1+1]$ and also inside $[theta_2,theta_2+1]$, then the likelyhoods are the same, because $f(x,theta_1) = f(x,theta_2)$ for those $x$ which lie in the intersection of the two intervals







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 22 '14 at 12:57









                fgpfgp

                17.8k22236




                17.8k22236






























                    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%2f722101%2flikelihood-function-for-the-uniform-density-theta-theta1%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

                    Wiesbaden

                    Image
                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                    Read more

                    Marschland

                    Image
                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more