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









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            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






























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