couples complexity theorem states that omega is complex. How? [closed]

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couples complexity theorem states that omega is complex. How?




probability-theory




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closed as off-topic by John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy Aug 25 at 11:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
















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    couples complexity theorem states that omega is complex. How?




    probability-theory




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    closed as off-topic by John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy Aug 25 at 11:32


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy
    If this question can be reworded to fit the rules in the help center, please edit the question.














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      couples complexity theorem states that omega is complex. How?




      probability-theory




      share|cite|improve this question














      couples complexity theorem states that omega is complex. How?





      probability-theory





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago

























      asked Aug 25 at 2:46









      mathnewbie

      274




      274




      closed as off-topic by John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy Aug 25 at 11:32


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy Aug 25 at 11:32


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Jendrik Stelzner, Brahadeesh, Pierre-Guy Plamondon, amWhy
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
          1






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          up vote
          0
          down vote













          There is no such probability measure.



          More generally, let $S$ be an uncountable sample space, and suppose $P$ is a probability measure such that all singleton subsets of $S$ have positive probability.



          For each integer $k > 1$, let
          $S_k=sin Smid P(s) > largefrac1k$.



          Then $S=displaystylebigcup_k=2^infty S_k$, hence $S_j$ is uncountable, for some $j$.



          In particular, $S_j$ has at least $j$ elements.



          But then, letting $A$ be a subset of $S_j$ with $j$ elements, we get
          $P(A) > (j)bigl(largefrac1jbigr) = 1$, contradiction.



          Thus, if $S$ is an uncountable sample space, there is no probability measure $P$ such that all singleton subsets of $S$ have positive probability.






          share|cite|improve this answer




















          • I have edited the question.
            – mathnewbie
            Aug 25 at 9:56










          • You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
            – quasi
            Aug 25 at 12:44











          • I apologise for the change.
            – mathnewbie
            Aug 25 at 12:46


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          There is no such probability measure.



          More generally, let $S$ be an uncountable sample space, and suppose $P$ is a probability measure such that all singleton subsets of $S$ have positive probability.



          For each integer $k > 1$, let
          $S_k=sin Smid P(s) > largefrac1k$.



          Then $S=displaystylebigcup_k=2^infty S_k$, hence $S_j$ is uncountable, for some $j$.



          In particular, $S_j$ has at least $j$ elements.



          But then, letting $A$ be a subset of $S_j$ with $j$ elements, we get
          $P(A) > (j)bigl(largefrac1jbigr) = 1$, contradiction.



          Thus, if $S$ is an uncountable sample space, there is no probability measure $P$ such that all singleton subsets of $S$ have positive probability.






          share|cite|improve this answer




















          • I have edited the question.
            – mathnewbie
            Aug 25 at 9:56










          • You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
            – quasi
            Aug 25 at 12:44











          • I apologise for the change.
            – mathnewbie
            Aug 25 at 12:46















          up vote
          0
          down vote













          There is no such probability measure.



          More generally, let $S$ be an uncountable sample space, and suppose $P$ is a probability measure such that all singleton subsets of $S$ have positive probability.



          For each integer $k > 1$, let
          $S_k=sin Smid P(s) > largefrac1k$.



          Then $S=displaystylebigcup_k=2^infty S_k$, hence $S_j$ is uncountable, for some $j$.



          In particular, $S_j$ has at least $j$ elements.



          But then, letting $A$ be a subset of $S_j$ with $j$ elements, we get
          $P(A) > (j)bigl(largefrac1jbigr) = 1$, contradiction.



          Thus, if $S$ is an uncountable sample space, there is no probability measure $P$ such that all singleton subsets of $S$ have positive probability.






          share|cite|improve this answer




















          • I have edited the question.
            – mathnewbie
            Aug 25 at 9:56










          • You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
            – quasi
            Aug 25 at 12:44











          • I apologise for the change.
            – mathnewbie
            Aug 25 at 12:46













          up vote
          0
          down vote










          up vote
          0
          down vote









          There is no such probability measure.



          More generally, let $S$ be an uncountable sample space, and suppose $P$ is a probability measure such that all singleton subsets of $S$ have positive probability.



          For each integer $k > 1$, let
          $S_k=sin Smid P(s) > largefrac1k$.



          Then $S=displaystylebigcup_k=2^infty S_k$, hence $S_j$ is uncountable, for some $j$.



          In particular, $S_j$ has at least $j$ elements.



          But then, letting $A$ be a subset of $S_j$ with $j$ elements, we get
          $P(A) > (j)bigl(largefrac1jbigr) = 1$, contradiction.



          Thus, if $S$ is an uncountable sample space, there is no probability measure $P$ such that all singleton subsets of $S$ have positive probability.






          share|cite|improve this answer












          There is no such probability measure.



          More generally, let $S$ be an uncountable sample space, and suppose $P$ is a probability measure such that all singleton subsets of $S$ have positive probability.



          For each integer $k > 1$, let
          $S_k=sin Smid P(s) > largefrac1k$.



          Then $S=displaystylebigcup_k=2^infty S_k$, hence $S_j$ is uncountable, for some $j$.



          In particular, $S_j$ has at least $j$ elements.



          But then, letting $A$ be a subset of $S_j$ with $j$ elements, we get
          $P(A) > (j)bigl(largefrac1jbigr) = 1$, contradiction.



          Thus, if $S$ is an uncountable sample space, there is no probability measure $P$ such that all singleton subsets of $S$ have positive probability.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 25 at 4:49









          quasi

          33.9k22461




          33.9k22461











          • I have edited the question.
            – mathnewbie
            Aug 25 at 9:56










          • You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
            – quasi
            Aug 25 at 12:44











          • I apologise for the change.
            – mathnewbie
            Aug 25 at 12:46

















          • I have edited the question.
            – mathnewbie
            Aug 25 at 9:56










          • You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
            – quasi
            Aug 25 at 12:44











          • I apologise for the change.
            – mathnewbie
            Aug 25 at 12:46
















          I have edited the question.
          – mathnewbie
          Aug 25 at 9:56




          I have edited the question.
          – mathnewbie
          Aug 25 at 9:56












          You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
          – quasi
          Aug 25 at 12:44





          You totally changed the question. Your previous title and question specified that the sample space was uncountable. My answer relates to that version.
          – quasi
          Aug 25 at 12:44













          I apologise for the change.
          – mathnewbie
          Aug 25 at 12:46





          I apologise for the change.
          – mathnewbie
          Aug 25 at 12:46



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