Discretization of normal distribution over a finite range











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If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?










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  • Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
    – Tunococ
    Aug 19 '14 at 8:16

















up vote
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down vote

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If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?










share|cite|improve this question
























  • Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
    – Tunococ
    Aug 19 '14 at 8:16















up vote
0
down vote

favorite









up vote
0
down vote

favorite











If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?










share|cite|improve this question















If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?







normal-distribution






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edited May 9 at 21:39









T J. Kim

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314217










asked Aug 19 '14 at 8:10









kutschkem

291310




291310












  • Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
    – Tunococ
    Aug 19 '14 at 8:16




















  • Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
    – Tunococ
    Aug 19 '14 at 8:16


















Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16






Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16












2 Answers
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If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.






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    If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).



    I know the thread is very old; I just couldn't resist.






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      2 Answers
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      2 Answers
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      If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
      $$ begin{align}sum p_i&=1\
      sum ip_i &= mu\
      sum i^2p_i &= sigma^2+mu^2end{align}$$
      These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
      $$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
      sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
      So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.






      share|cite|improve this answer

























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        If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
        $$ begin{align}sum p_i&=1\
        sum ip_i &= mu\
        sum i^2p_i &= sigma^2+mu^2end{align}$$
        These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
        $$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
        sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
        So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
          $$ begin{align}sum p_i&=1\
          sum ip_i &= mu\
          sum i^2p_i &= sigma^2+mu^2end{align}$$
          These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
          $$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
          sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
          So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.






          share|cite|improve this answer












          If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
          $$ begin{align}sum p_i&=1\
          sum ip_i &= mu\
          sum i^2p_i &= sigma^2+mu^2end{align}$$
          These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
          $$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
          sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
          So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 19 '14 at 8:29









          Hagen von Eitzen

          275k21267494




          275k21267494






















              up vote
              0
              down vote













              If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).



              I know the thread is very old; I just couldn't resist.






              share|cite|improve this answer

























                up vote
                0
                down vote













                If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).



                I know the thread is very old; I just couldn't resist.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).



                  I know the thread is very old; I just couldn't resist.






                  share|cite|improve this answer












                  If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).



                  I know the thread is very old; I just couldn't resist.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 '17 at 9:17









                  Vladimír Čunát

                  1011




                  1011






























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