Analytically computable functions with behavior similar to the skew normal pdf

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I'm looking for a (computationally cheap) way to compute a function that behaves similar to the skew normal distribution, i.e. it has a shape like the normal distribution when some parameter a = c, while allowing to set some skew by varying a.



The background is that I'm using a Gaussian model for a shape manipulation tool but there are some edge cases that would be better served by an asymmetric function. Ideally, the function should be able to fit a positively shifted Gaussian G(x,mu,sigma), mu > 0 for the range of positive x.




normal-distribution




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    up vote
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    down vote

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    I'm looking for a (computationally cheap) way to compute a function that behaves similar to the skew normal distribution, i.e. it has a shape like the normal distribution when some parameter a = c, while allowing to set some skew by varying a.



    The background is that I'm using a Gaussian model for a shape manipulation tool but there are some edge cases that would be better served by an asymmetric function. Ideally, the function should be able to fit a positively shifted Gaussian G(x,mu,sigma), mu > 0 for the range of positive x.




    normal-distribution




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm looking for a (computationally cheap) way to compute a function that behaves similar to the skew normal distribution, i.e. it has a shape like the normal distribution when some parameter a = c, while allowing to set some skew by varying a.



      The background is that I'm using a Gaussian model for a shape manipulation tool but there are some edge cases that would be better served by an asymmetric function. Ideally, the function should be able to fit a positively shifted Gaussian G(x,mu,sigma), mu > 0 for the range of positive x.




      normal-distribution




      share|cite|improve this question











      I'm looking for a (computationally cheap) way to compute a function that behaves similar to the skew normal distribution, i.e. it has a shape like the normal distribution when some parameter a = c, while allowing to set some skew by varying a.



      The background is that I'm using a Gaussian model for a shape manipulation tool but there are some edge cases that would be better served by an asymmetric function. Ideally, the function should be able to fit a positively shifted Gaussian G(x,mu,sigma), mu > 0 for the range of positive x.





      normal-distribution





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      share|cite|improve this question




      share|cite|improve this question









      asked Aug 2 at 15:43









      John Smith

      113




      113




















          1 Answer
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          The logistic curve
          $dfrac11+e^-ax
          $
          very closely matches
          the cumulative normal distribution
          especially if
          $a=dfrac4sqrt2pi$,
          which matches the slope
          at the origin.
          The maximum difference is
          at most $0.017$.



          Perhaps you can
          modify this
          to do what you want.






          share|cite|improve this answer





















          • Thanks, Ill definitely try that.
            – John Smith
            Aug 3 at 16:57










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          1 Answer
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          active

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          1 Answer
          1






          active

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          active

          oldest

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          up vote
          0
          down vote













          The logistic curve
          $dfrac11+e^-ax
          $
          very closely matches
          the cumulative normal distribution
          especially if
          $a=dfrac4sqrt2pi$,
          which matches the slope
          at the origin.
          The maximum difference is
          at most $0.017$.



          Perhaps you can
          modify this
          to do what you want.






          share|cite|improve this answer





















          • Thanks, Ill definitely try that.
            – John Smith
            Aug 3 at 16:57














          up vote
          0
          down vote













          The logistic curve
          $dfrac11+e^-ax
          $
          very closely matches
          the cumulative normal distribution
          especially if
          $a=dfrac4sqrt2pi$,
          which matches the slope
          at the origin.
          The maximum difference is
          at most $0.017$.



          Perhaps you can
          modify this
          to do what you want.






          share|cite|improve this answer





















          • Thanks, Ill definitely try that.
            – John Smith
            Aug 3 at 16:57












          up vote
          0
          down vote










          up vote
          0
          down vote









          The logistic curve
          $dfrac11+e^-ax
          $
          very closely matches
          the cumulative normal distribution
          especially if
          $a=dfrac4sqrt2pi$,
          which matches the slope
          at the origin.
          The maximum difference is
          at most $0.017$.



          Perhaps you can
          modify this
          to do what you want.






          share|cite|improve this answer













          The logistic curve
          $dfrac11+e^-ax
          $
          very closely matches
          the cumulative normal distribution
          especially if
          $a=dfrac4sqrt2pi$,
          which matches the slope
          at the origin.
          The maximum difference is
          at most $0.017$.



          Perhaps you can
          modify this
          to do what you want.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 2 at 17:38









          marty cohen

          69k446122




          69k446122











          • Thanks, Ill definitely try that.
            – John Smith
            Aug 3 at 16:57
















          • Thanks, Ill definitely try that.
            – John Smith
            Aug 3 at 16:57















          Thanks, Ill definitely try that.
          – John Smith
          Aug 3 at 16:57




          Thanks, Ill definitely try that.
          – John Smith
          Aug 3 at 16:57












           

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