What is the purpose of the constant in the probability density function












1












$begingroup$


I have been studying the probability density function...



$$frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu ))^2}{2sigma ^2}}$$



For now I remove the constant, and using the following proof, I prove that...



$$int_{-infty}^{infty}e^{frac{-x^2}{2}} = sqrt{2 pi }$$



The way I interpret this is that the area under the gaussian distribution is $sqrt{2 pi }$. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?






probability statistics probability-distributions gaussian-integral







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    so the integral of the probability density function over the entire space is equal to one
    $endgroup$
    – J. W. Tanner
    1 hour ago


















1












$begingroup$


I have been studying the probability density function...



$$frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu ))^2}{2sigma ^2}}$$



For now I remove the constant, and using the following proof, I prove that...



$$int_{-infty}^{infty}e^{frac{-x^2}{2}} = sqrt{2 pi }$$



The way I interpret this is that the area under the gaussian distribution is $sqrt{2 pi }$. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?






probability statistics probability-distributions gaussian-integral







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    so the integral of the probability density function over the entire space is equal to one
    $endgroup$
    – J. W. Tanner
    1 hour ago
















1












1








1





$begingroup$


I have been studying the probability density function...



$$frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu ))^2}{2sigma ^2}}$$



For now I remove the constant, and using the following proof, I prove that...



$$int_{-infty}^{infty}e^{frac{-x^2}{2}} = sqrt{2 pi }$$



The way I interpret this is that the area under the gaussian distribution is $sqrt{2 pi }$. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?






probability statistics probability-distributions gaussian-integral







share|cite|improve this question









$endgroup$




I have been studying the probability density function...



$$frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu ))^2}{2sigma ^2}}$$



For now I remove the constant, and using the following proof, I prove that...



$$int_{-infty}^{infty}e^{frac{-x^2}{2}} = sqrt{2 pi }$$



The way I interpret this is that the area under the gaussian distribution is $sqrt{2 pi }$. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?






probability statistics probability-distributions gaussian-integral




probability statistics probability-distributions gaussian-integral






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









BolboaBolboa

398516




398516








  • 1




    $begingroup$
    so the integral of the probability density function over the entire space is equal to one
    $endgroup$
    – J. W. Tanner
    1 hour ago
















  • 1




    $begingroup$
    so the integral of the probability density function over the entire space is equal to one
    $endgroup$
    – J. W. Tanner
    1 hour ago










1




1




$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago






$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago












2 Answers
2






active

oldest

votes


















2












$begingroup$

If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
    $endgroup$
    – John Doe
    1 hour ago





















2












$begingroup$

It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
      $endgroup$
      – John Doe
      1 hour ago


















    2












    $begingroup$

    If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
      $endgroup$
      – John Doe
      1 hour ago
















    2












    2








    2





    $begingroup$

    If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.






    share|cite|improve this answer









    $endgroup$



    If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    CyclotomicFieldCyclotomicField

    2,4681314




    2,4681314








    • 1




      $begingroup$
      (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
      $endgroup$
      – John Doe
      1 hour ago
















    • 1




      $begingroup$
      (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
      $endgroup$
      – John Doe
      1 hour ago










    1




    1




    $begingroup$
    (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
    $endgroup$
    – John Doe
    1 hour ago






    $begingroup$
    (In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
    $endgroup$
    – John Doe
    1 hour ago













    2












    $begingroup$

    It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)






        share|cite|improve this answer









        $endgroup$



        It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        GReyesGReyes

        2,39815




        2,39815






























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