What is the purpose of the constant in the probability density function The 2019 Stack...

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



The 2019 Stack Overflow Developer Survey Results Are InConfusion between probability distribution function and probability density functionProbability density function vs. probability mass functionBound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionCan a probability density function be used directly as probability function?Probability density of a function of a random variableis this function increasing or decreasing on what intervals?Homework: questions about probability distribution functions and probability density functionGaussian function constantDeriving the Covariance of Multivariate Gaussianprobability density function of a function of a random variable?












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









$endgroup$














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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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