What is the purpose of $frac{1}{sigma sqrt{2 pi}}$ in $frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu...












2












$begingroup$


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 normal-distribution gaussian-integral







share|cite|improve this question











$endgroup$








  • 2




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


















2












$begingroup$


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 normal-distribution gaussian-integral







share|cite|improve this question











$endgroup$








  • 2




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
















2












2








2





$begingroup$


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 normal-distribution 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 normal-distribution gaussian-integral




probability statistics probability-distributions normal-distribution gaussian-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









user21820

40.1k544162




40.1k544162










asked yesterday









BolboaBolboa

408616




408616








  • 2




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
















  • 2




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










2




2




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






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












3 Answers
3






active

oldest

votes


















12












$begingroup$

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









$endgroup$









  • 2




    $begingroup$
    (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
    yesterday





















6












$begingroup$

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












    $begingroup$

    As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ where the parameter space is $mathitTheta={(mu,sigma^2)inBbb R^2:sigma^2>0}$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.



    Now consider the simple case where $mu=0$ and $sigma^2=1$. Then the standard normal distribution has p.d.f. $$f(x)=frac1{sqrt{2pi}}expleft(-frac12x^2right).$$ If we integrate this in the interval $(-infty,infty)$, we will get $1$. This is by definition always the case as for all $xinmathit X$ (in this instance $mathit X=Bbb R$), $$int_{mathit X}f(x),dx=1.$$ That is, the sum of all the probabilities of $x$ being in each region in $mathit X$ is $1$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.



    A further example is the Beta distribution, with p.d.f. $$f(xmidalpha,beta)=frac{x^{alpha-1}(1-x)^{beta-1}}{text B(alpha,beta)}$$ where $1/text B(alpha,beta)$ is the normalising constant.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      12












      $begingroup$

      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









      $endgroup$









      • 2




        $begingroup$
        (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
        yesterday


















      12












      $begingroup$

      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









      $endgroup$









      • 2




        $begingroup$
        (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
        yesterday
















      12












      12








      12





      $begingroup$

      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 yesterday









      CyclotomicFieldCyclotomicField

      2,6331316




      2,6331316








      • 2




        $begingroup$
        (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
        yesterday
















      • 2




        $begingroup$
        (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
        yesterday










      2




      2




      $begingroup$
      (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
      yesterday






      $begingroup$
      (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
      yesterday













      6












      $begingroup$

      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









      $endgroup$


















        6












        $begingroup$

        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









        $endgroup$
















          6












          6








          6





          $begingroup$

          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 yesterday









          GReyesGReyes

          2,43815




          2,43815























              2












              $begingroup$

              As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ where the parameter space is $mathitTheta={(mu,sigma^2)inBbb R^2:sigma^2>0}$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.



              Now consider the simple case where $mu=0$ and $sigma^2=1$. Then the standard normal distribution has p.d.f. $$f(x)=frac1{sqrt{2pi}}expleft(-frac12x^2right).$$ If we integrate this in the interval $(-infty,infty)$, we will get $1$. This is by definition always the case as for all $xinmathit X$ (in this instance $mathit X=Bbb R$), $$int_{mathit X}f(x),dx=1.$$ That is, the sum of all the probabilities of $x$ being in each region in $mathit X$ is $1$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.



              A further example is the Beta distribution, with p.d.f. $$f(xmidalpha,beta)=frac{x^{alpha-1}(1-x)^{beta-1}}{text B(alpha,beta)}$$ where $1/text B(alpha,beta)$ is the normalising constant.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ where the parameter space is $mathitTheta={(mu,sigma^2)inBbb R^2:sigma^2>0}$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.



                Now consider the simple case where $mu=0$ and $sigma^2=1$. Then the standard normal distribution has p.d.f. $$f(x)=frac1{sqrt{2pi}}expleft(-frac12x^2right).$$ If we integrate this in the interval $(-infty,infty)$, we will get $1$. This is by definition always the case as for all $xinmathit X$ (in this instance $mathit X=Bbb R$), $$int_{mathit X}f(x),dx=1.$$ That is, the sum of all the probabilities of $x$ being in each region in $mathit X$ is $1$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.



                A further example is the Beta distribution, with p.d.f. $$f(xmidalpha,beta)=frac{x^{alpha-1}(1-x)^{beta-1}}{text B(alpha,beta)}$$ where $1/text B(alpha,beta)$ is the normalising constant.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ where the parameter space is $mathitTheta={(mu,sigma^2)inBbb R^2:sigma^2>0}$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.



                  Now consider the simple case where $mu=0$ and $sigma^2=1$. Then the standard normal distribution has p.d.f. $$f(x)=frac1{sqrt{2pi}}expleft(-frac12x^2right).$$ If we integrate this in the interval $(-infty,infty)$, we will get $1$. This is by definition always the case as for all $xinmathit X$ (in this instance $mathit X=Bbb R$), $$int_{mathit X}f(x),dx=1.$$ That is, the sum of all the probabilities of $x$ being in each region in $mathit X$ is $1$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.



                  A further example is the Beta distribution, with p.d.f. $$f(xmidalpha,beta)=frac{x^{alpha-1}(1-x)^{beta-1}}{text B(alpha,beta)}$$ where $1/text B(alpha,beta)$ is the normalising constant.






                  share|cite|improve this answer









                  $endgroup$



                  As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ where the parameter space is $mathitTheta={(mu,sigma^2)inBbb R^2:sigma^2>0}$. This is essentially saying that the mean is a value on the real line, and the variance is one on the positive real line.



                  Now consider the simple case where $mu=0$ and $sigma^2=1$. Then the standard normal distribution has p.d.f. $$f(x)=frac1{sqrt{2pi}}expleft(-frac12x^2right).$$ If we integrate this in the interval $(-infty,infty)$, we will get $1$. This is by definition always the case as for all $xinmathit X$ (in this instance $mathit X=Bbb R$), $$int_{mathit X}f(x),dx=1.$$ That is, the sum of all the probabilities of $x$ being in each region in $mathit X$ is $1$. In fact, the constant that makes this happen is so important in statistics (especially Bayesian statistics) that it is given a name: the normalising constant.



                  A further example is the Beta distribution, with p.d.f. $$f(xmidalpha,beta)=frac{x^{alpha-1}(1-x)^{beta-1}}{text B(alpha,beta)}$$ where $1/text B(alpha,beta)$ is the normalising constant.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 19 hours ago









                  TheSimpliFireTheSimpliFire

                  13.2k62464




                  13.2k62464






























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