What is the purpose of $frac1sigma sqrt2 pi$ in $frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Bound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionFind the standard deviation of $ fracgammasqrt2pisigmaexpleft(-fracgamma^2sigmafrac(x-mu)^22right)$How to compute normal integrals $int_-infty^inftyPhi(x)N(xmidmu,sigma^2),dx$ and $int_-infty^inftyPhi(x)N(xmidmu,sigma^2)x,dx$is this function increasing or decreasing on what intervals?Showing the expected value of $S_t^n$ where $S_t=S_0e^(r-fracsigma^22)t+sigma W_t$Deriving the Covariance of Multivariate GaussianGolden-Ratio Distribution - analogous to Normal distributiondelta method with $ frac1n sum (X_i - barX)^2 $What does determine if a distribution is Gaussian?Calculate $int_-infty^infty x^2e^-x^2 dx$ using Gaussian random variables and the properties of pdfs

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What is the purpose of $frac1sigma sqrt2 pi$ in $frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Bound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionFind the standard deviation of $ fracgammasqrt2pisigmaexpleft(-fracgamma^2sigmafrac(x-mu)^22right)$How to compute normal integrals $int_-infty^inftyPhi(x)N(xmidmu,sigma^2),dx$ and $int_-infty^inftyPhi(x)N(xmidmu,sigma^2)x,dx$is this function increasing or decreasing on what intervals?Showing the expected value of $S_t^n$ where $S_t=S_0e^(r-fracsigma^22)t+sigma W_t$Deriving the Covariance of Multivariate GaussianGolden-Ratio Distribution - analogous to Normal distributiondelta method with $ frac1n sum (X_i - barX)^2 $What does determine if a distribution is Gaussian?Calculate $int_-infty^infty x^2e^-x^2 dx$ using Gaussian random variables and the properties of pdfs










2












$begingroup$


I have been studying the probability density function...



$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$



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



$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$



The way I interpret this is that the area under the gaussian distribution is $sqrt2 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
    2 days ago
















2












$begingroup$


I have been studying the probability density function...



$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$



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



$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$



The way I interpret this is that the area under the gaussian distribution is $sqrt2 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
    2 days ago














2












2








2





$begingroup$


I have been studying the probability density function...



$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$



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



$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$



The way I interpret this is that the area under the gaussian distribution is $sqrt2 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...



$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$



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



$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$



The way I interpret this is that the area under the gaussian distribution is $sqrt2 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 2 days ago









user21820

40.1k544162




40.1k544162










asked 2 days ago









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
    2 days ago













  • 2




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








2




2




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





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











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 $sqrt2pi$ 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
    2 days ago



















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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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. $$f(x)=frac1sqrt2piexpleft(-frac12x^2right).$$ 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 $mathit X=Bbb R$), $$int_mathit Xf(x),dx=1.$$ 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. $$f(xmidalpha,beta)=fracx^alpha-1(1-x)^beta-1text B(alpha,beta)$$ 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 $sqrt2pi$ 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
        2 days ago
















      12












      $begingroup$

      If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ 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
        2 days ago














      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 $sqrt2pi$ 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 $sqrt2pi$ 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 2 days ago









      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
        2 days ago













      • 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
        2 days ago








      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
      2 days 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
      2 days ago












      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 2 days ago









          GReyesGReyes

          2,48315




          2,48315





















              2












              $begingroup$

              As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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. $$f(x)=frac1sqrt2piexpleft(-frac12x^2right).$$ 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 $mathit X=Bbb R$), $$int_mathit Xf(x),dx=1.$$ 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. $$f(xmidalpha,beta)=fracx^alpha-1(1-x)^beta-1text B(alpha,beta)$$ 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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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. $$f(x)=frac1sqrt2piexpleft(-frac12x^2right).$$ 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 $mathit X=Bbb R$), $$int_mathit Xf(x),dx=1.$$ 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. $$f(xmidalpha,beta)=fracx^alpha-1(1-x)^beta-1text B(alpha,beta)$$ 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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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. $$f(x)=frac1sqrt2piexpleft(-frac12x^2right).$$ 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 $mathit X=Bbb R$), $$int_mathit Xf(x),dx=1.$$ 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. $$f(xmidalpha,beta)=fracx^alpha-1(1-x)^beta-1text B(alpha,beta)$$ 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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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. $$f(x)=frac1sqrt2piexpleft(-frac12x^2right).$$ 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 $mathit X=Bbb R$), $$int_mathit Xf(x),dx=1.$$ 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. $$f(xmidalpha,beta)=fracx^alpha-1(1-x)^beta-1text B(alpha,beta)$$ 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 2 days ago









                  TheSimpliFireTheSimpliFire

                  13.2k62464




                  13.2k62464



























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