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

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

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  so the integral of the probability density function over the entire space is equal to one
  $endgroup$
  – J. W. Tanner
  Apr 10 at 1:53
  
  

  
  
  
  

add a comment | 

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

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  so the integral of the probability density function over the entire space is equal to one
  $endgroup$
  – J. W. Tanner
  Apr 10 at 1:53
  
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

$endgroup$

share|cite|improve this question

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

share|cite|improve this question

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

edited Apr 10 at 12:31

user21820

40.5k544163

edited Apr 10 at 12:31

user21820

40.5k544163

asked Apr 10 at 1:49

BolboaBolboa

408616

asked Apr 10 at 1:49

BolboaBolboa

408616

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

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

$endgroup$

• 
  
  
  

  
  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
  Apr 10 at 2:02
  
  

  
  
  
  

add a comment | 

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

answered Apr 10 at 1:53

GReyesGReyes

2,57315

$endgroup$

add a comment | 

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

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

$endgroup$

add a comment | 

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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

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

$endgroup$

• 
  
  
  

  
  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
  Apr 10 at 2:02
  
  

  
  
  
  

add a comment | 

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

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

$endgroup$

• 
  
  
  

  
  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
  Apr 10 at 2:02
  
  

  
  
  
  

add a comment | 

$begingroup$

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

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

$endgroup$

share|cite|improve this answer

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

answered Apr 10 at 1:58

CyclotomicFieldCyclotomicField

2,6531316

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

answered Apr 10 at 1:53

GReyesGReyes

2,57315

$endgroup$

add a comment | 

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

answered Apr 10 at 1:53

GReyesGReyes

2,57315

$endgroup$

add a comment | 

$begingroup$

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

answered Apr 10 at 1:53

GReyesGReyes

2,57315

$endgroup$

share|cite|improve this answer

answered Apr 10 at 1:53

GReyesGReyes

2,57315

answered Apr 10 at 1:53

GReyesGReyes

2,57315

answered Apr 10 at 1:53

GReyesGReyes

2,57315

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

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

$endgroup$

add a comment | 

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

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

$endgroup$

add a comment | 

$begingroup$

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

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

$endgroup$

share|cite|improve this answer

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765

answered Apr 10 at 7:55

TheSimpliFireTheSimpliFire

13.8k62765