What is the purpose of $frac{1}{sigma sqrt{2 pi}}$ in $frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu...
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What is the purpose of $frac{1}{sigma sqrt{2 pi}}$ in $frac{1}{sigma sqrt{2 pi}}e^{frac{(-(x - mu ))^2}{2sigma ^2}}$?
The 2019 Stack Overflow Developer Survey Results Are InBound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionFind the standard deviation of $ frac{gamma}{sqrt{2pisigma}}expleft(-frac{gamma^2}{sigma}frac{(x-mu)^2}{2}right)$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 $S_t=S_0e^{(r-frac{sigma^2}{2})t+sigma W_t}$Deriving the Covariance of Multivariate GaussianGolden-Ratio Distribution - analogous to Normal distributiondelta method with $ frac{1}{n} sum (X_{i} - bar{X})^{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
$begingroup$
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
edited 17 hours ago
user21820
40.1k544162
asked yesterday
BolboaBolboa
408616
$endgroup$
add a comment |
$begingroup$
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
edited 17 hours ago
user21820
40.1k544162
asked yesterday
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
yesterday
add a comment |
$begingroup$
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
edited 17 hours ago
user21820
40.1k544162
asked yesterday
BolboaBolboa
408616
$endgroup$
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
edited 17 hours ago
user21820
40.1k544162
asked yesterday
BolboaBolboa
408616
edited 17 hours ago
user21820
40.1k544162
asked yesterday
BolboaBolboa
408616
edited 17 hours ago
user21820
40.1k544162
edited 17 hours ago
user21820
40.1k544162
edited 17 hours ago
user21820
40.1k544162
40.1k544162
asked yesterday
BolboaBolboa
408616
asked yesterday
BolboaBolboa
408616
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
$endgroup$
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
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)
answered yesterday
GReyesGReyes
2,43815
$endgroup$
add a comment |
$begingroup$
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ 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.
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
$endgroup$
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
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 $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
$endgroup$
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
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 $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
$endgroup$
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.
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
answered yesterday
CyclotomicFieldCyclotomicField
2,6331316
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
add a comment |
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
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)
answered yesterday
GReyesGReyes
2,43815
$endgroup$
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)
answered yesterday
GReyesGReyes
2,43815
$endgroup$
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)
answered yesterday
GReyesGReyes
2,43815
$endgroup$
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)
answered yesterday
GReyesGReyes
2,43815
answered yesterday
GReyesGReyes
2,43815
answered yesterday
GReyesGReyes
2,43815
answered yesterday
GReyesGReyes
2,43815
2,43815
add a comment |
add a comment |
$begingroup$
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.
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
add a comment |
$begingroup$
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ 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.
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
add a comment |
$begingroup$
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1{sigmasqrt{2pi}}expleft(-frac12left(frac{x-mu}sigmaright)^2right)$$ 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.
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
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.
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 21 hours ago
TheSimpliFireTheSimpliFire
13.2k62464
13.2k62464
add a comment |
add a comment |
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$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
yesterday