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
$begingroup$
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
edited 2 days ago
user21820
40.1k544162
asked 2 days ago
BolboaBolboa
408616
$endgroup$
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
edited 2 days ago
user21820
40.1k544162
asked 2 days ago
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
2 days ago
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
edited 2 days ago
user21820
40.1k544162
asked 2 days ago
BolboaBolboa
408616
$endgroup$
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
edited 2 days ago
user21820
40.1k544162
asked 2 days ago
BolboaBolboa
408616
edited 2 days ago
user21820
40.1k544162
asked 2 days ago
BolboaBolboa
408616
edited 2 days ago
user21820
40.1k544162
edited 2 days ago
user21820
40.1k544162
edited 2 days ago
user21820
40.1k544162
40.1k544162
asked 2 days ago
BolboaBolboa
408616
asked 2 days ago
BolboaBolboa
408616
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
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
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
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 $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 2 days ago
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
2 days ago
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 2 days ago
GReyesGReyes
2,48315
$endgroup$
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.
answered 2 days 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 $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 2 days ago
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
2 days ago
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.
answered 2 days ago
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
2 days ago
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.
answered 2 days ago
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 $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 2 days ago
CyclotomicFieldCyclotomicField
2,6331316
answered 2 days ago
CyclotomicFieldCyclotomicField
2,6331316
answered 2 days ago
CyclotomicFieldCyclotomicField
2,6331316
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
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
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
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 2 days ago
GReyesGReyes
2,48315
$endgroup$
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 2 days ago
GReyesGReyes
2,48315
$endgroup$
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 2 days ago
GReyesGReyes
2,48315
$endgroup$
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 2 days ago
GReyesGReyes
2,48315
answered 2 days ago
GReyesGReyes
2,48315
answered 2 days ago
GReyesGReyes
2,48315
answered 2 days ago
GReyesGReyes
2,48315
2,48315
add a comment |
add a comment |
$begingroup$
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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.
answered 2 days 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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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.
answered 2 days 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 $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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.
answered 2 days ago
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
As you have correctly stated, the p.d.f. of the normal distribution is given by $$f(xmidmu,sigma^2)=frac1sigmasqrt2piexpleft(-frac12left(fracx-musigmaright)^2right)$$ 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.
answered 2 days ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 2 days ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 2 days ago
TheSimpliFireTheSimpliFire
13.2k62464
answered 2 days 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
2 days ago