What is the purpose of the constant in the probability density function The 2019 Stack...
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What is the purpose of the constant in the probability density function
The 2019 Stack Overflow Developer Survey Results Are InConfusion between probability distribution function and probability density functionProbability density function vs. probability mass functionBound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionCan a probability density function be used directly as probability function?Probability density of a function of a random variableis this function increasing or decreasing on what intervals?Homework: questions about probability distribution functions and probability density functionGaussian function constantDeriving the Covariance of Multivariate Gaussianprobability density function of a function of a random variable?
$begingroup$
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 gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
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 gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
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 gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
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 gaussian-integral
probability statistics probability-distributions gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
398516
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
add a comment |
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
1
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
add a comment |
2 Answers
2
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 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(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
1 hour 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 1 hour ago
GReyesGReyes
2,39815
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(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
1 hour 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 $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(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
1 hour 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 $sqrt{2pi}$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
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 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
2,4681314
1
$begingroup$
(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
1 hour ago
add a comment |
1
$begingroup$
(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
1 hour ago
1
1
$begingroup$
(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
1 hour 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
1 hour 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 1 hour ago
GReyesGReyes
2,39815
$endgroup$
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 1 hour ago
GReyesGReyes
2,39815
$endgroup$
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 1 hour ago
GReyesGReyes
2,39815
$endgroup$
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 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
2,39815
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
1 hour ago