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Normal distribution in an interval [closed]
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I am interested in the cumulative function of the normal distribution, because I want to get the distribution which has the support $[0,2pi]$, and it looks like a normal distribution. Therefore, if I get the closed form of $F(2pi)$ and $F(0)$, I can get the new distribution which has the pdf:
$$frac1F(2pi) -F(0)frac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2),~xin[0,2pi]$$
For example, can we calculate the following integral:
$$int_0^2pifrac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2)$$
Is there exists any method which can be used to get the closed form of it. Or can we get any similar distribution?
integration statistics definite-integrals normal-distribution
asked Jul 17 at 19:54
coolcat
1018
closed as off-topic by amWhy, callculus, Xander Henderson, max_zorn, Gibbs Jul 18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, callculus, Xander Henderson, max_zorn, Gibbs
add a comment |Â
up vote
-1
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I am interested in the cumulative function of the normal distribution, because I want to get the distribution which has the support $[0,2pi]$, and it looks like a normal distribution. Therefore, if I get the closed form of $F(2pi)$ and $F(0)$, I can get the new distribution which has the pdf:
$$frac1F(2pi) -F(0)frac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2),~xin[0,2pi]$$
For example, can we calculate the following integral:
$$int_0^2pifrac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2)$$
Is there exists any method which can be used to get the closed form of it. Or can we get any similar distribution?
integration statistics definite-integrals normal-distribution
asked Jul 17 at 19:54
coolcat
1018
closed as off-topic by amWhy, callculus, Xander Henderson, max_zorn, Gibbs Jul 18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, callculus, Xander Henderson, max_zorn, Gibbs
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
2
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
1
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am interested in the cumulative function of the normal distribution, because I want to get the distribution which has the support $[0,2pi]$, and it looks like a normal distribution. Therefore, if I get the closed form of $F(2pi)$ and $F(0)$, I can get the new distribution which has the pdf:
$$frac1F(2pi) -F(0)frac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2),~xin[0,2pi]$$
For example, can we calculate the following integral:
$$int_0^2pifrac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2)$$
Is there exists any method which can be used to get the closed form of it. Or can we get any similar distribution?
integration statistics definite-integrals normal-distribution
asked Jul 17 at 19:54
coolcat
1018
I am interested in the cumulative function of the normal distribution, because I want to get the distribution which has the support $[0,2pi]$, and it looks like a normal distribution. Therefore, if I get the closed form of $F(2pi)$ and $F(0)$, I can get the new distribution which has the pdf:
$$frac1F(2pi) -F(0)frac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2),~xin[0,2pi]$$
For example, can we calculate the following integral:
$$int_0^2pifrac1sqrt2pisigma^2exp(-frac(x-mu)^22sigma^2)$$
Is there exists any method which can be used to get the closed form of it. Or can we get any similar distribution?
integration statistics definite-integrals normal-distribution
asked Jul 17 at 19:54
coolcat
1018
edited Jul 17 at 20:21
asked Jul 17 at 19:54
coolcat
1018
asked Jul 17 at 19:54
coolcat
1018
asked Jul 17 at 19:54
coolcat
1018
1018
closed as off-topic by amWhy, callculus, Xander Henderson, max_zorn, Gibbs Jul 18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, callculus, Xander Henderson, max_zorn, Gibbs
closed as off-topic by amWhy, callculus, Xander Henderson, max_zorn, Gibbs Jul 18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, callculus, Xander Henderson, max_zorn, Gibbs
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
2
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
1
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02
add a comment |Â
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
2
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
1
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
2
2
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
1
1
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.
answered Jul 17 at 19:58
Adrian Keister
3,61721533
add a comment |Â
up vote
1
down vote
In fact the mostly used CDF in this case is $1-Q(dfracx-musigma)$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.
answered Jul 17 at 19:58
Adrian Keister
3,61721533
add a comment |Â
up vote
2
down vote
accepted
There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.
answered Jul 17 at 19:58
Adrian Keister
3,61721533
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.
answered Jul 17 at 19:58
Adrian Keister
3,61721533
There is no closed-form version of it, because the Gaussian does not have an elementary antiderivative. You use tables, or you compute it numerically.
answered Jul 17 at 19:58
Adrian Keister
3,61721533
answered Jul 17 at 19:58
Adrian Keister
3,61721533
answered Jul 17 at 19:58
Adrian Keister
3,61721533
answered Jul 17 at 19:58
Adrian Keister
3,61721533
3,61721533
add a comment |Â
add a comment |Â
up vote
1
down vote
In fact the mostly used CDF in this case is $1-Q(dfracx-musigma)$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
In fact the mostly used CDF in this case is $1-Q(dfracx-musigma)$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In fact the mostly used CDF in this case is $1-Q(dfracx-musigma)$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
In fact the mostly used CDF in this case is $1-Q(dfracx-musigma)$ where $Q(.)$ denotes the $Q$ function and has no close form. You can also refer to https://en.wikipedia.org/wiki/Q-function
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
answered Jul 17 at 21:10
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
Where did you get stuck? Please show us what work you've done. Or provide us with information about the class or course or text you are studying for/from, and some some definitions you've been taught, like "normal distribution", and then "cumulative normal distribution." If you don't know those definitions and so can't address those definitions in your post, search for them in your notes and/or in your text, and/or on-line. Please improve your post (edit it) by including your efforts and/or the definitions you're working with.
– amWhy
Jul 17 at 20:02
2
There exists no closed form. But you can calculate it numerically in a few step since $2pi$ is not far away from $0$
– callculus
Jul 17 at 20:10
1
It doesn´t help to edit you question several times without new information. You have to give a reply to the ansawer or the comments. Are you interested in a numerical method or not?
– callculus
Jul 17 at 20:24
Yes. I think the only way to get the answer is by some numerical methods. My goal is to find a distribution which looks like the normal distribution but has the support on a interval.
– coolcat
Jul 17 at 20:49
@coolcat You can look here for a nice and easy approximation (similar distribution). I think this is what you are looking for.
– callculus
Jul 17 at 21:02