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Bayes version for continuous case, what does the integral mean?
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In bayes version for continuous case, what does it mean to integrate with respect to $dtheta$ when $theta$ is a vector not a a scalar value?
$$p(theta|D) = fractheta)p(theta)p(D)$$
Where $D$ is a set of observed data points, and $theta$ is a vector of parameters to be estimated.
$$ p(D) = int p(D|theta)p(theta)dtheta $$
Is this integral a surface integral, component-wise integral or what ?
bayesian integral
asked 2 hours ago
Loai Ghoraba
1062
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up vote
1
down vote
favorite
In bayes version for continuous case, what does it mean to integrate with respect to $dtheta$ when $theta$ is a vector not a a scalar value?
$$p(theta|D) = fractheta)p(theta)p(D)$$
Where $D$ is a set of observed data points, and $theta$ is a vector of parameters to be estimated.
$$ p(D) = int p(D|theta)p(theta)dtheta $$
Is this integral a surface integral, component-wise integral or what ?
bayesian integral
asked 2 hours ago
Loai Ghoraba
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
1
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In bayes version for continuous case, what does it mean to integrate with respect to $dtheta$ when $theta$ is a vector not a a scalar value?
$$p(theta|D) = fractheta)p(theta)p(D)$$
Where $D$ is a set of observed data points, and $theta$ is a vector of parameters to be estimated.
$$ p(D) = int p(D|theta)p(theta)dtheta $$
Is this integral a surface integral, component-wise integral or what ?
bayesian integral
asked 2 hours ago
Loai Ghoraba
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In bayes version for continuous case, what does it mean to integrate with respect to $dtheta$ when $theta$ is a vector not a a scalar value?
$$p(theta|D) = fractheta)p(theta)p(D)$$
Where $D$ is a set of observed data points, and $theta$ is a vector of parameters to be estimated.
$$ p(D) = int p(D|theta)p(theta)dtheta $$
Is this integral a surface integral, component-wise integral or what ?
bayesian integral
bayesian integral
asked 2 hours ago
Loai Ghoraba
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Loai Ghoraba
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Loai Ghoraba
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Loai Ghoraba
1062
asked 2 hours ago
Loai Ghoraba
1062
1062
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Loai Ghoraba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
1
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago
add a comment |Â
You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
1
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago
You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
1
1
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago
add a comment |Â
1 Answer
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That will be the n-dimensional integral, like the 2D case below: $$p(D) = int_Thetaint p(D | theta_1,theta_2)p(theta_1,theta_2)dtheta_1dtheta_2$$
Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.
answered 1 hour ago
gunes
8108
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
That will be the n-dimensional integral, like the 2D case below: $$p(D) = int_Thetaint p(D | theta_1,theta_2)p(theta_1,theta_2)dtheta_1dtheta_2$$
Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.
answered 1 hour ago
gunes
8108
add a comment |Â
up vote
3
down vote
That will be the n-dimensional integral, like the 2D case below: $$p(D) = int_Thetaint p(D | theta_1,theta_2)p(theta_1,theta_2)dtheta_1dtheta_2$$
Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.
answered 1 hour ago
gunes
8108
add a comment |Â
up vote
3
down vote
up vote
3
down vote
That will be the n-dimensional integral, like the 2D case below: $$p(D) = int_Thetaint p(D | theta_1,theta_2)p(theta_1,theta_2)dtheta_1dtheta_2$$
Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.
answered 1 hour ago
gunes
8108
That will be the n-dimensional integral, like the 2D case below: $$p(D) = int_Thetaint p(D | theta_1,theta_2)p(theta_1,theta_2)dtheta_1dtheta_2$$
Because it does not matter if you put them in a vector or not; if you formulate your likelihood and joint PDF correctly. You just have $n$ random variables.
answered 1 hour ago
gunes
8108
answered 1 hour ago
gunes
8108
answered 1 hour ago
gunes
8108
answered 1 hour ago
gunes
8108
8108
add a comment |Â
add a comment |Â
Loai Ghoraba is a new contributor. Be nice, and check out our Code of Conduct.
Loai Ghoraba is a new contributor. Be nice, and check out our Code of Conduct.
Loai Ghoraba is a new contributor. Be nice, and check out our Code of Conduct.
Loai Ghoraba is a new contributor. Be nice, and check out our Code of Conduct.
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You are computing the marginal distribution of $p(D)$ over the range of your parameter space $theta in Theta$. The choice of integral depends on how your parameter space is defined.
– Maxtron
1 hour ago
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
1 hour ago
1
Say your parameter $theta$ is a 2-D vector and bounded, i.e., $theta_1 in [a_1, b_1]$ and $theta_2 in [a_2, b_2]$, then $Theta in [a_1, b_1] times [a_2, b_2] $, which is a surface.
– Maxtron
41 mins ago