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?
$$p(theta|D) = fracp(Dp(D)$$
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 Sep 22 at 16:42
Loai Ghoraba
1063
add a comment |Â
up vote
0
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?
$$p(theta|D) = fracp(Dp(D)$$
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 Sep 22 at 16:42
Loai Ghoraba
1063
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
Sep 22 at 16:48
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
2
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
Sep 22 at 18:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
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?
$$p(theta|D) = fracp(Dp(D)$$
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 Sep 22 at 16:42
Loai Ghoraba
1063
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?
$$p(theta|D) = fracp(Dp(D)$$
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 Sep 22 at 16:42
Loai Ghoraba
1063
asked Sep 22 at 16:42
Loai Ghoraba
1063
asked Sep 22 at 16:42
Loai Ghoraba
1063
asked Sep 22 at 16:42
Loai Ghoraba
1063
asked Sep 22 at 16:42
Loai Ghoraba
1063
1063
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
Sep 22 at 16:48
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
2
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
Sep 22 at 18:01
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
Sep 22 at 16:48
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
2
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
Sep 22 at 18:01
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
Sep 22 at 16:48
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
Sep 22 at 16:48
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
2
2
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
Sep 22 at 18:01
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
Sep 22 at 18:01
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
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 Sep 22 at 16:51
gunes
1,1308
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
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
accepted
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 Sep 22 at 16:51
gunes
1,1308
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
add a comment |Â
up vote
3
down vote
accepted
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 Sep 22 at 16:51
gunes
1,1308
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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 Sep 22 at 16:51
gunes
1,1308
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 Sep 22 at 16:51
gunes
1,1308
answered Sep 22 at 16:51
gunes
1,1308
answered Sep 22 at 16:51
gunes
1,1308
answered Sep 22 at 16:51
gunes
1,1308
1,1308
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
add a comment |Â
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Just to make sure, so it is not a surface integral, right? That is , $dtheta$ is just a compact representation of multiple deltas multiplied together ?
– Loai Ghoraba
Sep 22 at 21:03
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
Well, what would you call integrals with $dxdy$?
– gunes
Sep 22 at 21:12
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
You are right, my bad :)
– Loai Ghoraba
Sep 23 at 17:45
add a comment |Â
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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
Sep 22 at 16:48
Can you give an example with a specific parameter space? I'm not sure I fully understand your answer
– Loai Ghoraba
Sep 22 at 17:27
2
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
Sep 22 at 18:01