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 ?










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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

















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 ?










share|cite|improve this question





















  • 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













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 ?










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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

















  • 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











1 Answer
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up vote
3
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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.






share|cite|improve this answer




















  • 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










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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.






share|cite|improve this answer




















  • 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














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.






share|cite|improve this answer




















  • 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












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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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

















 

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