Can anyone help explain this basic example of posterior
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I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"
I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.
It is,
Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$
Then (From the book, page 24).
bayesian normal-distribution independence posterior identifiability
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up vote
3
down vote
favorite
I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"
I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.
It is,
Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$
Then (From the book, page 24).
bayesian normal-distribution independence posterior identifiability
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"
I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.
It is,
Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$
Then (From the book, page 24).
bayesian normal-distribution independence posterior identifiability
I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"
I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.
It is,
Consider one observation $x$, from a normal $$N(fractheta_1+theta_22,1)$$
Then (From the book, page 24).
bayesian normal-distribution independence posterior identifiability
bayesian normal-distribution independence posterior identifiability
edited 8 mins ago
Xi'an
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asked 18 mins ago
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Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
$$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
$$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
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
Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
$$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
$$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
add a comment |Â
up vote
3
down vote
accepted
Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
$$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
$$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
$$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
$$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.
Sorry for being confusing! The joint posterior distribution on $(xi_1,xi_2)$ is
$$pi(xi_1,xi_2|x)propto exp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2)$$
Therefore the marginal posterior on $xi_2$ is given by the marginal of the above, up to a constant, that is,
$$pi(xi_2|x)propto intexp-(x-xi_1)^2/2pi_1(2xi_1)pi_2(2xi_2),textdxi_1$$
which does not depend on $x$. This is a case, albeit an artificial case, when the posterior and the prior are equal.
answered 9 mins ago
Xi'an
51.4k686337
51.4k686337
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
add a comment |Â
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader.
â Learning
2 mins ago
add a comment |Â
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