Directed vs undirected graphs in Bishop's PRML
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In Bishop's PRML, chapter 8 is dedicated to graphical models. In figure 8.32, we have the following figure showing a directed and an undirected graph:
These two graphs are said to be "equivalent". What is meant by this? As far as I understand, the directed graph describes conditional probabilities, when an arrow from $x_1$ to $x_2$ means that $x_2$ is dependent on x1, or $p(x_2|x_1) neq p(x_2)$ in general but $p(x_1|x_2) =p(x_1)$. And in the directed graph, the conditional probabilities can go in any direction, so in general, $p(x_2|x_1) neq p(x_2)$ and $p(x_1|x_2) neq p(x_1)$.
If this is the case, the two graphs can not be equivalent since graph b) clearly doesn't express the fact that $x_1$ is independent of $x_2$. But why then are they said to be equivalent? And what is the point of converting a directed graph to an undirected graph if they do not express the same thing?
probability-theory graph-theory bayesian-network
asked Dec 3 '18 at 12:58
SandiSandi
255112
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add a comment |
$begingroup$
In Bishop's PRML, chapter 8 is dedicated to graphical models. In figure 8.32, we have the following figure showing a directed and an undirected graph:
These two graphs are said to be "equivalent". What is meant by this? As far as I understand, the directed graph describes conditional probabilities, when an arrow from $x_1$ to $x_2$ means that $x_2$ is dependent on x1, or $p(x_2|x_1) neq p(x_2)$ in general but $p(x_1|x_2) =p(x_1)$. And in the directed graph, the conditional probabilities can go in any direction, so in general, $p(x_2|x_1) neq p(x_2)$ and $p(x_1|x_2) neq p(x_1)$.
If this is the case, the two graphs can not be equivalent since graph b) clearly doesn't express the fact that $x_1$ is independent of $x_2$. But why then are they said to be equivalent? And what is the point of converting a directed graph to an undirected graph if they do not express the same thing?
probability-theory graph-theory bayesian-network
asked Dec 3 '18 at 12:58
SandiSandi
255112
$endgroup$
add a comment |
$begingroup$
In Bishop's PRML, chapter 8 is dedicated to graphical models. In figure 8.32, we have the following figure showing a directed and an undirected graph:
These two graphs are said to be "equivalent". What is meant by this? As far as I understand, the directed graph describes conditional probabilities, when an arrow from $x_1$ to $x_2$ means that $x_2$ is dependent on x1, or $p(x_2|x_1) neq p(x_2)$ in general but $p(x_1|x_2) =p(x_1)$. And in the directed graph, the conditional probabilities can go in any direction, so in general, $p(x_2|x_1) neq p(x_2)$ and $p(x_1|x_2) neq p(x_1)$.
If this is the case, the two graphs can not be equivalent since graph b) clearly doesn't express the fact that $x_1$ is independent of $x_2$. But why then are they said to be equivalent? And what is the point of converting a directed graph to an undirected graph if they do not express the same thing?
probability-theory graph-theory bayesian-network
asked Dec 3 '18 at 12:58
SandiSandi
255112
$endgroup$
In Bishop's PRML, chapter 8 is dedicated to graphical models. In figure 8.32, we have the following figure showing a directed and an undirected graph:
These two graphs are said to be "equivalent". What is meant by this? As far as I understand, the directed graph describes conditional probabilities, when an arrow from $x_1$ to $x_2$ means that $x_2$ is dependent on x1, or $p(x_2|x_1) neq p(x_2)$ in general but $p(x_1|x_2) =p(x_1)$. And in the directed graph, the conditional probabilities can go in any direction, so in general, $p(x_2|x_1) neq p(x_2)$ and $p(x_1|x_2) neq p(x_1)$.
If this is the case, the two graphs can not be equivalent since graph b) clearly doesn't express the fact that $x_1$ is independent of $x_2$. But why then are they said to be equivalent? And what is the point of converting a directed graph to an undirected graph if they do not express the same thing?
probability-theory graph-theory bayesian-network
probability-theory graph-theory bayesian-network
asked Dec 3 '18 at 12:58
SandiSandi
255112
asked Dec 3 '18 at 12:58
SandiSandi
255112
edited Dec 3 '18 at 13:12
Sandi
asked Dec 3 '18 at 12:58
SandiSandi
255112
asked Dec 3 '18 at 12:58
SandiSandi
255112
asked Dec 3 '18 at 12:58
SandiSandi
255112
255112
add a comment |
add a comment |
1 Answer
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They aren't equivalent. The caption for the second one should really say something like "the corresponding undirected graph" or "the underlying undirected graph".
The second picture as what you get if you forget the information of which way the edges are going. But this isn't an equivalence in any real sense, since you can't go backwards: if you start with an undirected graph there are many different ways to make it directed, which express different things.
The only time there is a point in converting a directed graph to an undirected graph is if the directions don't matter for a particular application.
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
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$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
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@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
add a comment |
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$begingroup$
They aren't equivalent. The caption for the second one should really say something like "the corresponding undirected graph" or "the underlying undirected graph".
The second picture as what you get if you forget the information of which way the edges are going. But this isn't an equivalence in any real sense, since you can't go backwards: if you start with an undirected graph there are many different ways to make it directed, which express different things.
The only time there is a point in converting a directed graph to an undirected graph is if the directions don't matter for a particular application.
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
$endgroup$
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
add a comment |
$begingroup$
They aren't equivalent. The caption for the second one should really say something like "the corresponding undirected graph" or "the underlying undirected graph".
The second picture as what you get if you forget the information of which way the edges are going. But this isn't an equivalence in any real sense, since you can't go backwards: if you start with an undirected graph there are many different ways to make it directed, which express different things.
The only time there is a point in converting a directed graph to an undirected graph is if the directions don't matter for a particular application.
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
$endgroup$
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
add a comment |
$begingroup$
They aren't equivalent. The caption for the second one should really say something like "the corresponding undirected graph" or "the underlying undirected graph".
The second picture as what you get if you forget the information of which way the edges are going. But this isn't an equivalence in any real sense, since you can't go backwards: if you start with an undirected graph there are many different ways to make it directed, which express different things.
The only time there is a point in converting a directed graph to an undirected graph is if the directions don't matter for a particular application.
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
$endgroup$
They aren't equivalent. The caption for the second one should really say something like "the corresponding undirected graph" or "the underlying undirected graph".
The second picture as what you get if you forget the information of which way the edges are going. But this isn't an equivalence in any real sense, since you can't go backwards: if you start with an undirected graph there are many different ways to make it directed, which express different things.
The only time there is a point in converting a directed graph to an undirected graph is if the directions don't matter for a particular application.
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
answered Dec 3 '18 at 13:11
Especially LimeEspecially Lime
21.9k22858
21.9k22858
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
add a comment |
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
It's strange though, because on page 394, section 8.4.1 Bishop says that they are equivalent and that they express the exact same set of conditional independence statements. What does he mean?
$endgroup$
– Sandi
Dec 4 '18 at 9:26
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
@Sandi In the specific context of dependence of events, you can't have a one-way dependence: if $p(x|y)=p(x)$ then it follows that $p(y|x)=p(y)$ (both are equivalent to $p(xwedge y)=p(x)p(y)$). But if this is what Bishop is talking about, it makes no sense to use a directed graph in the first place.
$endgroup$
– Especially Lime
Dec 4 '18 at 9:54
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
$begingroup$
I agree that dependence in one direction also means dependence in the other. Bishop uses directed graphs to express conditional probabilities in the "directions" expressed by a selected factorization of a joint probability into conditionals, not sure why.
$endgroup$
– Sandi
Dec 4 '18 at 9:57
add a comment |
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