Graphical models, independence
$begingroup$
I don't have any good picture to show, but hopefully I can explain what I have a problem with. Consider a directed graph where A is shaded (conditionally given to my understading) and we have the nodes A,B,C,D,E,F,G. B,C,D,E,F,G are NOT conditionally given and every single one of B,C,D,E,F,G are pointing towards A, there are no other arrows. I want to investigate whether B is independent of C. When I look at Bishops literature it says:
Any path is considered blocked if it includes a node such that either:
The arrows meet head-to-head at the node, and neither the node, nor any of its descendants, is in the set C.
Notice that the question formulation doesn't say it's conditionally given, I've only noticed that it's being shaded. When I look at what Bishop says, I would say my two nodes are independent.
When I try to do some calculation though, I end up with:
$p(B,C,D,E,F,G vert A) = frac{p(A,B,C,D,E,F,G)}{p(A)} = frac{p(B)p(C)p(D)p(E)p(F)p(G)P(Avert B,C,D,E,F,G)}{p(A)}$
Based on this, it doesn't seem like they are independent? Can someone provide me an intuitive picture of this, I'm not grasping the concept.
I actually forgot an important question, what would happen if I was to marginalize over A in the last expression, it seems to me like that would make me end up with the independence, but I haven't seen Bishop before any marginalization when he has conditional independence ?
probability
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
$endgroup$
add a comment |
$begingroup$
I don't have any good picture to show, but hopefully I can explain what I have a problem with. Consider a directed graph where A is shaded (conditionally given to my understading) and we have the nodes A,B,C,D,E,F,G. B,C,D,E,F,G are NOT conditionally given and every single one of B,C,D,E,F,G are pointing towards A, there are no other arrows. I want to investigate whether B is independent of C. When I look at Bishops literature it says:
Any path is considered blocked if it includes a node such that either:
The arrows meet head-to-head at the node, and neither the node, nor any of its descendants, is in the set C.
Notice that the question formulation doesn't say it's conditionally given, I've only noticed that it's being shaded. When I look at what Bishop says, I would say my two nodes are independent.
When I try to do some calculation though, I end up with:
$p(B,C,D,E,F,G vert A) = frac{p(A,B,C,D,E,F,G)}{p(A)} = frac{p(B)p(C)p(D)p(E)p(F)p(G)P(Avert B,C,D,E,F,G)}{p(A)}$
Based on this, it doesn't seem like they are independent? Can someone provide me an intuitive picture of this, I'm not grasping the concept.
I actually forgot an important question, what would happen if I was to marginalize over A in the last expression, it seems to me like that would make me end up with the independence, but I haven't seen Bishop before any marginalization when he has conditional independence ?
probability
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
$endgroup$
$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01
add a comment |
$begingroup$
I don't have any good picture to show, but hopefully I can explain what I have a problem with. Consider a directed graph where A is shaded (conditionally given to my understading) and we have the nodes A,B,C,D,E,F,G. B,C,D,E,F,G are NOT conditionally given and every single one of B,C,D,E,F,G are pointing towards A, there are no other arrows. I want to investigate whether B is independent of C. When I look at Bishops literature it says:
Any path is considered blocked if it includes a node such that either:
The arrows meet head-to-head at the node, and neither the node, nor any of its descendants, is in the set C.
Notice that the question formulation doesn't say it's conditionally given, I've only noticed that it's being shaded. When I look at what Bishop says, I would say my two nodes are independent.
When I try to do some calculation though, I end up with:
$p(B,C,D,E,F,G vert A) = frac{p(A,B,C,D,E,F,G)}{p(A)} = frac{p(B)p(C)p(D)p(E)p(F)p(G)P(Avert B,C,D,E,F,G)}{p(A)}$
Based on this, it doesn't seem like they are independent? Can someone provide me an intuitive picture of this, I'm not grasping the concept.
I actually forgot an important question, what would happen if I was to marginalize over A in the last expression, it seems to me like that would make me end up with the independence, but I haven't seen Bishop before any marginalization when he has conditional independence ?
probability
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
$endgroup$
I don't have any good picture to show, but hopefully I can explain what I have a problem with. Consider a directed graph where A is shaded (conditionally given to my understading) and we have the nodes A,B,C,D,E,F,G. B,C,D,E,F,G are NOT conditionally given and every single one of B,C,D,E,F,G are pointing towards A, there are no other arrows. I want to investigate whether B is independent of C. When I look at Bishops literature it says:
Any path is considered blocked if it includes a node such that either:
The arrows meet head-to-head at the node, and neither the node, nor any of its descendants, is in the set C.
Notice that the question formulation doesn't say it's conditionally given, I've only noticed that it's being shaded. When I look at what Bishop says, I would say my two nodes are independent.
When I try to do some calculation though, I end up with:
$p(B,C,D,E,F,G vert A) = frac{p(A,B,C,D,E,F,G)}{p(A)} = frac{p(B)p(C)p(D)p(E)p(F)p(G)P(Avert B,C,D,E,F,G)}{p(A)}$
Based on this, it doesn't seem like they are independent? Can someone provide me an intuitive picture of this, I'm not grasping the concept.
I actually forgot an important question, what would happen if I was to marginalize over A in the last expression, it seems to me like that would make me end up with the independence, but I haven't seen Bishop before any marginalization when he has conditional independence ?
probability
probability
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
edited Dec 10 '18 at 7:04
A.Maine
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
asked Dec 10 '18 at 6:52
A.MaineA.Maine
398
398
$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01
add a comment |
$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01
$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01
add a comment |
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$begingroup$
The subgraph $B to A gets C$ has head-to-head arrows at $A$, but $A$ belongs to the conditioning set, so this path is not blocked.
$endgroup$
– angryavian
Dec 10 '18 at 7:16
$begingroup$
I'm grateful for the response, I although have two questions: 1) Do you interpret A as the conditioning set even though it isn't mentioned as conditioned, it's only shaded? 2) If I had plate notation over three variables, let's say N,R and C, how do you take these into account when you use these graphical models?
$endgroup$
– A.Maine
Dec 10 '18 at 7:48
$begingroup$
Yes, I think variables in the conditioning set are marked on a graphical model by shading. This should have been made clear in whatever source you are reading. Plate notation is just shorthand for a usual graph to avoid repeatedly drawing a particular subgraph that appears many times in the graph.
$endgroup$
– angryavian
Dec 10 '18 at 7:53
$begingroup$
Okay, my "real" problem, is that I have plate notation over R,C for B,C,D,E,F,G and N,R,C for A, does this change anything? Is there any mathematical proof that I can use to see that it actually works or ?
$endgroup$
– A.Maine
Dec 10 '18 at 7:55
$begingroup$
I'm doing this for an assignment, and my next question is to answer whether $A_{r,c}$ is independent of $A_{r,c+1}$ conditionally given $B_{r,c}$ and $B_{r,c+1} where as I mentioned, A has plate notation over r,c, n while B only has over r,c. The question seems odd since my figure shows A already shaded, how do you interpret a question like this?
$endgroup$
– A.Maine
Dec 10 '18 at 8:01