Hello, I began seeing your lectures just a few days and I am really enjoying it. I have just a question, Why, in the example of d-separation, when we condition on M1, W2, W3, X2, and also on X1 we have independence? I am just wondering because even if we condition on X1 we are still conditioning on X2 which is the collider. Thank you
@@BradyNealCausalInference Oh, ok, so the X1 plays the role of a mediator from T to X2. So, even if we condition on the collider and unblock the path because we are also conditioning in X1, the path is blocked again, right? Thank you for your answer.
Thanks for the lecture. One quick question: On slide 31, I guess the path between the two nodes is blocked if both conditions are met, not either of them! Is that right?
In Theorem 3.1, we have ' If X and Y are d-separated in G conditioned on Z, then X and Y are independent in P conditioned on Z.' I wonder do we have " If X and Y are independent in P conditioned on Z, then X and Y are d-separated in G conditioned on Z"? Thanks in advance.
Consider the following M-shaped graph: T < - W1 - > W2 < - W3 - > Y Is T and Y d-separated if: (1) None of the nodes are conditioned on? (2) Either W1 or W3 are conditioned on? (3) W2 is conditioned on? Here are my answers with explanation. Are they correct? (1) Yes because W2 is a collider (2) Yes for conditioning on W2 because T < - W2 - > W3 is a fork and Yes for conditioning on W3 because W2 < - W3 - > Y is a fork (3) No because W2 is a collider which is conditioned which unblocks the W1 - > W2 < - W3 path
@@BradyNealCausalInference Thank you . Just a small clarification. You mentioned all are correct but I can see that your answer to (3) is "Yes" whereas my answer is "No". Can you please clarify whether T and Y are d-separated if W2 is conditioned on?
Would T and Y be d-separated in the below example assuming we only condition on W1: W1->W2->Y | v T Edit: I am guessing the answer is "No". The intuition behind my reasoning is that although W1 has been conditioned on (which means it no longer has an effect on W2) nonetheless W2 is still a random variable on its own which can increase or decrease and this change in W2 will have an effect on both Y and T making them dependent on each other ( i.e. T and Y are not independent )
In general no. But there are cases where they are. For example, if W2 is a deterministic function of W1, then you can show that conditioning on W1 d-separates T and Y.
Still have some confuse about slide 32's theorem after checking the answer below. Can you make a concrete example about G & P here? Or where can I find more detail about this theorem?
For every node in G, there is a corresponding variable in P (with the same symbol to refer to it). Then, if A and B are d-separated in G, then, A and B are independent in distribution P: p(a, b) = p(a) p(b). For examples, consider the examples of d-separated variables in slide 33. For every example, the variables in that example are independent in P. The Koller & Friedman book that I mention at some point in lecture 3 has a big focus on this stuff in their Chapter 3 (I think).
*conditioning nodes* not "intervening nodes" (important difference) Yes, the reason is that X and Y could be d-separated without conditioning on anything. Given the Markov assumption, that means that X and Y are marginally independent (as opposed to conditionally independent).
Brady Neal - Causal Inference Conditioning nodes, yes definitely an important difference 😬. Only case X & Y can be d-separated without conditioning will be in case of a collider and it’s descendant correct or are there any other cases?
@@nisargvp If there is a path, exactly; that path must have a collider on it. If there isn't a path, that also works. X and Y are d-separated (by the empty set) if there is no path between them.
That's right. That would be a different (seemingly less acceptable) assumption called *faithfulness*. We'll get to this in the causal discovery portion of the course, since this kind of assumption allows us to infer the graph from the independencies in the distribution. The usual example, other than the "intransitive" (CTRL + F for this term) cases that I mention in the book, is to think of two directed paths from X to Y. However, they "cancel out," leaving X and Y independent in distribution. So that would be an independence in distribution that isn't reflected with the corresponding lack of edges in the graph (more on this in the causal discovery week).
I'm currently following my third course on D-seperation, and watching this video finally makes me understand it fully. Amazing examplees!
Hello, I began seeing your lectures just a few days and I am really enjoying it. I have just a question, Why, in the example of d-separation, when we condition on M1, W2, W3, X2, and also on X1 we have independence? I am just wondering because even if we condition on X1 we are still conditioning on X2 which is the collider. Thank you
Check out the definition for a blocked path on slide 31. A path if blocked if *either* of those two considers are true. So X1 blocks the path.
@@BradyNealCausalInference Oh, ok, so the X1 plays the role of a mediator from T to X2.
So, even if we condition on the collider and unblock the path because we are also conditioning in X1, the path is blocked again, right?
Thank you for your answer.
@@reginaduarte6272 Right!
Thanks for the lecture. One quick question: On slide 31, I guess the path between the two nodes is blocked if both conditions are met, not either of them! Is that right?
In Theorem 3.1, we have ' If X and Y are d-separated in G conditioned on Z, then X and Y are independent in P conditioned on Z.' I wonder do we have " If X and Y are independent in P conditioned on Z, then X and Y are d-separated in G conditioned on Z"? Thanks in advance.
Consider the following M-shaped graph:
T < - W1 - > W2 < - W3 - > Y
Is T and Y d-separated if:
(1) None of the nodes are conditioned on?
(2) Either W1 or W3 are conditioned on?
(3) W2 is conditioned on?
Here are my answers with explanation. Are they correct?
(1) Yes because W2 is a collider
(2) Yes for conditioning on W2 because T < - W2 - > W3 is a fork and Yes for conditioning on W3 because W2 < - W3 - > Y is a fork
(3) No because W2 is a collider which is conditioned which unblocks the W1 - > W2 < - W3 path
1. Yes
2. Yes, and also the unconditioned collider still blocks, just like in 1.
3. Yes.
All are correct :)
@@BradyNealCausalInference Thank you . Just a small clarification. You mentioned all are correct but I can see that your answer to (3) is "Yes" whereas my answer is "No". Can you please clarify whether T and Y are d-separated if W2 is conditioned on?
@@farbodsafe Oops, I mean "Yes, it's correct"
Would T and Y be d-separated in the below example assuming we only condition on W1:
W1->W2->Y
|
v
T
Edit: I am guessing the answer is "No". The intuition behind my reasoning is that although W1 has been conditioned on (which means it no longer has an effect on W2) nonetheless W2 is still a random variable on its own which can increase or decrease and this change in W2 will have an effect on both Y and T making them dependent on each other ( i.e. T and Y are not independent )
In general no. But there are cases where they are. For example, if W2 is a deterministic function of W1, then you can show that conditioning on W1 d-separates T and Y.
why additionally condition on X_1 and then T & Y become d-separated?
Do you know the relation between d-separation and faithfulness, or between d-separation and minimality?
Still have some confuse about slide 32's theorem after checking the answer below. Can you make a concrete example about G & P here? Or where can I find more detail about this theorem?
For every node in G, there is a corresponding variable in P (with the same symbol to refer to it). Then, if A and B are d-separated in G, then, A and B are independent in distribution P: p(a, b) = p(a) p(b). For examples, consider the examples of d-separated variables in slide 33. For every example, the variables in that example are independent in P. The Koller & Friedman book that I mention at some point in lecture 3 has a big focus on this stuff in their Chapter 3 (I think).
@@BradyNealCausalInference Awesome explanation! I think I get it now! Thanks again~
Is there a reason we refer the intervening nodes between X & Y as potentially "empty" sets?
*conditioning nodes* not "intervening nodes" (important difference)
Yes, the reason is that X and Y could be d-separated without conditioning on anything. Given the Markov assumption, that means that X and Y are marginally independent (as opposed to conditionally independent).
Brady Neal - Causal Inference Conditioning nodes, yes definitely an important difference 😬. Only case X & Y can be d-separated without conditioning will be in case of a collider and it’s descendant correct or are there any other cases?
@@nisargvp If there is a path, exactly; that path must have a collider on it. If there isn't a path, that also works. X and Y are d-separated (by the empty set) if there is no path between them.
@@BradyNealCausalInference Makes sense, thanks :)
quick question: the reverse, which is: X indep Y | Z wrt P --> X indep Y | Z wrt G, is not true? (ref: ua-cam.com/video/yIwTzdwVz0Q/v-deo.html)
That's right. That would be a different (seemingly less acceptable) assumption called *faithfulness*. We'll get to this in the causal discovery portion of the course, since this kind of assumption allows us to infer the graph from the independencies in the distribution. The usual example, other than the "intransitive" (CTRL + F for this term) cases that I mention in the book, is to think of two directed paths from X to Y. However, they "cancel out," leaving X and Y independent in distribution. So that would be an independence in distribution that isn't reflected with the corresponding lack of edges in the graph (more on this in the causal discovery week).