All Questions
Tagged with conditional-independence bayesian-network
14
questions
0
votes
0
answers
9
views
Bayesian network extracting further conditional independence statements then just from d-separation theorem
Given a Bayesian network $(p,\mathcal{G})$, where $p$ is our joint distribution, and $\mathcal{G}$ is a DAG.
Then by the d-separation theorem we can deduce conditional independence statements, in ...
- 359
0
votes
0
answers
47
views
Why implied Conditional Independencies of mediator and confounder are the same?
I am trying to understand why the impliedConditionalIndependencies function of the rethinking package returns the same value for ...
- 389
1
vote
0
answers
24
views
Implications of violating Bayesian network independence assumptions during inference
Consider the example Bayesian network below where $X \perp \!\!\! \perp Y $ (X is independent of Y).
Assuming that this is the true independence structure of the process that is generating the data, ...
- 53
1
vote
1
answer
88
views
Conditional independence proof
I want to prove that
$\mathbb{P}(X|U,P) = \mathbb{P}(X|U) \implies \mathbb{P}(X|U,P,T) = \mathbb{P}(X|U,T)$
Where all the letters denote random variables. I'm not sure that this is right, but it seems ...
- 11
0
votes
0
answers
18
views
BayesNet Independence
For BayesNet, can anyone explain how we can check the independence between the set of random variables? e.g. $\{B, D\} \perp \{G, I\} | A?$
- 1
1
vote
1
answer
86
views
How to show mathematically whether the following conditional relationships hold?
In the following Bayesian network, the variables $ x_{i} $ are mutually independent (let's assume that these are the positions of $N$ boats). The variables $ y_{i,j} $ are distance measurements ...
- 204
0
votes
1
answer
1k
views
How can a random variable be independent of a member of its minimal Markov blanket?
Consider the following Bayes network of random variables on some probability space:
The local Markov property asserts that any variable is independent of its non-descendants given its parents. Here, $...
- 218
2
votes
3
answers
325
views
Bishop PRML Question 8.10: d-separation [closed]
I have trouble with solving the second part of question 8.10 from Bishop's PRML (attached as image).
I tried several things. Here's my latest attempt:
\begin{align}
p(a, b, d) &= \int p(a)p(b)p(...
- 122
2
votes
1
answer
167
views
bayesian network conditional independence test
In the book: Bayesian Networks With Examples in R, the author does this independence test:
As I see it, this works both ways, we test if travel is independent of education likewise if education is ...
- 789
2
votes
1
answer
282
views
Proving conditional independence using a bayesian belief network / factorization
I have a bayesian belief network with 4 binary variables $A, B, C, D$. I now need to proof that for joint probability distributions factorized according the Bayesian network given below the
...
2
votes
0
answers
188
views
Bayesian structure learning: how to identify z as a collider in x-z-y structure?
In BNSL(Bayesian Network Structure Learning) problem, we are asked to learn a DAG(Directed Acyclic Graph) over a randon variable set $U$, given samples of the underlying distribution of $U$. The ...
- 213
5
votes
2
answers
561
views
Order of Conditional Independence Tests
I'm studying the PC algorithm for learning the structure of a Bayesian Network.
One of the steps refers to performing several rounds of conditional independence tests of increasing order, zero, first,...
- 545
2
votes
0
answers
145
views
Building independence maps (I-maps) from data
I am just getting into Bayesian networks, and I am having a hard time understanding how this algorithm works: http://pgm.stanford.edu/Algs/page-79.pdf
(The algorithm is from Probabilistic Graphical ...
- 21
1
vote
1
answer
187
views
Assuming n variables are conditionally independent given y, how do I compute p(y | x_1,...,x_n)?
Referencing this question, I know that if $x_1$ and $x_2$ are conditionally independent given $y$ (big assumption), then
$$P(y | x_1,x_2) = \frac{P(x_1,x_2 | y)P(y)}{P(x_2 | x_1)P(x_1)}$$
$$ = \frac{...
- 11