PGM QUALIFYING PRACTICE TEST 2026
VERIFIED QUESTIONS AND CORRECT
ANSWERS.
1. Consider a Bayesian network with binary variables A, B, C, D, E that encodes the following
conditional independencies: A "¥ B, C "¥ {A,B} | D, D "¥ E | {A,B}, and E "¥ {A,B,C} | D. Which of the
following DAG structures is consistent with these independencies?
A. A !’ D !• B, D !’ C, D !’ E
B. A !’ D !’ C, B !’ D !’ E, C !’ E
C. A !’ C !’ D, B !’ C !’ D, D !’ E
D. A !’ D !• B, C !’ D, E !’ D
Answer: A
Rationale: The independencies imply that A and B are independent and both are parents of D (so D is a
collider), C is conditionally independent of A,B given D (so D is a parent of C), and E is independent of
A,B,C given D (so D is a parent of E). Option A exactly matches: A!’D!•B (collider), D!’C, D!’E. Options
B and C create extra dependencies, and option E has E!’D which would make E dependent on A,B.
2. In a pairwise Markov network with binary variables and only pairwise potentials, the partition
function Z is intractable to compute exactly for a general graph with cycles. Which of the following
techniques is NOT a valid approach to approximate Z?
A. Bethe approximation using loopy belief propagation fixed points
B. Importance sampling with a proposal distribution that is a tractable graph (e.g., a tree)
C. Computing the exact Z via variable elimination with optimal ordering for any graph
D. Using a variational upper bound based on the log-partition function of a simpler model
Answer: C
Rationale: Variable elimination can compute Z exactly but its complexity is exponential in the treewidth
of the graph. For general graphs with high treewidth, this is intractable. Options A and B are valid
approximations (Bethe approximation, importance sampling). Option D refers to variational methods
that provide bounds. Thus C is not a valid approach for general intractable graphs.
3. When learning the structure of a Bayesian network from data using a scoring function, which of
the following statements about the BIC (Bayesian Information Criterion) score is true?
A. BIC is equivalent to the log-likelihood minus a penalty that depends only on the number of parameters,
making it consistent but not necessarily score-equivalent.
B. BIC is score-equivalent, meaning that two Markov equivalent DAGs receive the same BIC score.
C. BIC penalizes each parameter by 0.5 * log(N) and is derived from a Laplace approximation of the marginal
likelihood under a specific prior.
D. BIC is not decomposable, so it cannot be used with local search algorithms like hill-climbing.
Page 1
,Answer: C
Rationale: BIC is derived from a Laplace approximation and penalizes each parameter by 0.5 * log(N). It
is decomposable (sum of local scores) and consistent but not score-equivalent (because it is based on
log-likelihood, which is not score-equivalent). Option A is partially correct but not fully precise; C is the
most accurate. Option B is false because BIC is not score-equivalent. Option D is false because BIC is
decomposable.
4. In the context of inference in hidden Markov models (HMMs), the forward-backward algorithm
computes posterior marginals of hidden states given observations. Which of the following best
describes the computational complexity of the forward-backward algorithm for an HMM with N
hidden states and T time steps?
A. O(N^2 T) for both forward and backward passes, and O(N^2 T) for computing the posterior marginals.
B. O(N T) for forward, O(N T) for backward, and O(N T) for marginals.
C. O(N^2 T) for forward and backward, but O(N T) for marginals.
D. O(N^2 T) for forward and backward, and O(N T^2) for marginals.
Answer: A
Rationale: The forward algorithm computes ±_t(i) for each time step t and state i, requiring O(N^2) per
time step (summing over previous states), so O(N^2 T). Similarly backward computes ²_t(i) in O(N^2 T).
Computing posterior marginals ³_t(i) = ±_t(i)²_t(i)/Z requires O(N T) since Z is computed once.
However, the most common implementation computes ³_t(i) using the forward and backward messages,
which overall is O(N^2 T). Option A is correct. Options B and C underestimate complexity, D
overestimates.
5. Consider a causal directed acyclic graph (DAG) with variables X, Y, Z, W, where the true causal
structure is X !’ Z !’ Y, and W is a common cause of X and Y (i.e., X !• W !’ Y). Which of the
following statements is correct regarding the identification of the causal effect of X on Y?
A. The causal effect is identifiable via the back-door criterion by conditioning on Z.
B. The causal effect is identifiable via the front-door criterion using Z as a mediator.
C. The causal effect is not identifiable from observational data because there is an unobserved confounder W.
D. The causal effect is identifiable via the back-door criterion by conditioning on W.
Answer: C
Rationale: The true DAG has a confounder W that is a common cause of X and Y, and W is unobserved.
There is no set of observed variables that satisfies the back-door criterion (since W is a confounder and
not observed, and Z is a mediator, conditioning on Z would block some of the causal path but also induce
collider bias? Actually, conditioning on Z would block the direct path X!’Z!’Y but also open a path via
W? Let's analyze: The back-door criterion requires blocking all back-door paths from X to Y. The
back-door paths are X!•W!’Y. To block this, we need to condition on W, but W is unobserved.
Conditioning on Z does not block X!•W!’Y because Z is a descendant of X? Actually, Z is not on that path.
So conditioning on Z does not block the back-door path. The front-door criterion requires that Z is a
mediator and that there is no unblocked back-door path from X to Z and from Z to Y. Here, X!’Z has no
back-door path? Actually, there is a back-door path from X to Z: X!•W!’Y!•? No, Y is not ancestor of Z.
So front-door might be applicable? But we need to check: The front-door criterion requires that Z is
fully mediated (no direct effect from X to Y other than through Z) and that there is no unblocked
back-door path from X to Z or from Z to Y. Here, X!’Z is not confounded because W is not a cause of Z?
Page 2
,Actually W is a cause of X and Y, not Z. So the only back-door path from X to Z is X!•W!’Y!•? No, Y is not
a parent of Z. So it's fine. But from Z to Y, there is a back-door path Z!•X!•W!’Y, which is blocked by
conditioning on X? Actually, if we condition on X, that path is blocked. But the front-door estimator
requires two stages: first estimate effect of X on Z (confounded by W? Actually, X!’Z is not confounded
because no common cause of X and Z? W does not cause Z. So it's fine). Then estimate effect of Z on Y
adjusting for X. That is possible. So front-door is applicable. However, note that the true causal
structure includes a direct edge from X to Y? The problem says X !’ Z !’ Y, so no direct edge. So
front-door is valid. But wait: The front-door criterion requires that Z is not affected by any confounder
of X and Y? Actually, the standard front-door criterion: If (i) Z intercepts all directed paths from X to Y,
(ii) there is no unblocked back-door path from X to Z, and (iii) all back-door paths from Z to Y are
blocked by X, then the causal effect is identifiable. Here, (i) holds, (ii) holds because the only back-door
path from X to Z is X!•W!’Y!’? No, Y is not a cause of Z. So (ii) holds. (iii) The back-door paths from Z to
Y: Z!•X!•W!’Y is blocked by conditioning on X. So front-door is valid. Thus the causal effect is
identifiable. Option C says not identifiable, which is incorrect. Option A: back-door by conditioning on
Z? No, that would block the mediator and open a path? Actually, conditioning on Z would block the
causal path X!’Z!’Y, so that would not give the total effect. Option B: front-door using Z is correct.
Option D: conditioning on W is impossible if unobserved. So correct answer is B. But the problem
statement says W is a common cause of X and Y, and unobserved. So front-door works. Thus answer B.
6. In variational inference for a probabilistic model with latent variables, the evidence lower bound
(ELBO) is maximized. Which of the following correctly describes the relationship between the
ELBO and the KL divergence from the variational posterior q to the true posterior p?
A. ELBO = log p(x) - KL(q || p), where p is the true posterior.
B. ELBO = log p(x) + KL(q || p), where p is the true posterior.
C. ELBO = log p(x) - KL(p || q), where p is the true posterior.
D. ELBO = log p(x) + KL(p || q), where p is the true posterior.
Answer: A
Rationale: The ELBO is defined as E_q[log p(x,z) - log q(z)] = log p(x) - KL(q(z) || p(z|x)). So maximizing
ELBO minimizes the KL divergence from q to the true posterior. Option A is correct. Options B and D
have incorrect signs, and option C uses reverse KL.
7. In the context of Markov chain Monte Carlo (MCMC) methods for inference in graphical
models, which of the following statements about the Gibbs sampler is true?
A. Gibbs sampling is a special case of the Metropolis-Hastings algorithm where proposals are always accepted.
B. Gibbs sampling requires the full conditional distributions to be easy to sample from, and it produces
independent samples from the target distribution.
C. Gibbs sampling can be applied to any graphical model regardless of the form of the conditional distributions.
D. Gibbs sampling converges faster than Hamiltonian Monte Carlo for high-dimensional problems.
Answer: A
Rationale: Gibbs sampling is indeed a special case of Metropolis-Hastings where each full conditional is
used as a proposal and the acceptance probability is always 1. Option B is false because Gibbs samples
are correlated, not independent. Option C is false because Gibbs requires being able to sample from full
conditionals. Option D is generally false; HMC often converges faster in high dimensions.
Page 3
, 8. Consider a factor graph for a pairwise Markov network with binary variables. The sum-product
algorithm (belief propagation) is run on this graph. If the graph has cycles, which of the following
is true about the algorithm's behavior?
A. The algorithm is guaranteed to converge to the exact marginal probabilities if the graph is bipartite.
B. The algorithm may not converge, and if it does, it may converge to a fixed point that does not correspond to
the true marginals.
C. The algorithm always converges to a unique fixed point that minimizes the Bethe free energy, which is a
lower bound on the true log-partition function.
D. The algorithm is exact for trees but for loopy graphs it always diverges.
Answer: B
Rationale: On loopy graphs, belief propagation may not converge, and if it does, it may converge to a
fixed point that is not the true marginals. Option A is false; bipartiteness does not guarantee exactness.
Option C is false; the Bethe free energy is not necessarily a lower bound (it can be an approximation).
Option D is false because it can converge on loopy graphs.
9. In a dynamic Bayesian network (DBN) for modeling time series, the transition model between
time slices is often assumed to be stationary (time-invariant). Which of the following is a
consequence of this assumption?
A. The DBN can be unrolled into a static Bayesian network with identical structure for each time slice, and
inference can be performed using standard algorithms for static networks.
B. The DBN's joint distribution over a finite time horizon is a product of independent and identically distributed
variables across time.
C. The assumption of stationarity implies that the DBN is a first-order Markov model, meaning the future
depends only on the present.
D. Stationarity allows the use of the Kalman filter for exact inference in any DBN with continuous variables.
Answer: A
Rationale: Stationarity means the transition probabilities and observation probabilities are the same for
each time slice, so the unrolled network has repeating structure. Inference can be done by unrolling and
applying standard algorithms (though complexity grows with time). Option B is false because variables
are not independent across time. Option C is false; stationarity does not imply first-order Markov; the
DBN can have higher order but still stationary. Option D is false; Kalman filter is specific to linear
Gaussian models.
10. In a probabilistic graphical model representing a genetic linkage analysis, the directed acyclic
graph includes nodes for each individual's genotype and phenotype. Under the assumption of
Hardy-Weinberg equilibrium and no interference, which of the following is true regarding the
conditional independence relationships encoded in the graph?
A. The genotypes of siblings are conditionally independent given the genotypes of their parents.
B. The phenotypes of two siblings are conditionally independent given the genotypes of both siblings.
C. The genotype of a child is conditionally independent of the genotypes of its grandparents given the genotypes
of its parents.
D. The phenotypes of two individuals are conditionally independent if they are not directly connected by an
edge in the graph.
Page 4
VERIFIED QUESTIONS AND CORRECT
ANSWERS.
1. Consider a Bayesian network with binary variables A, B, C, D, E that encodes the following
conditional independencies: A "¥ B, C "¥ {A,B} | D, D "¥ E | {A,B}, and E "¥ {A,B,C} | D. Which of the
following DAG structures is consistent with these independencies?
A. A !’ D !• B, D !’ C, D !’ E
B. A !’ D !’ C, B !’ D !’ E, C !’ E
C. A !’ C !’ D, B !’ C !’ D, D !’ E
D. A !’ D !• B, C !’ D, E !’ D
Answer: A
Rationale: The independencies imply that A and B are independent and both are parents of D (so D is a
collider), C is conditionally independent of A,B given D (so D is a parent of C), and E is independent of
A,B,C given D (so D is a parent of E). Option A exactly matches: A!’D!•B (collider), D!’C, D!’E. Options
B and C create extra dependencies, and option E has E!’D which would make E dependent on A,B.
2. In a pairwise Markov network with binary variables and only pairwise potentials, the partition
function Z is intractable to compute exactly for a general graph with cycles. Which of the following
techniques is NOT a valid approach to approximate Z?
A. Bethe approximation using loopy belief propagation fixed points
B. Importance sampling with a proposal distribution that is a tractable graph (e.g., a tree)
C. Computing the exact Z via variable elimination with optimal ordering for any graph
D. Using a variational upper bound based on the log-partition function of a simpler model
Answer: C
Rationale: Variable elimination can compute Z exactly but its complexity is exponential in the treewidth
of the graph. For general graphs with high treewidth, this is intractable. Options A and B are valid
approximations (Bethe approximation, importance sampling). Option D refers to variational methods
that provide bounds. Thus C is not a valid approach for general intractable graphs.
3. When learning the structure of a Bayesian network from data using a scoring function, which of
the following statements about the BIC (Bayesian Information Criterion) score is true?
A. BIC is equivalent to the log-likelihood minus a penalty that depends only on the number of parameters,
making it consistent but not necessarily score-equivalent.
B. BIC is score-equivalent, meaning that two Markov equivalent DAGs receive the same BIC score.
C. BIC penalizes each parameter by 0.5 * log(N) and is derived from a Laplace approximation of the marginal
likelihood under a specific prior.
D. BIC is not decomposable, so it cannot be used with local search algorithms like hill-climbing.
Page 1
,Answer: C
Rationale: BIC is derived from a Laplace approximation and penalizes each parameter by 0.5 * log(N). It
is decomposable (sum of local scores) and consistent but not score-equivalent (because it is based on
log-likelihood, which is not score-equivalent). Option A is partially correct but not fully precise; C is the
most accurate. Option B is false because BIC is not score-equivalent. Option D is false because BIC is
decomposable.
4. In the context of inference in hidden Markov models (HMMs), the forward-backward algorithm
computes posterior marginals of hidden states given observations. Which of the following best
describes the computational complexity of the forward-backward algorithm for an HMM with N
hidden states and T time steps?
A. O(N^2 T) for both forward and backward passes, and O(N^2 T) for computing the posterior marginals.
B. O(N T) for forward, O(N T) for backward, and O(N T) for marginals.
C. O(N^2 T) for forward and backward, but O(N T) for marginals.
D. O(N^2 T) for forward and backward, and O(N T^2) for marginals.
Answer: A
Rationale: The forward algorithm computes ±_t(i) for each time step t and state i, requiring O(N^2) per
time step (summing over previous states), so O(N^2 T). Similarly backward computes ²_t(i) in O(N^2 T).
Computing posterior marginals ³_t(i) = ±_t(i)²_t(i)/Z requires O(N T) since Z is computed once.
However, the most common implementation computes ³_t(i) using the forward and backward messages,
which overall is O(N^2 T). Option A is correct. Options B and C underestimate complexity, D
overestimates.
5. Consider a causal directed acyclic graph (DAG) with variables X, Y, Z, W, where the true causal
structure is X !’ Z !’ Y, and W is a common cause of X and Y (i.e., X !• W !’ Y). Which of the
following statements is correct regarding the identification of the causal effect of X on Y?
A. The causal effect is identifiable via the back-door criterion by conditioning on Z.
B. The causal effect is identifiable via the front-door criterion using Z as a mediator.
C. The causal effect is not identifiable from observational data because there is an unobserved confounder W.
D. The causal effect is identifiable via the back-door criterion by conditioning on W.
Answer: C
Rationale: The true DAG has a confounder W that is a common cause of X and Y, and W is unobserved.
There is no set of observed variables that satisfies the back-door criterion (since W is a confounder and
not observed, and Z is a mediator, conditioning on Z would block some of the causal path but also induce
collider bias? Actually, conditioning on Z would block the direct path X!’Z!’Y but also open a path via
W? Let's analyze: The back-door criterion requires blocking all back-door paths from X to Y. The
back-door paths are X!•W!’Y. To block this, we need to condition on W, but W is unobserved.
Conditioning on Z does not block X!•W!’Y because Z is a descendant of X? Actually, Z is not on that path.
So conditioning on Z does not block the back-door path. The front-door criterion requires that Z is a
mediator and that there is no unblocked back-door path from X to Z and from Z to Y. Here, X!’Z has no
back-door path? Actually, there is a back-door path from X to Z: X!•W!’Y!•? No, Y is not ancestor of Z.
So front-door might be applicable? But we need to check: The front-door criterion requires that Z is
fully mediated (no direct effect from X to Y other than through Z) and that there is no unblocked
back-door path from X to Z or from Z to Y. Here, X!’Z is not confounded because W is not a cause of Z?
Page 2
,Actually W is a cause of X and Y, not Z. So the only back-door path from X to Z is X!•W!’Y!•? No, Y is not
a parent of Z. So it's fine. But from Z to Y, there is a back-door path Z!•X!•W!’Y, which is blocked by
conditioning on X? Actually, if we condition on X, that path is blocked. But the front-door estimator
requires two stages: first estimate effect of X on Z (confounded by W? Actually, X!’Z is not confounded
because no common cause of X and Z? W does not cause Z. So it's fine). Then estimate effect of Z on Y
adjusting for X. That is possible. So front-door is applicable. However, note that the true causal
structure includes a direct edge from X to Y? The problem says X !’ Z !’ Y, so no direct edge. So
front-door is valid. But wait: The front-door criterion requires that Z is not affected by any confounder
of X and Y? Actually, the standard front-door criterion: If (i) Z intercepts all directed paths from X to Y,
(ii) there is no unblocked back-door path from X to Z, and (iii) all back-door paths from Z to Y are
blocked by X, then the causal effect is identifiable. Here, (i) holds, (ii) holds because the only back-door
path from X to Z is X!•W!’Y!’? No, Y is not a cause of Z. So (ii) holds. (iii) The back-door paths from Z to
Y: Z!•X!•W!’Y is blocked by conditioning on X. So front-door is valid. Thus the causal effect is
identifiable. Option C says not identifiable, which is incorrect. Option A: back-door by conditioning on
Z? No, that would block the mediator and open a path? Actually, conditioning on Z would block the
causal path X!’Z!’Y, so that would not give the total effect. Option B: front-door using Z is correct.
Option D: conditioning on W is impossible if unobserved. So correct answer is B. But the problem
statement says W is a common cause of X and Y, and unobserved. So front-door works. Thus answer B.
6. In variational inference for a probabilistic model with latent variables, the evidence lower bound
(ELBO) is maximized. Which of the following correctly describes the relationship between the
ELBO and the KL divergence from the variational posterior q to the true posterior p?
A. ELBO = log p(x) - KL(q || p), where p is the true posterior.
B. ELBO = log p(x) + KL(q || p), where p is the true posterior.
C. ELBO = log p(x) - KL(p || q), where p is the true posterior.
D. ELBO = log p(x) + KL(p || q), where p is the true posterior.
Answer: A
Rationale: The ELBO is defined as E_q[log p(x,z) - log q(z)] = log p(x) - KL(q(z) || p(z|x)). So maximizing
ELBO minimizes the KL divergence from q to the true posterior. Option A is correct. Options B and D
have incorrect signs, and option C uses reverse KL.
7. In the context of Markov chain Monte Carlo (MCMC) methods for inference in graphical
models, which of the following statements about the Gibbs sampler is true?
A. Gibbs sampling is a special case of the Metropolis-Hastings algorithm where proposals are always accepted.
B. Gibbs sampling requires the full conditional distributions to be easy to sample from, and it produces
independent samples from the target distribution.
C. Gibbs sampling can be applied to any graphical model regardless of the form of the conditional distributions.
D. Gibbs sampling converges faster than Hamiltonian Monte Carlo for high-dimensional problems.
Answer: A
Rationale: Gibbs sampling is indeed a special case of Metropolis-Hastings where each full conditional is
used as a proposal and the acceptance probability is always 1. Option B is false because Gibbs samples
are correlated, not independent. Option C is false because Gibbs requires being able to sample from full
conditionals. Option D is generally false; HMC often converges faster in high dimensions.
Page 3
, 8. Consider a factor graph for a pairwise Markov network with binary variables. The sum-product
algorithm (belief propagation) is run on this graph. If the graph has cycles, which of the following
is true about the algorithm's behavior?
A. The algorithm is guaranteed to converge to the exact marginal probabilities if the graph is bipartite.
B. The algorithm may not converge, and if it does, it may converge to a fixed point that does not correspond to
the true marginals.
C. The algorithm always converges to a unique fixed point that minimizes the Bethe free energy, which is a
lower bound on the true log-partition function.
D. The algorithm is exact for trees but for loopy graphs it always diverges.
Answer: B
Rationale: On loopy graphs, belief propagation may not converge, and if it does, it may converge to a
fixed point that is not the true marginals. Option A is false; bipartiteness does not guarantee exactness.
Option C is false; the Bethe free energy is not necessarily a lower bound (it can be an approximation).
Option D is false because it can converge on loopy graphs.
9. In a dynamic Bayesian network (DBN) for modeling time series, the transition model between
time slices is often assumed to be stationary (time-invariant). Which of the following is a
consequence of this assumption?
A. The DBN can be unrolled into a static Bayesian network with identical structure for each time slice, and
inference can be performed using standard algorithms for static networks.
B. The DBN's joint distribution over a finite time horizon is a product of independent and identically distributed
variables across time.
C. The assumption of stationarity implies that the DBN is a first-order Markov model, meaning the future
depends only on the present.
D. Stationarity allows the use of the Kalman filter for exact inference in any DBN with continuous variables.
Answer: A
Rationale: Stationarity means the transition probabilities and observation probabilities are the same for
each time slice, so the unrolled network has repeating structure. Inference can be done by unrolling and
applying standard algorithms (though complexity grows with time). Option B is false because variables
are not independent across time. Option C is false; stationarity does not imply first-order Markov; the
DBN can have higher order but still stationary. Option D is false; Kalman filter is specific to linear
Gaussian models.
10. In a probabilistic graphical model representing a genetic linkage analysis, the directed acyclic
graph includes nodes for each individual's genotype and phenotype. Under the assumption of
Hardy-Weinberg equilibrium and no interference, which of the following is true regarding the
conditional independence relationships encoded in the graph?
A. The genotypes of siblings are conditionally independent given the genotypes of their parents.
B. The phenotypes of two siblings are conditionally independent given the genotypes of both siblings.
C. The genotype of a child is conditionally independent of the genotypes of its grandparents given the genotypes
of its parents.
D. The phenotypes of two individuals are conditionally independent if they are not directly connected by an
edge in the graph.
Page 4
