OMSCS 7641 FINAL EXAM QUESTIONS &
ANSWERS!!
,1 of 34
Term
Assume we have 2 independent fair
coins: P(A) = P(B) = .5
P(A,B) =
P(A|B) =
H(A) =
H(B) =
H(A,B) =
H(A|B) =
I(A,B) =
Give this one a try later!
The NE for this game are (A,A) and (B,B)
(A,A) - Neither player can increase the payoff by selecting a different action.
This is a NE.
(A,B) - Player 1 can choose B to obtain a higher payoff. Player 2 can choose A to
obtain a higher payoff. This is not a NE.
(B,A) - Player 1 can choose A to obtain a higher payoff. Player 2 can choose B to
obtain a higher payoff. This is not a NE.
(B,B) - Neither player can increase the payoff by selecting a different action.
This is a NE.
P(A) = P(b) = .5
P(A,B) = Joint probability is given by product of P(A)*P(B) = .25 P(A|
B) = P(A) since they are independent of each other = .5 H(A) =
- Sum P(A) Log P(A) = ..5 log .5 - .5 log .5 = 1
H(B) = - Sum P(A) Log P(A) = ..5 log .5 - .5 log .5 = 1
H(A,B) = Sum P(A,B) log P(A,B) = -4 (.25 log .25) = 2 H(A|B)
= -Sum P (A,B) log P(A|B) = 1
I(A,B) = Mutual Information = H(A) - H(A|B) = 1 -1 = 0. Since the 2 coins are
independent they do not have any mutual information on each other.
,P(A) = P(B) = .5
P(A,B) = .5 since both can be either heads or tails.
P(A|B) = P (A,B) / P(B) = .5/.5 = 1
H(A) = 1 since still using fair coins
H (B) = 1 since still using fair coins
H(A,B) = 1
H(A|B) = 0
I(A,B) = 1
The NE for this game are (A,A) and (B,B).
(A,A) - Neither player can increase the payoff by selecting a different action.
This is a NE.
(A,B) - Player 1 can choose B to obtain a higher payoff. Player 2 can choose A to
obtain a higher
payoff. This is not a NE.
(B,A) - Player 1 can choose A to obtain a higher payoff. Player 2 can choose B to
obtain a higher
payoff. This is not a NE.
(B,B) - Neither player can increase the payoff by selecting a different action.
This is a NE.
Don't know?
, 2 of 34
Definition
As K-means requires that all objects need to be assigned to a
cluster; therefore, outliers have to be assigned to a particular
cluster, leading to non-descriptive centroids that no longer
capture the
characteristics of most objects, belonging to a particular cluster.
Some techniques that have some merit include:
-Remove outliers, prior to applying k-means
-Use the cluster medoid, instead of the cluster centroid as the
cluster summary
Give this one a try later!
K-means has difficulties clustering datasets which contain a lot of outliers.
Explain, why this is the case! What could be done to alleviate the problem?
Kmeans and EM clustering methods both require providing the value of K in
order to form clusters.
K-means only finds clusters that are a local (but not a global) maximum of the
objective function J it minimizes. What does this mean? What are the implications
of this fact for using k-means to cluster real world datasets?
Assume we have 2 independent fair
coins: P(A) = P(B) = .5
P(A,B) =
P(A|B) =
H(A) =
H(B) =
H(A,B) =
ANSWERS!!
,1 of 34
Term
Assume we have 2 independent fair
coins: P(A) = P(B) = .5
P(A,B) =
P(A|B) =
H(A) =
H(B) =
H(A,B) =
H(A|B) =
I(A,B) =
Give this one a try later!
The NE for this game are (A,A) and (B,B)
(A,A) - Neither player can increase the payoff by selecting a different action.
This is a NE.
(A,B) - Player 1 can choose B to obtain a higher payoff. Player 2 can choose A to
obtain a higher payoff. This is not a NE.
(B,A) - Player 1 can choose A to obtain a higher payoff. Player 2 can choose B to
obtain a higher payoff. This is not a NE.
(B,B) - Neither player can increase the payoff by selecting a different action.
This is a NE.
P(A) = P(b) = .5
P(A,B) = Joint probability is given by product of P(A)*P(B) = .25 P(A|
B) = P(A) since they are independent of each other = .5 H(A) =
- Sum P(A) Log P(A) = ..5 log .5 - .5 log .5 = 1
H(B) = - Sum P(A) Log P(A) = ..5 log .5 - .5 log .5 = 1
H(A,B) = Sum P(A,B) log P(A,B) = -4 (.25 log .25) = 2 H(A|B)
= -Sum P (A,B) log P(A|B) = 1
I(A,B) = Mutual Information = H(A) - H(A|B) = 1 -1 = 0. Since the 2 coins are
independent they do not have any mutual information on each other.
,P(A) = P(B) = .5
P(A,B) = .5 since both can be either heads or tails.
P(A|B) = P (A,B) / P(B) = .5/.5 = 1
H(A) = 1 since still using fair coins
H (B) = 1 since still using fair coins
H(A,B) = 1
H(A|B) = 0
I(A,B) = 1
The NE for this game are (A,A) and (B,B).
(A,A) - Neither player can increase the payoff by selecting a different action.
This is a NE.
(A,B) - Player 1 can choose B to obtain a higher payoff. Player 2 can choose A to
obtain a higher
payoff. This is not a NE.
(B,A) - Player 1 can choose A to obtain a higher payoff. Player 2 can choose B to
obtain a higher
payoff. This is not a NE.
(B,B) - Neither player can increase the payoff by selecting a different action.
This is a NE.
Don't know?
, 2 of 34
Definition
As K-means requires that all objects need to be assigned to a
cluster; therefore, outliers have to be assigned to a particular
cluster, leading to non-descriptive centroids that no longer
capture the
characteristics of most objects, belonging to a particular cluster.
Some techniques that have some merit include:
-Remove outliers, prior to applying k-means
-Use the cluster medoid, instead of the cluster centroid as the
cluster summary
Give this one a try later!
K-means has difficulties clustering datasets which contain a lot of outliers.
Explain, why this is the case! What could be done to alleviate the problem?
Kmeans and EM clustering methods both require providing the value of K in
order to form clusters.
K-means only finds clusters that are a local (but not a global) maximum of the
objective function J it minimizes. What does this mean? What are the implications
of this fact for using k-means to cluster real world datasets?
Assume we have 2 independent fair
coins: P(A) = P(B) = .5
P(A,B) =
P(A|B) =
H(A) =
H(B) =
H(A,B) =
