title: 斯坦福-随机图模型-week1.6
tags: note
notebook: 6- 英文课程-9-Probabilistic Graphical Models 1: Representation
---

斯坦福-随机图模型-week1.6

测试

1。第 1 个问题

Independencies in a graph.

Which pairs of variables are independent in the graphical model below, given that none of them have been observed? You may select 1 or more options.

A, B
正确
There are no active trails between A and B, so they are independent.

C, D

未选择的是正确的 

B, E

未选择的是正确的 

D, E

未选择的是正确的 

None - there are no pairs of independent variables.

未选择的是正确的 

第 2 个问题

正确
1 / 1 分
2。第 2 个问题
*Independencies in a graph. (An asterisk marks a question that is more challenging. Congratulations if you get it right!)

Now assume that the value of E is known. (E is observed. A, B, C, and D are not observed.) Which pairs of variables (not including E) are independent in the same graphical model, given E? You may select 1 or more options.

None - given E, there are no pairs of variables that are independent.

正确 
Observing E activates the V-structures around C and E, giving rise to active trails between every pair of variables in the network.

A, B

未选择的是正确的 

A, C

未选择的是正确的 

A, D

未选择的是正确的 

B, D

未选择的是正确的 

D, C

未选择的是正确的 

B, C

未选择的是正确的 

第 3 个问题

正确
1 / 1 分
3。第 3 个问题
I-maps. I-maps can also be defined directly on graphs as follows. Let I(G) be the set of independencies encoded by a graph G. Then G1 is an I-map for G2 if I(G1)⊆I(G2).

Which of the following statements about I-maps are true? You may select 1 or more options.

A graph K is an I-map for a graph G if and only if all of the independencies encoded by K are also encoded by G.

正确 
K is an I-map for G if K does not make independence assumptions that are not true in G. An easy way to remember this is that the complete graph, which has no independencies, is an I-map of all distributions.

A graph K is an I-map for a graph G if and only if K encodes all of the independences that G has and more.

未选择的是正确的 

An I-map is a function f that maps a graph G to itself, i.e., f(G)=G.

未选择的是正确的 

The graph K that is the same as the graph G, except that all of the edges are oriented in the opposite direction as the corresponding edges in G, is always an I-map for G, regardless of the structure of G.

未选择的是正确的 

I-maps are Apple's answer to Google Maps.

未选择的是正确的 

第 4 个问题

错误
0 / 1 分
4。第 4 个问题
*Naive Bayes.

Consider the following Naive Bayes model for flu diagnosis:

Assume a population size of 10,000. Which of the following statements are true in this model? You may select 1 or more options.

Say we observe that 1000 people have the flu, out of which 500 people have a headache (and possibly other symptoms) and 500 have a fever (and possibly other symptoms).

We would expect that approximately 250 people with the flu also have both a headache and fever.

正确 
Given that someone has the flu, whether he has a headache is independent of whether he has a fever. We can thus calculate:

P(Headache=1,Fever=1|Flu=1)=P(Headache=1|Flu=1)×P(Fever=1|Flu=1)≈0.5∗0.5=0.25.
Since 1000 people have the flu, we can estimate that 250 of these people will have both a headache and fever.

Note that this is only an estimate: we can assert with high confidence that P(Headache=1,Fever=1|Flu=1) is near to 0.25, but in general it will not be exactly 0.25. Moreover, even if it is exactly 0.25, the number of people with the flu, a headache and a fever need not be exactly 250 all the time. Think of this as analogous to flipping a fair coin: even though the probability of seeing a heads is exactly 0.5, in any given sequence of coin flips we need not see exactly half of the coins turning up heads.


Say we observe that 1000 people have a headache (and possibly other symptoms), out of which 500 people have the flu (and possibly other symptoms), and 500 people have a fever (and possibly other symptoms).

We would expect that approximately 250 people with a headache also have both the flu and a fever.

这个选项的答案不正确 
Even after observing the Headache variable, there is still an active trail from Flu to Fever. Thus, the probability of someone with a headache also having a flu is dependent on the probability of his having a fever as well. For example, if someone has a flu, he could be more likely to have a fever, irrespective of whether he has a headache or not.

We therefore cannot estimate P(Flu=1,Fever=1|Headache=1) from the conditional marginal probabilities P(Flu=1|Headache=1) and P(Fever=1|Headache=1).

Say we observe that 1000 people have the flu, out of which 500 people have a headache (and possibly other symptoms) and 500 have a fever (and possibly other symptoms).

We can conclude that exactly 250 people with the flu also have both a headache and fever.

未选择的是正确的 

Say we observe that 1000 people have the flu, out of which 500 people have a headache (and possibly other symptoms) and 500 people have a fever (and possibly other symptoms).

Without more information, we cannot estimate how many people with the flu also have both a headache and fever.

未选择的是正确的 

第 5 个问题

正确
1 / 1 分
5。第 5 个问题
I-maps.

Suppose (A⊥B)∈I(P), and G is an I-map of P, where G is a Bayesian network and P is a probability distribution. Is it necessarily true that (A⊥B)∈I(G)?

Yes

No

正确 
Since G is an I-map of P, all independencies in G are also in P. However, this doesn't mean that all independencies in P are also in G. An easy way to remember this is that the complete graph, which has no independencies, is an I-map of all distributions.