CISC 7700X Midterm Exam

1. c
2. b
3. a
4. d
5. c
6. b
7. a

8. We cannot train it by counting: cannot crate an array that large.
   Suppose each variable has 100 distinct values, then we would need an array
   with 100^1000 elements, which is more memory than observable universe. 

   Which means there is not enough training data in the observable universe
   to populate that probability table, even if we somehow manage to create it.
   
   One possibility is to assume some independence. Instead of creating 
   an array such as a[100][100][100][100][100][100][100]...[100][100][100]
   we could create many arrays such as: 
   a[100][100], b[100][100], c[100][100], d[100][100], where pairs of variables 
   are assumed to be dependent, but the pairs are independent. e.g.:
   P(A,B,C,D,E,F,G,H) is assumed to be equal P(A,B)P(C,D)P(E,F)P(G,H)

9. We canot marginalize over the full distribution. Same problem as in q8 above.

   If we go with the assume-independence route (explained above), then we can
   easily marginalize over the pairs (or triplets, or quadruplets).

   we cannot marginalize out P(A) out of P(A,B,C,D,E,F,G,H), but we can 
   marginalize out P(A) from P(A,B)P(C,D)P(E,F)P(G,H).

10. 0.8

   Given: P(A) = 0.6, P(-A) = 0.4,  
    P(q8|A) = 0.8, P(-q8|A) = 0.2, 
    P(q8|-A) = 0.3, P(-q8|-A) = 0.7,
   Question: P(A|q8) = P(q8|A)P(A) / P(q8) = P(q8|A)P(A) / (P(q8|A)P(A) + P(q8|-A)P(-A)) 
     = (0.8*0.6) / (0.8*0.6 + 0.3*0.4) =  0.8

11. 0.2;   1 - 0.8

12. 0.84

   Given: P(A) = 0.6, P(-A) = 0.4,  
    P(q10|A) = 0.7, P(-q10|A) = 0.3, 
    P(q10|-A) = 0.2, P(-q10|-A) = 0.8,
   Question: P(A|q10) = P(q10|A)P(A) / P(q10) = P(q10|A)P(A) / (P(q10|A)P(A) + P(q10|-A)P(-A)) 
     = (0.7*0.6) / (0.7*0.6 + 0.2*0.4) = 0.84

13. Not enough information. We were not given P(q8,q10|A) nor P(q8,q10). 

   Given: P(A) = 0.6, P(-A) = 0.4,  
    P(q8|A) = 0.8, P(-q8|A) = 0.2, 
    P(q8|-A) = 0.3, P(-q8|-A) = 0.7,
    P(q10|A) = 0.7, P(-q10|A) = 0.3, 
    P(q10|-A) = 0.2, P(-q10|-A) = 0.8,
   Question: P(A|q8,q10) = P(q8,q10|A)P(A) / P(q8,q10) 
   
14. 0.93333

   Given: P(A) = 0.6, P(-A) = 0.4,  
    P(q8|A) = 0.8, P(-q8|A) = 0.2, 
    P(q8|-A) = 0.3, P(-q8|-A) = 0.7,
    P(q10|A) = 0.7, P(-q10|A) = 0.3, 
    P(q10|-A) = 0.2, P(-q10|-A) = 0.8,
   Question: P(A|q8,q10) = P(q8,q10|A)P(A) / P(q8,q10) 

  Solution 1: 
   Assume: P(A|q8,q10) = P(q8|A)P(q10|A)P(A) / P(q8)P(q10) 
     = P(q8|A)P(q10|A)P(A) / P(q8)P(q10)
     = P(q8|A)P(q10|A)P(A) / ( P(q8|A)P(q10|A)P(A) + P(q8|-A)P(q10|-A)P(-A) ) 
     = 0.8*0.7*0.6 / ( 0.8*0.7*0.6 + 0.3*0.2*0.4 ) = 0.93333
     
  Solution 2: 
   We know P(A|q8) = 0.8. We can turn this into: 
    P(A|q8,q10) = P(q10|A)P(A|q8) / ( P(q10|A)P(A|q8) + P(q10|-A)P(-A|q8)) 
       = (0.7*0.8) / ( 0.7*0.8 + 0.2*0.2 ) = 0.93333

  Solution 3: 
   We know P(A|q10) = 0.84, We can turn this into: 
    P(A|q8,q10) = P(q8|A)P(A|q10) / ( P(q8|A)P(A|q10) + P(q8|-A)P(-A|q10)) 
       = (0.8*0.84) / (0.8 * 0.84 + 0.3*0.16) = 0.93333

15. d
16. c
17. b
18. a
19. d
20. b