![Count Regression
• Hurdle Model:
• Hierarchical approach.
• Level 1:
Treat count data as a Bernoulli process with p being the probability
of incurring a defect and 1 − p the probability of zero defects.
• Level 2: In the case of positive defects, the positive defect counts
are modeled as a zero-truncated count process.
• Expected defect count, E[Yi ] =
p1 · eβ1
, Tool i = 1
pi · eβ1+βi , Tool i = 1
Count Data
Zero Defect
Defect Count > 0
Zero-truncated
Poisson counts
Zero-truncated
NB counts
1-p
p
10](https://image.slidesharecdn.com/caf8b3be-3591-45a1-be9b-f0be5e4ca5d4-161117220522/85/INFORMS_2016_sg12-60-320.jpg)
INFORMS_2016_sg12
- 1. Ranking Routes in Semi-conductor Wafer Fabs Shreya Gupta John J. Hasenbein November 16, 2016 Operations Research and Industrial Engineering The University of Texas at Austin
- 2. Route Description a What does a route look like? 1
- 3. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk Tool 1 Tool 2 Tool 3 1
- 4. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 1
- 5. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 6. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 7. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 8. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 9. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 10. Route Description a What does a route look like? T1 T2 T3 T1 T2 T9 T1 T2 Tk 1
- 11. Data Description a What does defect data look like? 2
- 12. Data Description a What does defect data look like? Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Series of steps = Routes 2
- 13. Data Description a What does defect data look like? Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Wafers Series of steps = Routes 2
- 14. Data Description a What does defect data look like? Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Wafers 2
- 15. Data Description a What does defect data look like? Defect 4 Defect 1 Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Wafers 2
- 16. Data Description a What does defect data look like? Defect 4 Defect 1 Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Wafers Zero Defects - 2, 3 2
- 17. Objective Rank routes using defect count data. • Routes are comprised of a series of tools inside the fab. T1 T2 T3 T1 T2 T9 T1 T2 Tk Defect Data Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 3
- 18. Objective Rank routes using defect count data. • Routes are comprised of a series of tools inside the fab. • Defect data represents the number of defects of each type for the various routes. T1 T2 T3 T1 T2 T9 T1 T2 Tk Defect Data Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 3
- 19. Purpose Ranking routes using defect count data. • Exploratory adjustments on the best route (new recipes, or parameters, are tested on the best routes) T1 T2 T3 T1 T2 T9 T1 T2 Tk Defect Data Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 4
- 20. Purpose Ranking routes using defect count data. • Exploratory adjustments on the best route (new recipes, or parameters, are tested on the best routes) • Potential use in scheduling T1 T2 T3 T1 T2 T9 T1 T2 Tk Defect Data Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 4
- 21. Challenges
- 22. Challenges text text March 1, 2016 - April 19, 2016 Data Summary • 2 months of data 5
- 23. Challenges text text text Defects 1, 2, 3, 4 Data Summary • 2 months of data • 4 defect types 5
- 24. Challenges Step Number of Tools Step1 5 Step2 14 Step3 5 Step4 14 Step5 11 Step6 5 Step7 11 Step8 9 Step9 4 Step10 10 Step11 13 Posible Routes 13,873,860,000 Real Route 652 Data Summary • 2 months of data • 4 defect types • 11 steps 5
- 25. Challenges Step Number of Tools Step1 5 Step2 14 Step3 5 Step4 14 Step5 11 Step6 5 Step7 11 Step8 9 Step9 4 Step10 10 Step11 13 Posible Routes 13,873,860,000 Real Route 652 Data Summary • 2 months of data • 4 defect types • 11 steps • ≈ 14 billion possible routes 5
- 26. Challenges Step Number of Tools Step1 5 Step2 14 Step3 5 Step4 14 Step5 11 Step6 5 Step7 11 Step8 9 Step9 4 Step10 10 Step11 13 Posible Routes 13,873,860,000 Real Route 652 Data Summary • 2 months of data • 4 defect types • 11 steps • ≈ 14 billion possible routes • 652 routes represented 5
- 27. Challenges Data Summary • 2 months of data • 4 defect types • 11 steps • ≈ 14 billion possible routes • 652 routes represented • More than 85% zero defects 5
- 28. Challenges Objective: Build a statistical robust heuristic that can efficiently ranks ≈ 14 Billion routes. Data Summary • 2 months of data • 4 defect types • 11 steps • ≈ 14 billion possible routes • 652 routes represented • More than 85% zero defects 5
- 29. Solution
- 30. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Wafers Series of steps = Routes 6
- 31. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Counts / Positive Numbers / Positive Integers Defect 4 Defect 1 Wafers Zero Defects - 2, 3 6
- 32. Model Counts / Positive Numbers / Positive Integers Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 0 4 53 2 59 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 19 2 0 0 21 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 6
- 33. Model 14 billion routes Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 6
- 34. Model Is there a better way? Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . 6
- 35. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 6
- 36. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 6
- 37. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 6
- 38. Model Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 6
- 39. Count Regression • n: Number of tools Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 7
- 40. Count Regression • n: Number of tools • Xi : Dummy variable for ith tool, Xi = 1, Tool i 0, otherwise Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 7
- 41. Count Regression • n: Number of tools • Xi : Dummy variable for ith tool, Xi = 1, Tool i 0, otherwise • Yi : Expected number of defects incurred by the ith tool Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 7
- 42. Count Regression • n: Number of tools • Xi : Dummy variable for ith tool, Xi = 1, Tool i 0, otherwise • Yi : Expected number of defects incurred by the ith tool • log(Yi ) = β1 + n i=2 βi Xi Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 7
- 43. Count Regression • n: Number of tools • Xi : Dummy variable for ith tool, Xi = 1, Tool i 0, otherwise • Yi : Expected number of defects incurred by the ith tool • log(Yi ) = β1 + n i=2 βi Xi • Yi = eβ1 , Tool i = 1 eβ1+βi , Tool i = 1 Step 1 Step 2 … Step N Route Defect 1 Defect 2 Defect 3 Defect 4 Total Defects T1,1 T2,1 … TN,3 route 1 2 0 0 2 4 T1,2 T2,4 … TN,3 route 2 0 0 0 0 0 T1,3 T2,1 TN,7 route 3 0 4 53 2 59 . . . . . . . . . . . . . . . . . . . . … . . . . . . . Count Regression 7
- 44. Count Regression Our Approach We begin modeling the defect count data set from the most basic and proceed forward until we find the model that best fits our data. 8
- 45. Count Regression • Poisson Regression: 9
- 46. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. 9
- 47. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ 9
- 48. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. 9
- 49. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: 9
- 50. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ 9
- 51. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ • May not fix overdispersion. 9
- 52. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ • May not fix overdispersion. • Negative Binomial Regression: 9
- 53. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ • May not fix overdispersion. • Negative Binomial Regression: • If overdispersion is due to excess zeros. 9
- 54. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ • May not fix overdispersion. • Negative Binomial Regression: • If overdispersion is due to excess zeros. • Negative Binomial accounts for excess zeros well. 9
- 55. Count Regression • Poisson Regression: • Distribution of count data is assumed to be Poisson. • σ2 = µ • However, it maybe Poisson overdispersed if σ2 > µ. • Quasipoisson Regression: • Assume σ2 > φ · µ • May not fix overdispersion. • Negative Binomial Regression: • If overdispersion is due to excess zeros. • Negative Binomial accounts for excess zeros well. • Negative Binomial overdispersion or a bad fit may occur due to excess zeros beyond what the NB fit can account for. 9
- 56. Count Regression • Hurdle Model: 10
- 57. Count Regression • Hurdle Model: • Hierarchical approach. Count Data Zero Defect Defect Count > 0 Zero-truncated Poisson counts Zero-truncated NB counts 10
- 58. Count Regression • Hurdle Model: • Hierarchical approach. • Level 1: Treat count data as a Bernoulli process with p being the probability of incurring a defect and 1 − p the probability of zero defects. Count Data Zero Defect Defect Count > 0 Zero-truncated Poisson counts Zero-truncated NB counts 1-p p 10
- 59. Count Regression • Hurdle Model: • Hierarchical approach. • Level 1: Treat count data as a Bernoulli process with p being the probability of incurring a defect and 1 − p the probability of zero defects. • Level 2: In the case of positive defects, the positive defect counts are modeled as a zero-truncated count process. Count Data Zero Defect Defect Count > 0 Zero-truncated Poisson counts Zero-truncated NB counts 1-p p 10
- 60. Count Regression • Hurdle Model: • Hierarchical approach. • Level 1: Treat count data as a Bernoulli process with p being the probability of incurring a defect and 1 − p the probability of zero defects. • Level 2: In the case of positive defects, the positive defect counts are modeled as a zero-truncated count process. • Expected defect count, E[Yi ] = p1 · eβ1 , Tool i = 1 pi · eβ1+βi , Tool i = 1 Count Data Zero Defect Defect Count > 0 Zero-truncated Poisson counts Zero-truncated NB counts 1-p p 10
- 61. Count Regression Models Snapshot: Best Count Model Fit Defect Step Regression Type P-value Dispersion AIC Best Fit def1 Step1 Poisson 0.00 3.73 6577.11 No def1 Step1 Quasipoisson 0.00 3.73 1e+07 No def1 Step1 Negative Binomial 0.02 1.09 5029.06 No def1 Step1 Hurdle - Poisson NA NA 6312.13 No def1 Step1 Hurdle - Negative Binomial NA NA 5012.51 Yes def2 Step3 Poisson 0.00 2.52 6525.1 No def2 Step3 Quasipoisson 0.00 2.52 1e+07 No def2 Step3 Negative Binomial 1.00 0.77 5026.4 No* def2 Step3 Hurdle - Poisson NA NA 6293.64 No def2 Step3 Hurdle - Negative Binomial NA NA 5017.27 Yes * This model was not considered a best fit in spite of having a significant p-value > α(= 0.5) because the dispersion was ≈ 1.25. Also, another model (Hurdle - Negative Binomial) yielded a lower AIC statistic. NA: Could not be extracted using R or model does not have this statistic 11
- 63. Count Regression Algorithm Count Regression Procedure text Poisson Reg. Good Fit Yes Quasipoisson Reg. No Negative Binomial Reg. Yes Hurdle model (Poisson)Hurdle model (Neg Bin) Extract Coefficients Generate average defect rates for all equipment under this defect-step pair Good Fit No Good Fit No Data Set Proceed to Ranking AIC Statistic Choose model with smallest AIC statistic 12
- 64. Ranking Algorithm Snapshot: Defect-1 Tool Ranks Defect Step Tools (i) Model Yi Rank def1 Step1 EQP 31 Hurdle - NB 23.21 5 def1 Step1 EQP 32 Hurdle - NB 3.42 3 def1 Step1 EQP 35 Hurdle - NB 2.71 1 def1 Step1 EQP 36 Hurdle - NB 3.03 2 def1 Step11 EQP 60 Hurdle - NB 2.80 4 def1 Step11 EQP 61 Hurdle - NB 2.87 5 def1 Step11 EQP 62 Hurdle - NB 3.59 11 def1 Step11 EQP 50 Hurdle - NB 23.96 12 Step 1: Tool Ranks Assign ranks from 1 to n to the tools under this step. - Highest rank, 1, for the tool generating the smallest number of defects. - Lowest rank, n, for the tool generating the largest number of defects. 13
- 65. Ranking Algorithm Snapshot: Defect-3 Specific Route Ranks Step 1 Tool Rank Step 2 Tool Rank Defect Specific Route Score Defect Specific Route Rank EQP 31 5 EQP 57 6 11 2 EQP 32 1 EQP 58 1 2 1 EQP 35 6 EQP 59 7 13 3 EQP 36 4 EQP 60 9 13 3 Step 2: Defect Specific Ranks - For a particular defect generate the route score of the, say R, routes by summing up the ranks of the tools falling under the steps of that route. - Rank these scores from 1 to R with 1 corresponding to the smallest score and R to the largest. 14
- 66. Ranking Algorithm Snapshot: Global Route Ranks Route Defect Specific Route Ranks Route Statistics Step1 Step2 Step3 Defect 1 (w1 = 1) Defect 2 (w2 = 1) Defect 3 (w3 = 1) Defect 4 (w4 = 1) Weighted Global Score Global Rank EQP 35 EQP 16 EQP 49 8 5 24 18 55 4 EQP 38 EQP 16 EQP 48 6 10 19 17 52 2 EQP 32 EQP 10 EQP 48 14 7 8 13 42 1 EQP 31 EQP 16 EQP 49 12 6 21 15 54 3 Step 3: Global Route Ranks - Generate the global score for each route by taking the weighted sum of their defect specific scores, weighted by the importance of the defect. - Rank these scores from 1 to R with 1 corresponding to the smallest score and R to the largest. 15
- 67. Conclusions
- 68. Future Work 1. The methodology could be incorporated into scheduling algorithms. 16
- 69. Future Work 1. The methodology could be incorporated into scheduling algorithms. 2. Statistically significant differences between routes may be evaluated. 16
- 70. Future Work 1. The methodology could be incorporated into scheduling algorithms. 2. Statistically significant differences between routes may be evaluated. 3. Validation against out-of-sample data. 16
- 71. Additional Work Score-based Ranking For output data like yield (greater the better), we have developed a ranking technique using ANCOVA and Tukey HSD pair-wise difference techniques that rank routes based on the significant differences between their output levels. 17
- 72. Additional Work (ii) Target-based Ranking For output metrics like thickness, which have upper and lower specifications bounding the target to be achieved, we have designed ranking techniques that rank routes based on the accuracy and precision of their output. 18
- 73. Questions 19
- 74. Thank You
- 75. Appendix
- 76. Appendix 1 Akaike Information Criteria (AIC) • the AIC statistic is used to compare models that do not generate a p-value (i.e., in our algorithm we use it to compare hurdle models with Poisson and NB-2 count distributions,as well as to compare these hurdle models to all the other models.) 20
- 77. Appendix 1 Akaike Information Criteria (AIC) • the AIC statistic is used to compare models that do not generate a p-value (i.e., in our algorithm we use it to compare hurdle models with Poisson and NB-2 count distributions,as well as to compare these hurdle models to all the other models.) • AIC = −2l(θ) + 2s, where: s is the number of model parameters, and θ is a vector representing the MLE parameter estimates that maximize the log-likelihood, l(θ), of the obtaining the data with the distribution (model) under consideration. 20
- 78. Appendix 1 Akaike Information Criteria (AIC) • the AIC statistic is used to compare models that do not generate a p-value (i.e., in our algorithm we use it to compare hurdle models with Poisson and NB-2 count distributions,as well as to compare these hurdle models to all the other models.) • AIC = −2l(θ) + 2s, where: s is the number of model parameters, and θ is a vector representing the MLE parameter estimates that maximize the log-likelihood, l(θ), of the obtaining the data with the distribution (model) under consideration. • Thus, AIC is a conservative statistic for measuring the model fit, as quantified by l(θ), and model complexity, as quantified by s. 20
- 79. Appendix 1 Akaike Information Criteria (AIC) • the AIC statistic is used to compare models that do not generate a p-value (i.e., in our algorithm we use it to compare hurdle models with Poisson and NB-2 count distributions,as well as to compare these hurdle models to all the other models.) • AIC = −2l(θ) + 2s, where: s is the number of model parameters, and θ is a vector representing the MLE parameter estimates that maximize the log-likelihood, l(θ), of the obtaining the data with the distribution (model) under consideration. • Thus, AIC is a conservative statistic for measuring the model fit, as quantified by l(θ), and model complexity, as quantified by s. • The quasipoisson model does not generate the AIC statistic because it is not derived using Maximum Likelihood Estimation (MLE). In stead we have QAIC = −2 l(θ) φ + 2s 20
- 80. Appendix 2 Bayesian Information Criterion (BIC) • BIC is analogous to AIC except 2s is replaced with s log n. BIC imposes a stronger penalty on model complexity than AIC for n ≥ 8,i.e., when sample size is large (Zheng and Loh 1995). 21
- 81. Appendix 2 Bayesian Information Criterion (BIC) • BIC is analogous to AIC except 2s is replaced with s log n. BIC imposes a stronger penalty on model complexity than AIC for n ≥ 8,i.e., when sample size is large (Zheng and Loh 1995). • So, BIC = −2l(θ) + s log n 21
- 82. Appendix 2 Bayesian Information Criterion (BIC) • BIC is analogous to AIC except 2s is replaced with s log n. BIC imposes a stronger penalty on model complexity than AIC for n ≥ 8,i.e., when sample size is large (Zheng and Loh 1995). • So, BIC = −2l(θ) + s log n • See Burnham and Anderson 2002 for all variants of AIC. 21
- 83. Appendix 3 Quasipoisson Fit • If Pearson chi-square statistic given by (??) below: Pχ2 = n i=1 (yi − µi )2 (νi ) , (where n is the sample size) is not approximately distributed χ2 n−p, where p is the number of estimated parameters, then the statistic provides evidence of lack of fit. 22
- 84. Appendix 3 Quasipoisson Fit • If Pearson chi-square statistic given by (??) below: Pχ2 = n i=1 (yi − µi )2 (νi ) , (where n is the sample size) is not approximately distributed χ2 n−p, where p is the number of estimated parameters, then the statistic provides evidence of lack of fit. • A convenient adjustment is to assume var(Yi ) = kνi . 22
- 85. Appendix 3 Quasipoisson Fit • If Pearson chi-square statistic given by (??) below: Pχ2 = n i=1 (yi − µi )2 (νi ) , (where n is the sample size) is not approximately distributed χ2 n−p, where p is the number of estimated parameters, then the statistic provides evidence of lack of fit. • A convenient adjustment is to assume var(Yi ) = kνi . • P∗ χ2 = n i=1 (yi − µi )2 k(νi ) , = Pχ2 k , ⇒ k = φ = Pχ2 n − p . 22
- 86. Appendix 3 Quasipoisson Fit • The exponential family f (y; ψ, φ) (where ψ is the mean) may no longer integrate to unity and is should be simply considered a useful modification of the likelihood function. 23
- 87. Appendix 3 Quasipoisson Fit • The exponential family f (y; ψ, φ) (where ψ is the mean) may no longer integrate to unity and is should be simply considered a useful modification of the likelihood function. • Only the variance changes with an adjustment factor of k estimated by φ and this is accounted for by: ∂l (β; y) ∂βj = 0) j = 1, . . . , p; ⇒ 1 φ n i=1 ∂µi ∂βj (yi − µi ) νi = 1 φ ∂l(β; y) ∂βj j = 1, . . . , p. 23
- 88. Appendix 3 Quasipoisson Fit • The exponential family f (y; ψ, φ) (where ψ is the mean) may no longer integrate to unity and is should be simply considered a useful modification of the likelihood function. • Only the variance changes with an adjustment factor of k estimated by φ and this is accounted for by: ∂l (β; y) ∂βj = 0) j = 1, . . . , p; ⇒ 1 φ n i=1 ∂µi ∂βj (yi − µi ) νi = 1 φ ∂l(β; y) ∂βj j = 1, . . . , p. • Thus, the MLE estimates remain unchanged. 23