Estimation
Pr  X ni
The Model
exp  x 
 x ; ,   
ni
ni
n
n
  i 
1  exp  n   i 
i
Pr  X ni  1; n ,  i  
exp  n   i 
1  exp  n   i 
Probability
1
0.8
0.6
0.4
0.2
0
-4
-3
-2
-1
0
n  i
1
2
3
4
The Model for N Items — 1
Let X n   X n1 , X n 2 ,
, X nI 
T
denote a response vector for person n
I
Pr  X n  x n ; n , δ    Pr  X ni  xni ; n ,  i 
i 1
I

i 1
exp  xni  n   i 
1  exp  n   i 
The vector probability takes this form if we assume
independence
The Model for N Items — 2
I
Pr  X n  x n ; n , δ   
exp  xni  n   i 
i 1
I
rn   xni
i 1
the raw score of person n
1  exp  n   i 
 I

exp   xni  n   i   

 I  i 1
 1  exp  n   i 
i 1
I


exp  rn n   xni i  
i 1



 I
 1  exp  n   i 
i 1
Probabilities of scoring 2 for different response
patterns
0.15
All modes at same
location
0.13
Mode is ability that
makes this pattern
most likely
Probability
0.11
0.09
0.07
Likelihood principle: Which ability
maximises the probability of what was
obtained?
0.05
0.03
0.01
-0.01
n
Graphical Display of the likelihoods for a five item
test
0.06
Probability
0.05
0.04
Pr  Rn  0;n , δ 
Pr  Rn  2; n , δ 
Pr  Rn  3; n , δ 
Pr  Rn  4; n , δ 
Pr  Rn  1;n , δ 
0.03
0.02
0.01
0
All items with difficulty parameter zero
Pr  Rn  5;n , δ 
n
Estimation Methods
•
Approximate Methods
– PROX
– UFORM
•
•
•
Pair-wise
Minimum Chi-square
Maximum likelihood
Maximum Likelihood Methods–1
• Joint maximum likelihood
–
–
–
also called unconditional maximum likelihood
(UCON)
method used in Quest, WinSteps, Facets,
ConQuest, TAM
Ability and difficulty estimates
Maximum Likelihood Methods–2
•
Marginal Maximum Likelihood
– really a new model that invokes a population
distribution assumption
– Works well with more general models
– ConQuest, TAM, default method
Joint Maximum Likelihood
N
L   X    Pr  X n  x n ; n , δ 
n 1
N
I
 
exp  xni  n   i 
n 1 i 1
1  exp  n   i 
N I

exp   xni  n   i   

 N  nI 1 i 1
 1  exp  n   i 
n 1 i 1

I
 1  exp 
n 1 i 1
rn   xni score of person n
i 1
N
I
 N

exp   rn n   si i  
i 1

 n 1
N
I
n
  i 
si   xni score of item i
n 1
Maximising the Joint Likelihood — 1
   X   log L   X 
I

 N
 
 exp   rn n   si i   

i 1
 n 1
 
 log  N I

  1  exp  n   i  
 n 1 i 1



I
 N

   rn n   si i  
i 1

 n 1
N
I
  log 1  exp  n   i 
n 1 i 1
Maximising the Joint Likelihood — 2
   X 
 t


 t
I
 N

  rn n   si i  
i 1

 n 1
 N I


log
1

exp



 n i 
 
 t  n 1 i 1

I
 rt  
i 1
0
exp  t   i 
1  exp  t   i 
Maximising the Joint Likelihood — 3
   X 
 u


 u
I
 N

  rn n   si i  
i 1

 n 1
 N I


 log 1  exp  n   i 
 u  n 1 i 1

N
exp  n   u 
 su  
n 1
1  exp  n   u 
0
A total of N+I equations to be solved
At the solution — 1
First derivatives are zero (called scores)
What the student
scored
I
rt  
i 1
exp  t   i 
1  exp  t   i 
N
su  
n 1
exp  n   u 
1  exp  n   u 
What the model
predicts
It happens to be the
expected score
We only need the marginals!
Student 1
Student 2
Student 3
…
Item Score
Item 1 Item 2 Item 3
0
1
1
1 We do not1 need to use1 this
0
0
1
…
..
…
56
49
89
…
…
…
…
…
…
Student Score
23
34
15
…
• Many sets of patterns will satisfy a given set of marginals
• Estimates, errors, reliability do not depend on the
patterns
• Parameter estimates do not depend upon fit
• Implications for the order debate
Further Implications
– Student and item scores are sufficient statistics for
Rasch estimation.
– Students with the same score will have the same
ability estimate.
– One-to-one match between raw score and Rasch
ability estimate (when no missing data).
– Use of score equivalence table.
– So why (and when) do we need Rasch scores?
At the solution: What must the distribution of
estimates look like?
First derivatives are zero (called scores)
I
rt  
i 1
exp  t   i 
1  exp  t   i 
N
su  
n 1
exp  n   u 
1  exp  n   u 
The Resulting Ability Distribution
Score 3
Score 4
Score 2
Score 5
Score 1
Score 0
Score 6
Proficiency on Logit Scale