# HLM by xiuliliaofz

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```									                             Running HLM using Proc Mixed
Lan Xue 1

This work focuses on fitting hierarchical linear models with PROC MIXED (SAS). It is
based on the SAS help menu and a paper by Sawako Suzuki and Ching-Fang Sheu (see
reference). It contains three parts: the first part discusses the notations used in PROC
MIXED, especially the understandings of RANDOM and SUBJECT options in PROC
MIXED; the second part discusses several examples illustrated in Sawako Suzuki and
Ching-Fang Sheu, we further discuss why those SAS code are used; the third part
summaries some helpful simple steps to write the SAS code for a HLM.

Part 1: Notations used in SAS proc Mixed.

Y  X  Z  e

Here,

 is an unknown vector of fixed-effects parameters with known design matrix X,
 is an unknown vector of random-effects parameters with known design matrix Z,
is an unknown random error vector .

In Proc Mixed,
(1) Model part defines the fixed effects, so the design matrix X will be determined.
(2) Random part defines the random effects, so Z will be determined.
(3) Subject gives the classifying (Blocking) variable. We will see the difference
between with Subject command and without Subject command in the next
example.

Example: The data consists of the test scores and IQ score of the students from 3
classes. There are 10 students in each class. Our total samples size is 30.
Code1, With subject command:
proc mixed data=rc;
class class;
model Score = IQ / s;
random intercept IQ / type=un sub=Class s;
run;
Code2, without subject command:
proc mixed data=rc;
class class;
model Score = IQ / s;
random intercept IQ / type=un s;

1
The author was supported by the assistantship from Statistics Consulting Service and under the
supervision of Prof. Connie Page.
run;

Since the two codes have the same model statement, they will have the same design
matrix for the fixed effects: X which is a 30  2 matrix with 1 and the IQ score for i-th
student as its i-th row. But the Z matrix will be different.

Z matrix for the first code, Z1 is a 30  6 matrix defined as follows:

 1 q1,1     0    0      0    0    
                                  
.    .      0    0      0    0    
1 q         0    0      0    0    
    1,10

0 0         1   q 2,1   0    0    
                             0 
Z1 =  0 0        .    .      0         
0 0         1 q 2,10    0    0 
                                  
0 0         0   0       1   q3,1 
0 0         0   0       .     . 

0 0                               
            0    0      1   q3,10 


But the Z matrix for the second code, Z2 is a 30  2 matrix same as X.

1 q1,1 
         
1    . 
1 q 
    1,10

1 q 2,1 
     . 
Z2= 1        
1 q 2,10 
         
1 q3,1 
1    . 

1 q     
    3,10 

Part 2: SAS codes for HLM.

Level                             Name                         Factor and Covariate
1 ----i                           Student                      Pretest Score
2 ----j                           Class                        NA
3 ----k                           School                        Condition (4 levels)

Example1:

HLM: we consider 3-level HLM with only random coefficient for interceptor.
y ijk =  0 jk +  1 * PS +  i ( j ( k ))
Student level, coefficient of PS is fixed, but coefficient of interceptor is allowed to be
varying across the classes (School).
 0 jk =  0 k +  j (k )
Class level, no predictor available at this level and the coefficient of interceptor is
allowed to be varying across School.

 0k =  0 +          2 * CC   3 * TV   4 * CCTV +  k

School level, with factor: Condition (4 levels).

So, put those three level models together, we have,

y ijk =  0 +  1 * PS +                2 * CC   3 * TV   4 * CCTV +  k +  j (k ) +  i ( j ( k ))

Here the fixed coefficients are:  0 ,  1 ,  2 ,  3 ,  4
the random are:  k ,  j (k )
As what is given in the paper.

SAS code:

In the above example, we have fixed effects: Intercept, PS, CC, TV, CCTV. So the
determined corresponding design matrix for fixed effects X is a N  5 matrix, with (1, i-th
student’s pretest score, arrangement of the conditions on the student) as its i-th row.
Then our model statement in Proc Mixed, should be:
Model Y = PS CC TV CCTV;

Secondly, in the above example, we have K+J*K random effects:
 k k 1,...,K  j ( k ) j 1,..., J
k 1,... K

So, if we put them together in the vector form, define
I k be the indicator variable for k-th school. k=1,…K.
I j{k ) be the indicator variable for j(k) class. j=1,…J, k=1,…,K
                                                                           K
Y =  0 * I +  1 * PS +  2 * CC   3 * TV   4 * CCTV +                    
k 1
k   * Ik +        j {k )   * I j (k ) + E
j ,k
Then we can see that,
Z= ( I 1 , I 2 ,…., I k ,I 1(1) ,…, I J ( K ) )
r= (  1 , …,  k ,  1(1) , ….,  J ( K ) )

So to generate the indicators I k k=1,…K, the random statement with Sub= School is
needed.
Random Intercept / Sub = School ;
Another option is:

Random School ;

Since School is defined in Class statement, the indicator variables at each level of school
will be automatically generated. To generate the indicators I j{k ) , the random statement
with Sub= Class (School) needed.

Random Intercept / Sub= Class(School);

Similarly, another option is:

Random Class(School);

Finally, our SAS code for this 3 level HLM with only random coefficient for interceptor
is:

proc Mixed;
class class school condition;
model Y= PS CC TV CCTV ;
random Intercept / Sub = School;
random Intercept / Sub = Class(School);
run;

Or
proc Mixed;
class class school condition;
model Y= PS CC TV CCTV ;
random School;
random Class(School);
run;

Example 2:
The same data as above, if we allow the coefficient of Pretest Score (PS) is random too.
Consider the following 3 level HLM:

y ijk =  0 jk +  1 jk * PS +  i ( j ( k ))

 0 jk =  0 k +  j (k )
 1 jk =  1k +  j (k )

 0 k =  0 +  02 * CC   03 * TV   04 * CCTV +  k
 1k =  1 +  k
After putting them together, we have
y ijk =  0 +  1 *PS +  02 * CC   03 * TV   04 * CCTV                       +  k +  j (k ) +  j (k ) *PS +
 k *PS +  i ( j ( k ))

Here the fixed coefficients are:  0 ,  1 ,  02 ,  03 ,  04
The random are :  k ,  k ,  j (k ) ,  j (k )
The design matrix for fix effects: X should be a N  5 matrix as,
X= (1, PS, CC, TV,CCTV)
And the design matrix for random effects Z of the form:
Z= ( I 1 , I 1 *PS, I 2 ,I 2 *PS,…., I k ,I k *PS, I 1(1) , I 1(1) *PS,…, I J ( K ) ,I J ( K ) *PS)
r= (  1 ,  1 ,…,  k , ,  k ,  1(1) ,  1(1) ,….,  J ( K ) ,  J (k ) )

So the SAS code should be:

Proc Mixed;
class class school condition;
model Y= PS CC TV CCTV ;
random Intercept PS / Sub = School;
random Intercept PS/ Sub = Class(School);
run;

Part3: Discussion
Because of the nested structure of the HLM, in the random statement, the subject
command is very useful to generate the needed design matrix. There are some simple
rules we can apply to use subject command for the HLM.
(1). Write the models for different level together

(2). Put the random effects in the same level together, for example

y ijk =  0 +  1 *PS +  02 * CC   03 * TV   04 * CCTV
+  k +  k *PS           random effects at school level
+  j (k ) +  j (k ) *PS random effects at class level
+  i ( j ( k ))

(3). Then use the Random statement for each level separately, using subject equals to the
corresponding level and the effects are the effects exist in that level.

random Intercept PS/ subject = school;
random Intercept PS/ subject = class;
References