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Re: mixed model residuals computation and plotting (Revised)

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Re: mixed model residuals computation and plotting (Revised)

Maguin, Eugene

Thanks, Ryan.

 

It looks like Carol (second message) was asking about a single random term model, maybe an intercept only model.

Let’s take a specific example.

 

mixed y with xc w z/fixed=xc w z xc*w xc*z/

   method=ml/print solution/random intercept xc | subjectid(clus) covtype(un)/

   save=fixpred(fixpred) pred(pred) resid(resid). (probably not syntactically correct but will facilitate communication).

 

I write the composite equation as

y = (g00 + g01*w + g02*z + u0) + xc*(g10 + g11*w + g12*z + u1) + e

 

I did some computations on the save command terms.

I found that variance of (y-pred) equaled the variance of resid, which I figure must be ‘e’ in the equation.

If that’s true then I figure that variance of (y-fixpred) would seem to be the variance of ((u0+xc*u1)+e).

However, I notice that in the your reply to Tino (March 19) you said

COMPUTE residual_level_2 = y - FXPRED_1.

Or, (u0+xc*u1).

 

The last difference number is (pred-fixpred). What would this variance be?

 

If resid is ‘e’, and either (y-fixpred) or (pred-fixpred) is (u0+xc*u1), is there any way to extract/compute u0 and u1?

 

Gene Maguin

 

 

 

 

 

 

From: GMAIL [mailto:[hidden email]]
Sent: Tuesday, June 17, 2014 5:38 PM
To: Maguin, Eugene
Cc: [hidden email]
Subject: Re: mixed model residuals computation and plotting

 



 

On Jun 17, 2014, at 5:17 PM, "Maguin, Eugene" <[hidden email]> wrote:

I know that mixed provides fixed predicted values,  predicted (fitted = fixed+random) values  and residuals (observed-predicted). Is it possible to compute the level 2 residuals from these predicted and residuals data? How would it be done computationally? Singer and Willet mention this in their book and may have offered directions for computing these on their supplementary website but that seems to be gone. UCLA/ats-idre has examples but not those details, so far as I can tell.

Thanks, Gene Maguin
Ryan
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Re: mixed model residuals computation and plotting (Revised)

Ryan
Gene,

You pose an interesting question which I had not thought of previously. I *think* one way to obtain U0J and U1J requires the use of the distributive property of multiplication on the RANDOM statement. 

Suppose we have the following random coefficient model:

Level 1 Equation:

y = B0J + B1J*time + eij

Level 2 Equations:

B0J = Gamma00 + u0j
B1J = Gamma10 + u1j

Full Equation:

y = Gamma00
    + Gamma10*time
    + ( u1j*time + u0j + eij )

Assuming the random components are uncorrelated, the standard MIXED model code would be:

MIXED y WITH time
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM intercept time | subject(subject_ID).

However, using the distributive property of multiplication, the following model is mathematically equivalent to:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time.

Using the second parameterization of the model, I *think* one can take advantage of the TEST sub-command to estimate B0J, U0J, B1J, and U1J.

Assuming there are a total of N=10 subjects and the time variable is a whole number that starts at 0 and increases by one unit as 0,1,2,3,... etc., then the code to obtain the parameter estimates of interest for the first subject would be:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time
  /TEST 'B01' intercept 1 | subject_ID 1 0 0 0 0 0 0 0 0 0 
  /TEST 'U01' | subject_ID 1 0 0 0 0 0 0 0 0 0
  /TEST 'B11' time 1 | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /TEST 'U11' | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /SAVE=PRED.

To test this out, make sure that the following is true:

B01 = Gamma00 + U01
B11 = Gamma10 + U11

where,

Gamma00 = fixed intercept
Gamma10 = fixed slope

If the above is correct (which it should be), then make sure the following is also true (for subject 1),

PRED_1 = B01 + B11*("value of "time")

HTH,

Ryan


On Wed, Jun 18, 2014 at 5:07 PM, Maguin, Eugene <[hidden email]> wrote:

Thanks, Ryan.

 

It looks like Carol (second message) was asking about a single random term model, maybe an intercept only model.

Let’s take a specific example.

 

mixed y with xc w z/fixed=xc w z xc*w xc*z/

   method=ml/print solution/random intercept xc | subjectid(clus) covtype(un)/

   save=fixpred(fixpred) pred(pred) resid(resid). (probably not syntactically correct but will facilitate communication).

 

I write the composite equation as

y = (g00 + g01*w + g02*z + u0) + xc*(g10 + g11*w + g12*z + u1) + e

 

I did some computations on the save command terms.

I found that variance of (y-pred) equaled the variance of resid, which I figure must be ‘e’ in the equation.

If that’s true then I figure that variance of (y-fixpred) would seem to be the variance of ((u0+xc*u1)+e).

However, I notice that in the your reply to Tino (March 19) you said

COMPUTE residual_level_2 = y - FXPRED_1.

Or, (u0+xc*u1).

 

The last difference number is (pred-fixpred). What would this variance be?

 

If resid is ‘e’, and either (y-fixpred) or (pred-fixpred) is (u0+xc*u1), is there any way to extract/compute u0 and u1?

 

Gene Maguin

 

 

 

 

 

 

From: GMAIL [mailto:[hidden email]]
Sent: Tuesday, June 17, 2014 5:38 PM
To: Maguin, Eugene
Cc: [hidden email]
Subject: Re: mixed model residuals computation and plotting

 



 

On Jun 17, 2014, at 5:17 PM, "Maguin, Eugene" <[hidden email]> wrote:

I know that mixed provides fixed predicted values,  predicted (fitted = fixed+random) values  and residuals (observed-predicted). Is it possible to compute the level 2 residuals from these predicted and residuals data? How would it be done computationally? Singer and Willet mention this in their book and may have offered directions for computing these on their supplementary website but that seems to be gone. UCLA/ats-idre has examples but not those details, so far as I can tell.

Thanks, Gene Maguin


===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD
Ryan
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Re: mixed model residuals computation and plotting (Revised)

Ryan
All,

Below is a simulated example verifying what I stated in the previous post.

Best,

Ryan
--

/*Generate Data*/.
/*seed for random  generator*/.
set seed 987879546.
 
new file.
input program.
compute subject_ID = -99.
compute Gamma00 = -99.
compute Gamma10 = -99.
compute u0j = -99.
compute u1j = -99.
compute B0J = -99.
compute B1J = -99.
compute eij = -99.
compute time = -99.
leave subject_ID to time.
 
    /*100 subjects*/.
    loop subject_ID = 1 to 100.
    /*fixed intercept*/.
    compute Gamma00 = 0.50. 
    /*fixed slope*/.
    compute Gamma10 = 0.30. 
    /*random intercept error term*/.
    compute u0j = sqrt(.30)*rv.normal(0,1).
    /*random slope error term*/.
    compute u1j = sqrt(.30)*rv.normal(0,1).
    /*random intercept term*/.
    compute B0J = Gamma00 + u0j.
    /*random slope term*/.
    compute B1J = Gamma10 + u1j.
    /*30 time points*/.
    loop time = 0 to 29.
    /*error term*/.
    compute eij = rv.normal(0,1).
    /*full equation*/.
    compute y = B0J + B1J*time + eij.
    end case.
   end loop.
  end loop.
 end file.
end input program.
execute.

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time
  /TEST 'B01' intercept 1 | subject_ID 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'U01' | subject_ID 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                           0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'B11' time 1 | subject_ID*time 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'U11' | subject_ID*time 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /SAVE=PRED.


On Fri, Jun 20, 2014 at 11:32 PM, Ryan Black <[hidden email]> wrote:
Gene,

You pose an interesting question which I had not thought of previously. I *think* one way to obtain U0J and U1J requires the use of the distributive property of multiplication on the RANDOM statement. 

Suppose we have the following random coefficient model:

Level 1 Equation:

y = B0J + B1J*time + eij

Level 2 Equations:

B0J = Gamma00 + u0j
B1J = Gamma10 + u1j

Full Equation:

y = Gamma00
    + Gamma10*time
    + ( u1j*time + u0j + eij )

Assuming the random components are uncorrelated, the standard MIXED model code would be:

MIXED y WITH time
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM intercept time | subject(subject_ID).

However, using the distributive property of multiplication, the following model is mathematically equivalent to:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time.

Using the second parameterization of the model, I *think* one can take advantage of the TEST sub-command to estimate B0J, U0J, B1J, and U1J.

Assuming there are a total of N=10 subjects and the time variable is a whole number that starts at 0 and increases by one unit as 0,1,2,3,... etc., then the code to obtain the parameter estimates of interest for the first subject would be:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time
  /TEST 'B01' intercept 1 | subject_ID 1 0 0 0 0 0 0 0 0 0 
  /TEST 'U01' | subject_ID 1 0 0 0 0 0 0 0 0 0
  /TEST 'B11' time 1 | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /TEST 'U11' | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /SAVE=PRED.

To test this out, make sure that the following is true:

B01 = Gamma00 + U01
B11 = Gamma10 + U11

where,

Gamma00 = fixed intercept
Gamma10 = fixed slope

If the above is correct (which it should be), then make sure the following is also true (for subject 1),

PRED_1 = B01 + B11*("value of "time")

HTH,

Ryan


On Wed, Jun 18, 2014 at 5:07 PM, Maguin, Eugene <[hidden email]> wrote:

Thanks, Ryan.

 

It looks like Carol (second message) was asking about a single random term model, maybe an intercept only model.

Let’s take a specific example.

 

mixed y with xc w z/fixed=xc w z xc*w xc*z/

   method=ml/print solution/random intercept xc | subjectid(clus) covtype(un)/

   save=fixpred(fixpred) pred(pred) resid(resid). (probably not syntactically correct but will facilitate communication).

 

I write the composite equation as

y = (g00 + g01*w + g02*z + u0) + xc*(g10 + g11*w + g12*z + u1) + e

 

I did some computations on the save command terms.

I found that variance of (y-pred) equaled the variance of resid, which I figure must be ‘e’ in the equation.

If that’s true then I figure that variance of (y-fixpred) would seem to be the variance of ((u0+xc*u1)+e).

However, I notice that in the your reply to Tino (March 19) you said

COMPUTE residual_level_2 = y - FXPRED_1.

Or, (u0+xc*u1).

 

The last difference number is (pred-fixpred). What would this variance be?

 

If resid is ‘e’, and either (y-fixpred) or (pred-fixpred) is (u0+xc*u1), is there any way to extract/compute u0 and u1?

 

Gene Maguin

 

 

 

 

 

 

From: GMAIL [mailto:[hidden email]]
Sent: Tuesday, June 17, 2014 5:38 PM
To: Maguin, Eugene
Cc: [hidden email]
Subject: Re: mixed model residuals computation and plotting

 



 

On Jun 17, 2014, at 5:17 PM, "Maguin, Eugene" <[hidden email]> wrote:

I know that mixed provides fixed predicted values,  predicted (fitted = fixed+random) values  and residuals (observed-predicted). Is it possible to compute the level 2 residuals from these predicted and residuals data? How would it be done computationally? Singer and Willet mention this in their book and may have offered directions for computing these on their supplementary website but that seems to be gone. UCLA/ats-idre has examples but not those details, so far as I can tell.

Thanks, Gene Maguin



===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD
Ryan
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Re: mixed model residuals computation and plotting (Revised)

Ryan
For clarity, the TEST statements I wrote previously for this simulation example:
 
 
...provides estimates for subject 1. If one wanted estimates for subject 2, the following TEST statements would need to be added:
 
 
/TEST 'B02' intercept 1 | subject_ID 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
/TEST 'U02' | subject_ID 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                         0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
/TEST 'B12' time 1 | subject_ID*time 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
/TEST 'U12' | subject_ID*time 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                              0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 
Clearly one could do this for each subject.
 
Ryan


On Sun, Jun 22, 2014 at 12:22 AM, Ryan Black <[hidden email]> wrote:
All,

Below is a simulated example verifying what I stated in the previous post.

Best,

Ryan
--

/*Generate Data*/.
/*seed for random  generator*/.
set seed 987879546.
 
new file.
input program.
compute subject_ID = -99.
compute Gamma00 = -99.
compute Gamma10 = -99.
compute u0j = -99.
compute u1j = -99.
compute B0J = -99.
compute B1J = -99.
compute eij = -99.
compute time = -99.
leave subject_ID to time.
 
    /*100 subjects*/.
    loop subject_ID = 1 to 100.
    /*fixed intercept*/.
    compute Gamma00 = 0.50. 
    /*fixed slope*/.
    compute Gamma10 = 0.30. 
    /*random intercept error term*/.
    compute u0j = sqrt(.30)*rv.normal(0,1).
    /*random slope error term*/.
    compute u1j = sqrt(.30)*rv.normal(0,1).
    /*random intercept term*/.
    compute B0J = Gamma00 + u0j.
    /*random slope term*/.
    compute B1J = Gamma10 + u1j.
    /*30 time points*/.
    loop time = 0 to 29.
    /*error term*/.
    compute eij = rv.normal(0,1).
    /*full equation*/.
    compute y = B0J + B1J*time + eij.
    end case.
   end loop.
  end loop.
 end file.
end input program.
execute.

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time
  /TEST 'B01' intercept 1 | subject_ID 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'U01' | subject_ID 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                           0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'B11' time 1 | subject_ID*time 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /TEST 'U11' | subject_ID*time 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
                                0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  /SAVE=PRED.


On Fri, Jun 20, 2014 at 11:32 PM, Ryan Black <[hidden email]> wrote:
Gene,

You pose an interesting question which I had not thought of previously. I *think* one way to obtain U0J and U1J requires the use of the distributive property of multiplication on the RANDOM statement. 

Suppose we have the following random coefficient model:

Level 1 Equation:

y = B0J + B1J*time + eij

Level 2 Equations:

B0J = Gamma00 + u0j
B1J = Gamma10 + u1j

Full Equation:

y = Gamma00
    + Gamma10*time
    + ( u1j*time + u0j + eij )

Assuming the random components are uncorrelated, the standard MIXED model code would be:

MIXED y WITH time
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM intercept time | subject(subject_ID).

However, using the distributive property of multiplication, the following model is mathematically equivalent to:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time.

Using the second parameterization of the model, I *think* one can take advantage of the TEST sub-command to estimate B0J, U0J, B1J, and U1J.

Assuming there are a total of N=10 subjects and the time variable is a whole number that starts at 0 and increases by one unit as 0,1,2,3,... etc., then the code to obtain the parameter estimates of interest for the first subject would be:

MIXED y WITH time BY subject_ID
  /FIXED=time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION G 
  /RANDOM=subject_ID subject_ID*time
  /TEST 'B01' intercept 1 | subject_ID 1 0 0 0 0 0 0 0 0 0 
  /TEST 'U01' | subject_ID 1 0 0 0 0 0 0 0 0 0
  /TEST 'B11' time 1 | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /TEST 'U11' | subject_ID*time 1 0 0 0 0 0 0 0 0 0
  /SAVE=PRED.

To test this out, make sure that the following is true:

B01 = Gamma00 + U01
B11 = Gamma10 + U11

where,

Gamma00 = fixed intercept
Gamma10 = fixed slope

If the above is correct (which it should be), then make sure the following is also true (for subject 1),

PRED_1 = B01 + B11*("value of "time")

HTH,

Ryan


On Wed, Jun 18, 2014 at 5:07 PM, Maguin, Eugene <[hidden email]> wrote:

Thanks, Ryan.

 

It looks like Carol (second message) was asking about a single random term model, maybe an intercept only model.

Let’s take a specific example.

 

mixed y with xc w z/fixed=xc w z xc*w xc*z/

   method=ml/print solution/random intercept xc | subjectid(clus) covtype(un)/

   save=fixpred(fixpred) pred(pred) resid(resid). (probably not syntactically correct but will facilitate communication).

 

I write the composite equation as

y = (g00 + g01*w + g02*z + u0) + xc*(g10 + g11*w + g12*z + u1) + e

 

I did some computations on the save command terms.

I found that variance of (y-pred) equaled the variance of resid, which I figure must be ‘e’ in the equation.

If that’s true then I figure that variance of (y-fixpred) would seem to be the variance of ((u0+xc*u1)+e).

However, I notice that in the your reply to Tino (March 19) you said

COMPUTE residual_level_2 = y - FXPRED_1.

Or, (u0+xc*u1).

 

The last difference number is (pred-fixpred). What would this variance be?

 

If resid is ‘e’, and either (y-fixpred) or (pred-fixpred) is (u0+xc*u1), is there any way to extract/compute u0 and u1?

 

Gene Maguin

 

 

 

 

 

 

From: GMAIL [mailto:[hidden email]]
Sent: Tuesday, June 17, 2014 5:38 PM
To: Maguin, Eugene
Cc: [hidden email]
Subject: Re: mixed model residuals computation and plotting

 



 

On Jun 17, 2014, at 5:17 PM, "Maguin, Eugene" <[hidden email]> wrote:

I know that mixed provides fixed predicted values,  predicted (fitted = fixed+random) values  and residuals (observed-predicted). Is it possible to compute the level 2 residuals from these predicted and residuals data? How would it be done computationally? Singer and Willet mention this in their book and may have offered directions for computing these on their supplementary website but that seems to be gone. UCLA/ats-idre has examples but not those details, so far as I can tell.

Thanks, Gene Maguin




===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD
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