Gelman & Hill (pp. 252-259) discuss "no-pooling" (single-level), and "partial-pooling regression" (multi-level) with no predictor ($section~ 12.2$).

In almost all mixed-effects models (i.e, partial-pooling) texts (e.g., pp. 4-6 this book), one of the advantages of such methods is said to be their larger $SE$ (standard error) for regression coefficient estimates compared to those from their NON-multi-level peers.

Question: Below, I'm comparing partial_ and no_pooling models. However, I see that the partial_pooling model has a far smaller $SE$. I wonder why I'm seeing the opposite?

set.seed(0)                            # Make the following reproducible
groups <- gl(20, 10)                   # 20 grouping indicators each of length 10 (20 classes each with 10 students)
design <- model.matrix(~groups-1)      # Design matrix
   U0j <- rnorm(20, 0, 20)             # Random intercept deviations each for a classroom
   eij <- rnorm(length(groups), 0, 30) # Common error term for observations
     y <- 1629 + design%*%U0j + eij    # Response variable


no_pooling <- lm(y~groups-1)
(SE_no_pooling <- sqrt(diag(vcov(no_pooling))))

#> 8.864905 # for all groups

partial_pooling <- lmer(y~ 1 + (1|groups))
(SE_partial_pooling <- sqrt(diag(vcov(partial_pooling))))
#>  0.2443936 # for intercept

1 Answer 1


I think that you might be confusing "no pooling" and "complete pooling." The former is represented by the no_pooling model and is an alternative way to deal with multilevel data by treating the clusters as a fixed population rather than as a random sample of similar clusters, which is the case in partial_pooling. In a complete pooling model, cluster membership is ignored. Such a model would be as follows:

lm(formula = y ~ 1, data = df)

    Min      1Q  Median      3Q     Max 
-73.903 -23.997   0.006  21.714  98.714 

            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 1628.976      2.383   683.6   <2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 33.7 on 199 degrees of freedom

The standard error for the intercept is 2.383. In contrast, the standard error for the intercept in the partial_pooling model is 4.716:

Linear mixed model fit by REML ['lmerMod']
Formula: y ~ 1 + (1 | groups)
   Data: df

REML criterion at convergence: 1929.7

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-2.1039 -0.7621 -0.1037  0.6983  2.8887 

Random effects:
 Groups   Name        Variance Std.Dev.
 groups   (Intercept) 366.2    19.14   
 Residual             785.9    28.03   
Number of obs: 200, groups:  groups, 20

Fixed effects:
            Estimate Std. Error t value
(Intercept) 1628.976      4.716   345.4

Thus the complete pooling model, by ignoring correlations of y-values within clusters, assumes that all individuals are independent. In so doing, it estimates a standard error consistent with such an assumption. The partial_pooling model is designed for this problem and such, appropriately adjusts the standard error estimate by differentially weighting the sample size. I will try to come back and put in the different standard errors calculations for the three models.

Edit: The three standard errors, as promised. These are for the balanced case where $n_j=n$ and $J$ is the number of clusters. $\hat\psi$ is the level 2 between-cluster variance and $\hat\theta$ is the level 1 within-cluster variance. The mixed model $\widehat{SE}$ will vary slightly for unbalanced group sizes:

$\widehat{SE}(\hat{\beta}^{OLS}) \approx \sqrt{\dfrac{\hat\psi + \hat\theta}{Jn}}$

$\widehat{SE}(\hat{\beta}^{Mixed}) = \sqrt{\dfrac{\hat\psi + \dfrac{\hat\theta}{n}}{J}}$

$\widehat{SE}(\hat{\beta}^{NoPool}) = \sqrt{\dfrac{\hat\theta}{Jn}}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.