Suppose I have a model

$$\begin{bmatrix}y(1)\\y(2)\end{bmatrix} = \begin{bmatrix}\mu \\ \mu \end{bmatrix} + \begin{bmatrix}\Lambda \\ \Lambda\end{bmatrix} + \begin{bmatrix}\varepsilon(1)/\rho_1\\ \varepsilon(2)/\rho_2\end{bmatrix}$$

with 100 observations, where $\Lambda \sim iiN(0, \lambda^2)$, $\varepsilon \sim iiN(0, \sigma^2)$ and also $\rho_1, \rho_2, \sigma^2$ are known

If I'm not mistaken, this can be viewed as Linear Mixed Model, with $\mu$ being fixed effect and $\Lambda$ being random effect (I'm learning those right now).


1) Are parameters $\mu,~\lambda^2$ estimable/identifiable?

2) I think that the answer to the first question is yes. But I'm finding hard on how to translate this into R: specificaly, lme function from nlme package.

For example, if I run

lme(y0 ~ 1 , random = list(groups = ~1), data = ylmm)

where groups is identifier for $y(1),~y(2)$ I get the output

  Linear mixed-effects model fit by REML
  Data: ylmm 
  Log-restricted-likelihood: -13342.28
  Fixed: y0 ~ 1 

Random effects:
 Formula: ~1 | groups
        (Intercept) Residual
StdDev:  0.01312919 191.2608

Number of Observations: 2000
Number of Groups: 2 

My question is - are the estimates $\mu, \lambda^2$ correspondingly 102.8827 and 191.2608 ?


1 Answer 1


I don't know if I understand you correctly, but if you take a general linear mixed model:

$$\boldsymbol{y}_i = X_i\boldsymbol{\beta} + Z_i\boldsymbol{b}_i + \epsilon_i$$

where $X$ and $Z$ are design matrices, $\beta$ are fixed effects parameters and $b$ are random effects parameters. Then (Intercept) from the Fixed part of the output is $\mu$ (in notation you used). And Residual's from Random part of the output are $\lambda^2$. summary method for lme provides more information. So $\mu$ and $\lambda^2$ are estimable.

You could check the book by authors of nlme: Pinheiro, J.C. & Bates, D.M. (2000). Mixed-effects models in S and S-Plus. New York: Springer-Verlag.

BTW, check also lme4 package - a more recent approach to lmm in R by Bates et al.

However, for random effects you have to use more than two groups, for two groups this approach is improper.


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