I am interested in the following model : ($1 \leq i \leq p$, $1 \leq j \leq n_{i}$)

$$y_{i,j} = A (1+a_{i})(t_{i,j}+\gamma_{i}) + \varepsilon_{i,j}$$

where $A \in \mathbb{R}$, $(a_{i})_{1 \leq i \leq p}$ is a $p$-random sample from $\mathcal{N}(0,\sigma_{a}^{2})$ and $(\gamma_{i})_{1 \leq i \leq p}$ is a $p$-random sample from the distribution $\mathcal{N}(0,\sigma_{\gamma}^{2})$, independent from $(a_{i})_{1 \leq i \leq p}$. $(\varepsilon_{i,j})_{i,j}$ is a random sample from $\mathcal{N}(0,\sigma^{2})$, independent from $(a_{i})_{i}$ and $(\gamma_{i})_{i}$.

Given some observations $(y_{i,j},t_{i,j})_{i,j}$, I would like to estimate the model parameters ($A,\sigma_{a}^{2},\sigma_{\gamma}^{2},\sigma^{2})$. I am wondering whether this is possible. I created some synthetical data using the model in Matlab and tried to estimate the parameters of the model using nlmefit(for nonlinear mixed-effects models) but the estimated values for $A$ and $\sigma_{a}$ were (respectively) three times smaller (and three times) bigger than the ones I used to create the data. This makes me wonder whether the model is identifiable. Can you see another reason to this ?

  • $\begingroup$ p-random? Is that an abbreviation for pseudo-random, or something else? $\endgroup$ – Glen_b May 12 '14 at 8:48
  • $\begingroup$ No, sorry! I just meant that it is a random sample of length $p$. $\endgroup$ – pitchounet May 12 '14 at 9:02
  • $\begingroup$ Sorry, I deleted the question because I thought I found the answer. As a matter of fact, the nlmefit function from Matlab only applies to nonlinear models of this form : $y_{i}=f_{i}(\phi_{i},x_{i})+\varepsilon_{i}$ where $\phi_{i}=A_{i,j}\beta + B_{i,j}b_{i}$ and $b_{i} \sim \mathcal{N}(0,\sigma^{2}D)$. But here, the model I'm interested cannot be written like this because of the product $\gamma_{i}\tau_{i}$. I thought this was the reason to my problem. How can you prove that the model is not identifiable ? $\endgroup$ – pitchounet May 12 '14 at 9:54

Actually, with my understanding corrected, I think the model is strictly speaking algebraically identifiable, but in some circumstances may go close to unidentifiability.

If we look at $\log(E(y))$ we have $\log(A) +\log(1+a_{i})+\log(t_{i,j}+\gamma_{i})$.

In this form, $\log(A)$ has the role of a intercept and $\log(1+a_{i})$ and $\log(t_{i,j}+\gamma_{i})$ are terms that in some circumstances can be close to collinear. If $t_{i,j}$ doesn't vary much across $j$ compared to its variation across $i$ (e.g. if $t_{i,j}$ turns out to be approximately $r_i+s_j$ and $s_j$ doesn't vary much compared to $r_j$) then there'd be near-collinearity on the log scale.


I'm leaving this here for the present for context. I'll probably delete it eventually.

Unless I missed something, your model seems to be not identifiable.

For example, consider a second version of the model with $A^* = A/k$, $t^*_{i,j}=kt_{i,j}$ and $\gamma^*_{i}=k\gamma_{i}$.

Then $E(y)$ is the same under both models.

Since $k$ is arbitrary, the model won't be identifiable.

  • $\begingroup$ You're right but $t_{i,j}$ are given. $t_{i,j}$ represents a time at which the observation $y_{i,j}$ has been obtained. As a consequence, only $(A,\sigma_{a},\sigma_{\gamma},\sigma)$ are to be estimated. $\endgroup$ – pitchounet May 12 '14 at 10:03
  • $\begingroup$ Ah, sorry, that should have been obvious; you say that $t$ is observed later. $\endgroup$ – Glen_b May 12 '14 at 10:05
  • $\begingroup$ If you'd rather delete I can try deleting my answer, that might enable you to do so. But you might still end up with a good answer. $\endgroup$ – Glen_b May 12 '14 at 10:10
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    $\begingroup$ I'll leave the question. I still don't know whether the model is identifiable but I am quite convinced that nlmefit is not suited for this model because it is not linear in $\gamma_{i}$ and $a_{i}$. $\endgroup$ – pitchounet May 12 '14 at 10:14

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