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Suppose my theory tells me that some outcome $y^{}_{it}$ for individual $i$ in time $t$ is generated by the following process:

\begin{align} y_{it} =& \ \alpha^{}_{i} + \beta^{}_{i}x^{}_{it}+\varepsilon^{}_{it} \\ \beta^{}_{i} =& \ \beta + \gamma^{}_{i} \end{align}

Here $\alpha^{}_{i}$ is an individual fixed effect, $x^{}_{it}$ is a time-varying observable, $\beta$ is the average impact of $x$ on $y$, and $\gamma^{}_{i}$ is an additional effect of $x$ on $y$ which varies by $i$.

As I understand it, I could model the uncertainty from $\gamma$ in one of two ways: (i) using a random effects model where $\gamma^{}_{i}$ is the random coefficient; or (ii) as a finite mixture model where the mixture is over the distribution of $\gamma$.

Questions:

  1. What are the trade-offs with each approach and how do I choose between the two of them? I am interested in the values of $\gamma^{}_{i}$; in particular I am interested in the correlation between the two unobserved parameters $\gamma^{}_{i}$ and $\alpha^{}_{i}$. Is one approach better-suited for this?
  2. I understand that with the finite mixture model, $\gamma^{}_{i}$ can only take a finite number of values. This may be a con. Whereas with random effects, one usually has to make a parametric assumption about the distribution that $\gamma^{}_{i}$ is drawn from. This is a con for the RE model. Is this correct?
  3. It appears there is also a difference in interpretation. With the RE model, we can predict $\gamma^{}_{i}$ for each $i$. Whereas for the finite mixture model, we can only assign some probability $\pi^{}_{ik}$ that $\gamma^{}_{i}$ takes the $k^{th}$ type value for $i$. Is this correct?
  4. Lastly, if we specify a distribution (say gaussian) for $\gamma$ in the mixture model thereby allowing for infinite values of $\gamma^{}_{i}$, are we back to a random effects world? Is the difference between FMM and RE then that in FMM $\gamma$ is finite and we do not make distributional assumptions on $\gamma$?
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  • $\begingroup$ Why do you believe that you can use a mixture model, and what techniques to analyse it do you envision? Possibly a Bayesian approach? $\endgroup$
    – Sextus Empiricus
    Commented Dec 22, 2022 at 13:21
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    $\begingroup$ I was envisioning using expectation maximization to estimate this. This paper covers finite mixture model with individual fixed effects: (york.ac.uk/media/economics/documents/herc/wp/11_03.pdf). From what I understand, it involves fixing the number of unobserved types, taking an initial guess for the $\gamma$'s, updating the type probabilities, evaluating the likelihood, and then iterating on this until convergence. I thought this could be applied to my problem, but I am not sure about this. Do you see any issues with using a mixture model here $\endgroup$
    – lasoon
    Commented Dec 22, 2022 at 13:48
  • $\begingroup$ A complication in my setting is that the unobserved $\gamma^{}_{i}$ might be correlated with $x^{}_{it}$ e.g., an individual might be more likely to actions to increase the value of $x^{}_{it}$ if $\gamma^{}_{i}$ is large and vice versa. Ignoring this dependence (i.e., setting $\gamma^{}_{i}=0$) would bias the estimate for $\beta$ since $x^{}_{it}$ is more likely to be non-zero for those with higher returns $\gamma^{}_{i}$. This would inflate upwards our estimate of $\beta$ if we set $\gamma^{}_{i}=0$. Can either the mixture model or RE model deal with this type of dependence? $\endgroup$
    – lasoon
    Commented Dec 22, 2022 at 14:49
  • $\begingroup$ I still have to digest that article, but it seems to be relating to a case when your data generating process is not the way as you describe. $\endgroup$
    – Sextus Empiricus
    Commented Dec 22, 2022 at 16:14
  • $\begingroup$ Section 5.1.1 in Page 11 uses almost the same DGP. There they write it as $\mu^{}_{jit}=\beta^{}_{j}x^{}_{it}+\alpha^{}_{ij}$, where $j \in \{1,2\}$ indexes the two types. This is equivalent to what I have written if you impose the restriction that $\beta^{}_{i} = \beta+\gamma^{}_{i} \in \{\beta^{}_{1},\beta^{}_{2}\}$ i.e., if you restrict $\gamma$ to take only two possible values (the finite type restriction). $\endgroup$
    – lasoon
    Commented Dec 22, 2022 at 17:45

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