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Let our model be: $X_{ij}=\mu+\alpha_i+\epsilon_{ij}$, with $\epsilon_{ij}\sim^{iid}N(0,\sigma^2)$. The $\alpha_i$ is the fixed-effect, and $i \in \{1,...,m\}$, $j\in \{1,...,n_i\}$.

If we have $m+1$ parameters for $m$ equations, we need one more equation to determine the parameters, but why should we have that $\sum _i n_i\alpha_i=0$, or $\sum _i \alpha_i=0$ if $\forall i \ n_i=n$?

Is it just to make it easier to compute the expected value of the estimator for $\sigma^2$ that uses the sum of squares between groups, when the null is not true? Or would changing this extra condition change many things?

Any help would be appreciated.

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  • $\begingroup$ en.wikipedia.org/wiki/Identifiability $\endgroup$
    – Glen_b
    Jul 13, 2016 at 23:40
  • $\begingroup$ @Glen_b could please explain me how identifiability relates to this question? It could help me understand your comment... $\endgroup$
    – An old man in the sea.
    Jul 13, 2016 at 23:49
  • $\begingroup$ Compare the opening sentence of the second paragraph at the link with the opening sentence of whuber's comment (which - besides describing the particular kind of non-identifiability here - contains the word "identify" being used in exactly this sense). The link is for additional context for anyone unfamiliar with the notion. $\endgroup$
    – Glen_b
    Jul 13, 2016 at 23:54
  • $\begingroup$ @Glen_b so what you're saying is that without an extra condition the model would not be identifiable. Thanks for the extra info. ;) Could you also help me with the reason why we should use this precise condition? $\endgroup$
    – An old man in the sea.
    Jul 13, 2016 at 23:59
  • $\begingroup$ If your model is not identifiable you'll have a set of parameter values that all have the same fit. If the parameters are continuous there will be an infinity of exactly equally-good fits. Consider trying to estimate the parameters of a model like $Y_i\sim(\alpha_1-\alpha_2,\sigma^2)$ from a set of observations $y_i,\,i=1,...,n$. You can estimate the difference $\alpha_1-\alpha_2$ just fine, but you have no hope of telling what the individual parameters are, since $(\alpha_1+\delta)-(\alpha_2+\delta)$ would have the same fit, for any $\delta$. $\endgroup$
    – Glen_b
    Jul 14, 2016 at 0:47

1 Answer 1

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If you don't apply some such constraint, you could not identify any of the parameters, because you could add an arbitrary $\delta$ to each $\alpha_i$ and compensate by subtracting $\delta$ from $\mu$.

You are free to impose any set of constraints that will lead to identifiable parameters. The sum-to-zero constraints shown in the question are arbitrary but convenient ways to pin down a particular set of solutions. Usually such constraints are chosen to be linear and to have simple, relevant interpretations. When the $m$ equations are linearly independent, a single linear constraint (that is independent of those) will suffice. Instead of the sum-to-zero constraint, you could elect to set any single coefficient to zero (thereby making its variable the "baseline"): such constraints are also popular and easy to interpret.

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