Is the following model parameterization identifiable? Let $X_i's$ be independent $i=1,2...n$ with $X_i\sim N(\mu+\alpha_i, \sigma^2)$ for each $i$. Let $\theta=(\alpha_1,...\alpha_p,\mu,\sigma)$ and $P_\theta$ be the joint pdf of the $X_i's$.
So, $P_\theta=(\frac{1}{\sqrt{2\pi}\sigma})^n e^{\frac{-1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu-\alpha_i)^2}$
From what I understand, a model $P_\theta$ is identifiable if $\theta_1=\theta_2$ implies that $P_{\theta_1}=P_{\theta_2}$ and $\theta_1\not=\theta_2$ implies that $P_{\theta_1}\not=P_{\theta_2}$
Proof by counterexample:
Let $\theta_1=(\alpha_1,...\alpha_p,\mu,\sigma)$ and  $\theta_2=(\alpha_p,...\alpha_1,\mu,\sigma)$
$\theta_1\not=\theta_2 $ but $P_{\theta_1}=P_{\theta_2}$
So the model is unidentifiable?
 A: proof by counterexample
Let $\theta=(\alpha_1,...\alpha_p, \mu.\sigma)$ s.t. $\mu=-\alpha_i$ for all $i$.
This means that $P_{\theta}=(\frac{1}{\sqrt{2\pi}\sigma})^n e^{\frac{-1}{2\sigma^2}\sum_{i=1}^n (x_i)^2}$
But this $\theta$ is not unique. We can have an infinite number of $\theta$ such that $\mu=-\alpha_i$ for all $i$. All of them would have the same $P_{\theta}$. Thus, the model is not identifiable.
A: One way to deal with a question like this is to "tweak" one of the parameters and see if you can derive how you need to modify the other parameters to compensate and get back to where you started from.
For example, set $\mu' = \mu - \Delta$, for some real $\Delta$. Then:
$$\begin{align}
P_\theta&=\left(\frac{1}{\sqrt{2\pi}\sigma}\right)^n \exp\left({\frac{-1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu-\alpha_i)^2}\right)\\
&=\left(\frac{1}{\sqrt{2\pi}\sigma}\right)^n \exp\left({\frac{-1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu' + \Delta-\alpha_i)^2}\right)\\
&=\left(\frac{1}{\sqrt{2\pi}\sigma}\right)^n \exp\left({\frac{-1}{2\sigma^2}\sum_{i=1}^n (x_i-\mu' - \alpha'_i)^2}\right)\\
\end{align}$$
where $\alpha'_i = \alpha_i + \Delta$ for each $i$. That is, the distribution is the same for any parameter vector $(\alpha'_1,...,\alpha'_n,\mu',\sigma)=(\alpha_1 + \Delta, ..., \alpha_n + \Delta, \mu - \Delta, \sigma)$, no matter the value of $\Delta$, so the model is not identifiable.
This also shows you exactly where the symmetry is: the parameter $\mu$ defines a global reference level that you measure the $\alpha_i$ from, but it has no real meaning and you could raise it or lower it without any impact by compensating with the $\alpha_i$. You can simply set $\Delta = \mu$ and measure them from zero instead to get an identified model.
