# Identifiability of normal distribution

I am working on an exercise problem and am stuck in this problem:

Suppose that $X_1,\dots,X_n$ are independent with $X_i\sim\mathrm{N}(\alpha_i + \nu, \sigma^2)$. Let $\theta = (\alpha_1, . . . , \alpha_n, \nu, \sigma^2)$ and consider the family of sampling distributions $P_θ$. Is the above parameterization identifiable?

• What's the actual question? – shadowtalker Sep 9 '14 at 5:32
• @ssdecontrol: oops..edited – Mohit Sep 9 '14 at 7:38
• What happens when, say, you sum the number 2 to all alphas and subtract 2 to nu? What distribution do you get? – a.arfe Sep 9 '14 at 12:45
• Write the likelihood function $L_x(\theta)=\prod_{i=1}^n f_i(x_i\mid\theta)$. If you can find different $\theta$ and $\theta'$ such that $L_x(\theta)=L_x(\theta')$, then the model is nonidentifiable. Hint: consider the permutation of two $\alpha_i$'s keeping the rest fixed. – Zen Sep 9 '14 at 18:33

You should give a close look at the definition of identifiability, that is that the application $\theta \rightarrow P_{\theta}$ is injective. What does it mean here ?
Yes, it is. Identifiability means that if you had an infinite amount of data you could estimate the parameters of interest. Consider taking infinite draws from each marginal distribution $X_1, . . . , X_p$. You could clearly identify $σ^2$ and each of the summations $\ α_i + ν$. Next take infinite draws from each of the possible bivariate distributions. This will separately identify $ν$ and each of the $α_i$. If the elements of $θ$ are the only objects of interest then you've identified everything.
• But you can never figure out what $\nu$ actually is. – AdamO Sep 9 '14 at 15:40
• Yes, good point, that is true. If you were to estimate the parameters you would get an answer, but $ν$ would be estimated as the grand mean of the independent random variables, and $a_i$ would be estimated as the mean distribution-specific deviation from the grand mean. In that sense the model is only identified with respect to the scaling parameter $ν$. – user1189728 Sep 9 '14 at 15:51
• You can estimate $\nu$ as the grand mean only if $\sum \alpha_i =0$. – a.arfe Sep 9 '14 at 16:01