1
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ What's the actual question? $\endgroup$
    – shadowtalker
    Commented Sep 9, 2014 at 5:32
  • $\begingroup$ @ssdecontrol: oops..edited $\endgroup$
    – frazman
    Commented Sep 9, 2014 at 7:38
  • 2
    $\begingroup$ What happens when, say, you sum the number 2 to all alphas and subtract 2 to nu? What distribution do you get? $\endgroup$
    – a.arfe
    Commented Sep 9, 2014 at 12:45
  • 1
    $\begingroup$ 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. $\endgroup$
    – Zen
    Commented Sep 9, 2014 at 18:33

2 Answers 2

Reset to default
2
$\begingroup$

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 ?

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I'm not sure this is entirely correct (consider my comment to the original question and Aerandal's answer). Could you provide some more details of your reasoning? $\endgroup$
    – a.arfe
    Commented Sep 9, 2014 at 14:57
  • $\begingroup$ But you can never figure out what $\nu$ actually is. $\endgroup$
    – AdamO
    Commented Sep 9, 2014 at 15:40
  • $\begingroup$ 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 $ν$. $\endgroup$
    – user1189728
    Commented Sep 9, 2014 at 15:51
  • $\begingroup$ You can estimate $\nu$ as the grand mean only if $\sum \alpha_i =0$. $\endgroup$
    – a.arfe
    Commented Sep 9, 2014 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.