Is the following parametrizations identifiable?

Question

$X$ and $Y$ are independent $N(\mu_1, \sigma^2)$ and $N(\mu_2, \sigma^2)$, $\theta = (\mu_1, \mu_2)$ and we observe $Y − X$.

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 $P_{\theta_1}\not=P_{\theta_2}$.

How can I choose ${\theta}$ to prove or disprove this?

$$Y-X \sim N(\mu_2-\mu_1, 2\sigma^2)$$

If I try both $\mu_2,\mu_1$ with a shift - I think I will get the same answer? Does that mean it's Identifiable?

Answer 1

If you are trying to prove that the model

$$ \mathcal{P} = \Big\{\ f_\theta(x) = \tfrac{1}{\sqrt{2\pi}2\sigma^2} e^{ -\frac{1}{4\sigma^4}(x-(\mu_2-\mu_1))^2 }\ \Big|\ \theta=(\mu_2-\mu_1): \mu_1,\mu_2 \in\mathbb{R}, \,\sigma\!>0 \ \Big\}. $$

is identifiable. Then you need to check that if $f_{\theta_1} = f_{\theta_2} \implies \theta_1 = \theta_2$. Suppose that $\theta_1 = \mu_2-\mu_1$ and $\theta_2 = \mu'_2 - \mu'_1$. Then

$$ f_{\theta_1} = f_{\theta_2} \Longleftrightarrow {} \frac 1 {2\sigma^2}(x-(\mu_2-\mu_1))^2 + \ln 2\sigma = \frac 1 {2\sigma^2}(x-(\mu'_2-\mu'_1))^2 + \ln 2\sigma. $$

Thus, $$(x-(\mu_2-\mu_1))^2 = (x-(\mu'_2-\mu'_1))^2 \implies \mu_2-\mu_1 = \mu_2'-\mu_1'.$$

Answer 2

It is not identifiable from $\theta$, because as you said, a shift in both $\mu_1$ and $\mu_2$ can produce the same model for Y - X. For example, $\theta = (2,3)$ and $\theta = (3,4)$ would both give you the same $P_{\theta}(Y-X) \sim N(1, 2\sigma^2)$