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Suppose

$$ Y_i = \alpha \beta^2 X_i^3 \ \ \text{for $i$ in $1, \ldots, N$} $$

Both $Y_i$ and $X_i$ are positive real numbers. If we were to put a uniform prior on $\alpha$ and $\beta$, why would the model be non-identifiable? Thanks.

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Identifiability is determined by the sampling distribution, not the prior. It means that different parameter combinations lead to the exact same sampling distribution, so they cannot be distinguished based on the data. (Bayesian analysis for a non-identifiable parameters is explained in this related answer; short version is that adding a prior distribution to the parameters does not fix the identifiability problem.)

In this case, the fact that the parameters enter the model only through the term $\alpha \beta^2$ means that neither $\alpha$ or $\beta$ are identifiable parameters. The minimum sufficient parameter in the model is $\phi \equiv \alpha \beta^2$. This parameter can be estimated from the data, but it is not possible to use the data to decompose this into its parts based on the other two parameters.

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I would look at this problem by assuming the error distribution associated with the model:

$$ Y_i = \alpha \beta^2 X_i^3 \ \ \text{for $i$ in $1, \ldots, N$} $$

has a multiplicative error (so lognormal). Then, upon apply a log transform, the error distribution is normal.

So, now you have a standard linear model (in the transformed variables and coefficients) that can be viewed in the context of Bayesian linear regression. In particular, the conditional prior distribution is Normal and the posterior can be expressed as the Normal distribution times an Inverse-gamma distribution (see https://en.wikipedia.org/wiki/Bayesian_linear_regression ).

Note, when you derive the Bayesian counterpart to a confidence interval for the mean expected value of ln(Beta), be mindful on applying the inverse transform (Exp(ln(Beta)), the interval now relates to the median of the original log-normal error distribution.

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