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I have been reading Tony Lancaster's 2004 book "An Introduction to Modern Bayesian Econometrics." On pages 116-117, Lancaster derives a conditional joint distribution for the data $p(y,X|\boldsymbol{\beta})$ for a linear model $y=X\boldsymbol{\beta}+\epsilon$ using a joint distribution of $X$ and the residuals, $p(\epsilon, X)$. But the substitutions he uses en route to his result don't always seem justified. Are they justified?

He starts out with the goal: posterior inference about $\boldsymbol{\beta}$, given $X$ and $y$, using Bayes Theorem

$$p(\boldsymbol{\beta}|y,X)\propto p(y,X|\boldsymbol{\beta})p(\boldsymbol{\beta})$$

Which requires deriving a conditional joint distribution for the data

$$p(y,X|\boldsymbol{\beta})$$

His strategy is to first obtain a joint distribution for $\epsilon$ and $X$, under the assumption that $\epsilon$ and $X$ are independent, not just uncorrelated

$$p(\epsilon, X)=p_{\epsilon}(\epsilon)p(X)$$

His next step is to make substitutions into this joint distribution using $y=X\boldsymbol{\beta}+\epsilon$, and then condition on $\boldsymbol{\beta}$, yielding the required conditional joint distribution

$$p(y,X|\boldsymbol{\beta})=p_{\epsilon}(y-X\boldsymbol{\beta})p(X|\boldsymbol{\beta})$$

I understand the right hand side of this result, simply using $\epsilon=y-X\boldsymbol{\beta}$. But what about the left hand side? It can't be a substitution simply of $y=\epsilon$, as that doesn't make sense. Can someone help me better understand how this result is justified? Am I missing something basic?

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The left-hand side of your final equation, $p(y,X|\beta$), can be written as $p_y(y|X,\beta)p(X|\beta)$, and writing the right-hand side more explicitly then gives $$ p_y(y|X,\beta)p(X|\beta) = p_\epsilon(y - X\beta|X,\beta)p(X|\beta) $$ Your intuition is right in that generally $p_y(y|X,\beta) \neq p_\epsilon(\epsilon|X,\beta)$. Instead, you have $$ p_y(y|X,\beta) = p_\epsilon(\epsilon|X,\beta) \left| \frac{\mathrm{d} \epsilon}{\mathrm{d} y} \right| $$ where $\left| \frac{\mathrm{d} \epsilon}{\mathrm{d} y} \right|$ is known as the Jacobian of the transformation. But in linear regression, $\epsilon = y - X\beta$ and therefore $\frac{\mathrm{d} \epsilon}{\mathrm{d} y} =1$, which means the Jacobian can be ignored.

I think that's pretty much all there is to say about this. Things would look different if your model was non-linear (for instance, a Box-Cox regression), or if your model was autoregressive, but in linear regression that transformation of variables is trivial.

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