I was trying to read this book on Bayesian Statistic and have I have trouble understanding the part in orange in the bottom image:

I think I get what is defined in equation (3.9) and how having $R = X-AX$ could give us $X-AX \sim \pi_{mod.error}(r)$ but I don't get how we can have the prior distribution of $x$ from that, maybe knowing what is "hidden" behind proportional term could help.

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Source : (Calvetti and E. Somersalo, Introduction to Bayesian Scientific Computing, Springer, 2007)

It is actually a type of issue I encountered several times in this type of examples ( like in interpolation noise-free data ) and I never know how to deal with those cases when we have an $Ax=R$ and we want to express the prior of $x$ given that the distribution of $R$ and the value of $A$ are known. I think the example above is quite representative. Any advice on how I could understand that?


$X$ follows the following distribution by definition : $X \sim \pi_{prior}(x)$. Also by definition, $R \sim \pi_{mod.error}(r)$.

We also defined $X = AX+R $ so $R = X-AX$ implying that $\pi_{mod.error}(r) = \pi_{mod.error}(x-Ax)$.

maybe rewriting $R = X-AX$ as $R = X (I -A)$ allows you to see that $R$ and $X$ are proportionnal up to the term $I - A$.

so we find effectively that $\pi_{mod.error}(x-Ax)$ and $\pi_{prior}(x)$ are proportionnal up to the term $I - A$.

Hope this helps.

  • 1
    $\begingroup$ Thank you very much for this answer, it's actually helped me a lot to understand, I think I was a bit confused by the fact the "distribution name" changing. $\endgroup$
    – Etienne
    Apr 27 '20 at 11:45

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