Calculating the posterior distribution - missing dependency

Question

I am reading an old paper from https://www.jmlr.org/papers/volume1/tipping01a/tipping01a.pdf. In that paper, specifically, Equation 10 says $$ p(w|t,\alpha,\sigma^2) = \frac{p(t|w,\sigma^2)p(w|\alpha)}{p(t|\alpha,\sigma^2)}. $$ However, it seems that in order for the above equality to hold, we need to assume that $$ (*) \quad \frac{p(t,\sigma^2|w,\alpha)}{p(\sigma^2|\alpha)} = p(t|w,\sigma^2). $$ This is because \begin{align} p(w|t,\alpha,\sigma^2) &= \frac{p(w,t,\alpha,\sigma^2)}{p(t,\alpha,\sigma^2)} = \frac{p(t,\sigma^2|w,\alpha)p(w,\alpha)}{p(t|\alpha,\sigma^2)p(\alpha,\sigma^2)} \\ &=\frac{p(t,\sigma^2|w,\alpha)}{p(\sigma^2|\alpha)}\frac{p(w|\alpha)}{p(t|\alpha,\sigma^2)}. \end{align} However, I am not sure why (*) should hold. I feel there must be some hidden assumptions but I couldn't figure this out. Any help would be highly appreciated. Thanks!

Answer 1

I figured this out. Note that the following relations are assumed: $$ p(t|w,\alpha,\sigma^2) = p(t|w,\sigma^2), \quad p(\alpha,\sigma^2) = p(\alpha)p(\sigma^2), \quad p(w,\alpha,\sigma^2) = p(w,\alpha)p(\sigma^2), $$ which also gives $p(\sigma^2|\alpha) = p(\sigma^2)$. Hence, it can be checked that $$ \frac{p(t,\sigma^2|w,\alpha)}{p(\sigma^2|\alpha)} = p(t|w,\alpha,\sigma^2)\frac{p(w,\alpha,\sigma^2)}{p(w,\alpha)p(\sigma^2)} = p(t|w,\alpha,\sigma^2) = p(t|w,\sigma^2). $$