I saw people using such equivalence $$P(X|\mu) P(\mu | D) = P(X,\mu|D)$$ how to prove it is valid?
My attempt: \begin{align} P(X|\mu) P(\mu | D) &= P(X|\mu) \frac{P(\mu,D)}{P(D)}\\ &= P(X|\mu) \frac{P(D|\mu) P(\mu)}{P(D)}\\ &=\frac{P(X,\mu) P(D|\mu)}{P(D)} \end{align} where $P(X,\mu|D) = \frac{P(X,\mu,D)}{P(D)}$. I have stuck here, couldn't figure out why $P(X,\mu,D) = P(X,\mu) P(D|\mu)$.
Edited Additional:
If instead we have $P(X|\mu,D) P(\mu | D)$, then \begin{align} P(X|\mu,D) P(\mu | D) = \frac{P(X,\mu,D)}{P(\mu,D)} \frac{P(\mu,D)}{P(D)}=P(X,\mu|D) \end{align} seems like in order to obtain such equivalence it implicitly assumes $P(X|\mu,D)=P(X|\mu)$, correct me if I am wrong.