Confused about conditional counterparts to traditional probability laws

Question

I'm self-studying probability and have seen the following in various readings.

The "conditional counterpart" $$P(x,y|\theta) = P(x|y,\theta)P(y|\theta)$$ to the traditional conditional probability definition below. $$P(x, y) = P(x|y)P(y)$$

I have also seen a "marginal pdf counterpart" $$p(\tilde{y}|y) = \int p(\tilde{y}, \theta|y)d\theta$$ to the traditional definition of marginal pdf below. $$p(\tilde{y}) = \int p(\tilde{y},\theta)d\theta$$

My question is the following: I understand equations #2 and #4 above, i.e. the traditional definitions. But why is it that equations #1 and #3 hold?

Answer 1

One key to understanding this is that conditional probability is probability. So results holding for probability should also hold for conditional probability. Your formula $$ P(x, y) = P(x|y)P(y) $$ is general, so must also hold for a conditional probability. Denote the corresponding conditional probability given $\theta$ by $Q$, that is $Q(\cdot)= P(\cdot \mid \theta)$. Then use the formula above for $Q$, use the definition of $Q$, and see what you get. The same should work for the other question.

EDIT

Trying to answer the extra question in comments:

The formula above for $Q(⋅)=P(⋅|θ)$ makes sense for $Q(x,y)$ and $Q(y)$ as those terms would translate to $P(x,y|θ)$ and $P(y|θ)$ respectively, but what about for $Q(x|y)$? Wouldn't that term translate to $P(x|y|θ)$? Why does $P(x|y|θ)=P(x|y,θ)$?

Lets go back to definitions. Let $P$ be a probability on $\Omega$ and $B\subset\Omega$ with $0\lt P(B) \lt 1$. Then $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}=Q(A) $$ and $Q$ is a probability measure on $\Omega'=\Omega\cap B$. Then let $C\subset \Omega'$ with $0< Q(C) <1$ and look at \begin{align} Q(A\mid C) &=& \frac{Q(A\cap C)}{Q(C)} \\ &=& \frac{P(A\cap C \cap B)/P(B)}{P(C\cap B)/P(B)} \\ &=& \frac{P(A\cap C\cap B)}{P(C\cap B)} \\ &=& \frac{P(A\mid B\cap C)P(B\cap C)}{P(B\cap C)} \\ &=& P(A\mid B\cap C) ~~ \text{"$= P(A \mid B \mid C$)"} \end{align} and that completes the proof in the case of events.