A dilettante here trying to better understand the basic definition/properties of conditional probabilities.
In my setting I have a probability triple $(S, \Sigma, P)$ and a random vector $(X,Y)$ living on that space and mapping to a measurable space with a product space structure $(\textbf{X $\times$ Y}, \mathcal{X} \otimes \mathcal{Y})$.
Let $\mu$ be the "law"/"push-forward measure" associated with $(X,Y)$ and let $\mu_X$, $\mu_Y$ be the (marginal) laws of $X$ and $Y$ defined through "projection", e.g. $\mu_X(B) = \mu(B\times \textbf{Y})$ for $B \in \mathcal{X}$. Suppose conditional probability measures denoted by $\mu_{X | Y=y}$ are well-defined (regular) for $y \in \textbf{Y}$.
My question: Is it true that $\mu_{X | Y=y}$ is absolutely continuous w.r.t. $\mu_X$ for almost all $y$? Intuitively, it would make sense to me that if X being in some range is "impossible", getting more information through $Y$ cannot suddenly make this "possible".
I have been attempting a contradiction argument and assume there is a set $A \in \mathcal{Y}$ with $\mu_Y(A) >0$ s.t. for all $y \in A$ there is a $B_y$ with $\mu_X(B_y) = 0$ but $\mu_{X|Y=y}(B_y) >0$. I have tried to get a contradiction by using the definition of conditional probabilities, in particular that for $C \in \mathcal{X}$ we have $\int_A \mu_{X | Y=y}(C)dy = \mu(C \times A)$, but this has not been fruitful for me so far.
Best Answer
This is not true. Consider $(X, Y) = (Z,Z) \in \mathbb{R}^2$ where $Z$ is a $\mathcal{N}(0,1)$ random variable. Then $\mu_X = \mathcal{N}(0,1)$ but $\mu_{X \mid Y=y} = \delta_{y}$, the Dirac measure at point $y$.
The problem is that absolute continuity is not always about having the same range, but rather about having a density.