Mixed case joint probability distribution and absolute continuity with respect to the product measure

Question

Let $X$ and $Y$ be absolutely continuous and discrete random variables, respectively. Does random vector $(X,Y)$ have to be absolutely continuous with respect to the product measure on the respective supports of $X$ and $Y$?

I think this is true because the following equality holds

$$f_{X,Y}(x,y) = f_{X|Y}(x)\mathbb{P}(Y=y),$$

so $f_{X,Y}(x,y)$ is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

Answer 1

assuming $$f_{X,Y}=f_X f_Y$$

product $f_{X,Y}(x,y)$ can't be absolutely continuous on $Y$ axis - since $Y$ is not absolutely continuous then from https://en.wikipedia.org/wiki/Absolute_continuity#Definition

$$\forall x_0,\exists y_0\ne y_1, \exists\epsilon_0:|f(y_0)-f(y_1)| > \epsilon_0 \ =>\ |f_{X,Y}(x_0, y_0)-f_{X,Y}(x_0, y_1)| > \epsilon_0$$

example: $X$ is normal and $Y \sim Bernoulli(0.5)$ then $f_{X,Y}(0,0) = f_X(0) * 0.5$ and $f_{X,Y}(\epsilon, 0) = 0$

Answer 2

Let $X$ be a random variable taking values in a measurable space $(E,\mathcal{E})$ and let $Y$ be a discrete random variable taking values in a measurable space $(F,\mathcal{F})$. I will prove that the joint distribution $P_{(X,Y)}$ is absolutely continuous wrt. the product measure $P_X\otimes P_Y$.

For $B\in\mathcal{E}$ we find that $$ P_X(B)=\int_{\mathrm{supp}(Y)}P_{X|Y=y}(B)\,P_Y(\mathrm{d}y). $$ Let $y\in\mathrm{supp}(Y)$ and suppose that $P_X(B)=0$. Then, according to the above, $P_{X|Y=y}(B)=0$. This shows that $P_{X|Y=y}$ is absolutely continuous wrt. $P_X$. By the Radon-Nikodym theorem there exists a non-negative measurable function $f_y\colon (E,\mathcal{E})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ such that $P_{X|Y=y}(\mathrm{d}x)=f_y(x)P_X(\mathrm{d}x)$.

Now, for $A\in\mathcal{E}\otimes\mathcal{F}$ we can write $$ P_{(X,Y)}(A)=\int_{\mathrm{supp}(Y)}\int_E 1_A(x,y)\,P_{X|Y=y}(\mathrm{d}x)P_Y(\mathrm{d}y)=\int_{\mathrm{supp}(Y)}\int_E1_A(x,y)f_y(x)\,P_X(\mathrm{d}x)P_Y(\mathrm{d}y). $$ This shows that $P_{(X,Y)}$ is absolutely continuous wrt. $P_X\otimes P_Y$ with density $(x,y)\mapsto f_y(x)$.