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Consider the discrete random variables $Y,X$ with supports $\mathcal{Y}, \mathcal{X}$.

Suppose that the probability measure of $X$ conditional on $Y=y$ is itself a discrete random variable with support $\mathcal{A}_{y}$ for any $y\in \mathcal{Y}$ (random probability measure).

Let $P_{X|y}(x)$ be the probability measure of $X$ conditional on $Y=y$ evaluated at $X=x$.

Let $P_{Y, P_{X|y}}(y_1,a)$ be the joint probability measure of $Y,P_{X|y}$ evaluated at $Y=y_1$, $P_{X|y}(x)=a$.

Questions:

(1) Should $P_{X|y}(x)$, $X$ and $Y$ be random variables defined on two different probability spaces (one for $P_{X|y}(x)$ and one for $X$ and $Y$)? Could you explain what happens if we use the same probability space $(\Omega, \mathbb{P}, \mathcal{F})$ to define all the three random variables?

What I am confused about is the following: let $X,Y$ be defined on the probability space $(\Omega, \mathbb{P}, \mathcal{F})$.

$X$ induces the probability measure $P_X$ such that $P_X(B)=\mathbb{P}(\omega \in \Omega \text{ s.t. } X(\omega)\in B)$ $\forall B\in \mathcal{B}$ where $\mathcal{B}$ is the Borel sigma algebra

Similarly, $Y$ induces the probability measure $P_Y$ such that $P_Y(B)=\mathbb{P}(\omega \in \Omega \text{ s.t. } Y(\omega)\in B)$ $\forall B\in \mathcal{B}$

Lastly, $P_{X|y}(B)=\mathbb{P}(\omega \in \Omega_y \text{ s.t. } X(\omega)\in B)$ $\forall B\in \mathcal{B}$, where $\Omega_y:=\{\omega \in \Omega \text{ s.t. } Y(\omega)=y\}$.

How can $P_{X|y}(B)$ be defined on $(\Omega, \mathbb{P}, \mathcal{F})$ too?

(2) Is $P_{Y, P_{X|y}}(y_1,a)=0$ by definition?

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  • $\begingroup$ Why is $P_{X|y}$ a random variable in your opinion ? It seems to be that it is a function $\mathcal{X} \to \mathbb{R}^+$. Furthermore, why do you say that $\mathcal{X}$ is the support of $X$ ? The support of a function is something else, isn't $X$ a function $\Omega \to \mathcal{X} $ ? $\endgroup$ – Robin Vogel Apr 20 '17 at 19:39
  • $\begingroup$ Is an assumption (see definition.of random probability measure) $\endgroup$ – STF Apr 20 '17 at 20:44
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If a random measure is to be thought of as the conditional distribution of a random variable $X$, then in a way $X$ is a 'random' random variable, and it's hard to separate the different layers of randomness when we talk about $X$.

In your context we are discussing the joint distribution of $Y$ and $P_{X|y}$. This means that necessarily these two objects are defined on the same probability space. As for $X$, notice that $X$ is not explicitly defined here except in terms of its distribution. For example consider a discrete probability space $(\Omega, {\mathbb P}, {\cal F})$ with three objects:

$$ \begin{array}{cccl} \omega&{\mathbb P}(\omega)&Y(\omega)&P_{X|y}(\omega)\\ \hline \omega_1&1/4&0&\text{Bernoulli}(p=0)\\ \omega_2&1/4&0&\text{Bernoulli}(p=1)\\ \omega_3&1/2&1&\text{Bernoulli}(p=1/2)\\ \end{array} $$ This specification is enough to determine $P_{X|y}(x)$, which is a random variable defined by $$P_{X|y}(x)(\omega):=P_{X|y}(\omega)(\{x\});$$ to emphasize that $P_{X|y}(x)$ is a random variable for each $x$, we could fill out the above table with two additional columns $P_{X|y}(0)(\omega)$ and $P_{X|y}(1)(\omega)$. Similarly the random measure $P_{X|y}(\cdot)$ is defined by $$ P_{X|y}(B)(\omega):=\sum_{x\in B}P_{X|y}(x)(\omega); $$ to answer your first question, that's how we define $P_{X|y}(B)$; note that $P_{X|y}(B)$ is a random variable for each $B$.

To answer your second question, no, $P_{Y, P_{X|y}}(y_1,a)$ is not zero by definition. In the discrete probability space defined above, $Y$ has two possible values (namely $0$ and $1$), and $P_{X|y}$ has three possible values (namely, three flavors of the Bernoulli distribution). The joint probabilities of all possible combinations of these will be $0$, or $1/4$, or $1/2$, depending on how you match up $y_1$ and $a$.

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  • $\begingroup$ Thanks, I have one doubt that I really do not get. In the table: since $\mathcal{Y}:=\{0,1\}$ in your example, should there be one column for $P_{X|y=0}(\omega)$ and one column for $P_{X|y=1}(\omega)$? Or I should read it in the following way: the entry (2,4) of the table is $P_{X|Y(\omega_1)}(\omega_1)$ i.e. $P_{X|0}(\omega_1)$, the entry (3,4) of the table is $P_{X|Y(\omega_2)}(\omega_2)$ i.e. $P_{X|0}(\omega_2)$, the entry (4,4) of the table is $P_{X|Y(\omega_3)}(\omega_3)$ i.e. $P_{X|1}(\omega_3)$? $\endgroup$ – STF Apr 21 '17 at 9:21
  • $\begingroup$ Lastly, and possibly related to my first question, how does $P_{X|y}$ varies with $y$ in the example? $\endgroup$ – STF Apr 21 '17 at 9:30
  • $\begingroup$ And also, how do you get $0,\frac{1}{4}, \frac{1}{2}$ in the last line of your answer. $\endgroup$ – STF Apr 21 '17 at 9:37
  • $\begingroup$ I think here I have a second layer of difficulties that is the condition on $Y$. I have asked a simpler version of this question here math.stackexchange.com/questions/2244089/… $\endgroup$ – STF Apr 21 '17 at 10:06
  • $\begingroup$ @STF Good point, in $P_{Y, P_{X|y}}(y_1,a)$ it's unclear what is the connection between $y_1$ and $P_{X|y}$, i.e., whether it's true that the $y$ in $P_{X|y}$ is supposed to equal $y_1$. That was my assumption (so your second reading is what I understood), but maybe that's not what the author intended. Under my interpretation, $P_{X|y}(\omega)$ is locked to the value of $Y(\omega)$, so in my example, only three combinations of the possible values for $Y$ and $P_{X|y}$ have nonzero probability; the other three have zero prob. $\endgroup$ – grand_chat Apr 21 '17 at 16:51

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