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 $Y=y_1$, $P_{X|y}(x)=a$.


(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?

  • $\begingroup$ 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

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 $a$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.