Consider a probabilty space $(\Omega ,\mathcal F, \mathbb P$) and two measurable random variables $X,Y:\Omega \rightarrow S$.

Define $\mathcal A :=\sigma(Y)\subset \mathcal F$ and $\mathcal B :=\sigma(X)\subset \mathcal F$.

Question: How are the conditional probabilities $\mathbb P \{X\in B|\mathcal A\}=\mathbb P \{X\in B|Y\}$ and $\mathbb P \{X\in B|Y=s\}$ connected?

My idea is, that for $A,B\subset S$ measurable, we have $\int_A \mathbb P \{X\in B|Y=y\}\mathbb P\{Y^{-1}(dy)\}=\mathbb P \{X\in B|Y\in A\}$, but how do we know there exists some (measurable?) $\mathbb P \{X\in B|Y=y\}$ as function of $y$? And how do I know how it looks (if I have $\mathbb P \{X \in B|\mathcal A\}$)?

Usually there is the connection $\mathbb P \{X\in B|Y \in A\}=\frac{\mathbb P \{X \in B,Y\in A\}}{\mathbb P \{Y\in A\}}$, but this doesn't help if $A=\{y\}$ with $\mathbb P \{Y=y\}=0$.

  • $\begingroup$ $\mathbb{P}(X \mid \mathcal{A})$ doesn't make sense since $X$ is not an event. You probably mean $\mathbb{E}[X \mid \mathcal{A}]$? $\endgroup$ – Nate Eldredge Apr 21 '15 at 14:46
  • $\begingroup$ @NateEldredge I see the problems, I will think I have to edit the question... $\endgroup$ – Paul Apr 21 '15 at 14:54

If $\mathbb{P}(Y=s) > 0$ then $\mathbb{P}(X \in B \mid Y)(\omega) = \mathbb{P}(X \in B \mid Y = s) = \frac{\mathbb{P}(X \in B, Y = s)}{\mathbb{P}(Y=s)}$ for $\mathbb{P}$-almost every $\omega \in \{Y = s\}$. If $\mathbb{P}(Y=s) = 0$ then $\mathbb{P}(X \in B \mid Y = s)$ is undefined so the question has no meaning.

One statement that may be helpful is the Doob-Dynkin lemma, which says that for every random variable $Z$ which is $\sigma(Y)$ measurable, there exists a Borel function $f : \mathbb{R} \to \mathbb{R}$ such that $Z = f(Y)$. (It's easy to prove: start with $Z = 1_A$ where $A = \{Y \in B\}$ for $B$ Borel, then extend to simple functions by linearity, and extend to measurable functions by monotone limits.) So there exists $f$ such that $\mathbb{P}(X \in B \mid Y) = f(Y)$ almost surely. You might then think of defining "$\mathbb{P}(X \in B \mid Y=y)$" as $f(y)$. However, the function $f$ is typically not unique, so for each real number $y$ the value might depend on which $f$ you choose, and so this "definition" is not well defined.

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  • $\begingroup$ Well, as I see it, $f$ is well defined up to sets of $\mathbb P \{Y^{-1}\}$-measure zero, which is what I was looking for. $\endgroup$ – Paul Apr 21 '15 at 15:52
  • $\begingroup$ @Paul: Yes, that is true. $\endgroup$ – Nate Eldredge Apr 21 '15 at 15:58

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