Find the probability distribution, knowing the conditional probability distribution.

Question

I have been working on some physics problem that I ‘translated’ into the following mathematical problem for which I need help to solve.

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Suppose that $ \alpha $ is a random variable uniformly distributed on the interval $ [0,2 \pi] $. Then for each random (there are technicalities here) subset $ A $ of $ [0,2 \pi] $, we have $$ \textbf{Pr}(\alpha \in A \mid \lambda(A) = x) = \frac{x}{2 \pi}, $$ where $ \lambda $ denotes the Lebesgue measure on $ [0,2 \pi] $.

Further suppose that we can randomly choose random subsets of $ [0,2 \pi] $ so that $$ \{ A \mid \text{$ A $ is a random subset of $ [0,2 \pi] $ and $ \lambda(A) \in [a,b] $} \} $$ is a random event for $ 0 \leq a \leq b \leq 2 \pi $ and that the random variable $ X $ defined by $$ X \stackrel{\text{df}}{=} \left\{ \begin{matrix} \text{Random subsets of $ [0,2 \pi] $} & \to & [0,2 \pi] \\ A & \mapsto & \lambda(A) \end{matrix} \right\} $$ has a probability density function $ p $. Then I thought: $$ \textbf{Pr}(\alpha \in A) = \int_{0}^{2 \pi} \frac{x}{2 \pi} \cdot p(x) ~ \mathrm{d}{x}. $$

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Now follow some questions that I have:

• Is the above formula actually correct?
• Is it possible to randomly ‘choose’ a random subset of $ [0,2 \pi] $ (or any other interval)?

Answer 1

The second question is tantamount to asking if there exist a $ \sigma $-algebra $ \Sigma $ on the Borel $ \sigma $-algebra $ \mathscr{B}([0,2 \pi]) $ on $ [0,2 \pi] $ itself and a probability measure $ \mu: \Sigma \to [0,1] $ such that $$ \forall x \in [0,2 \pi]: \quad \{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \} \in \Sigma, $$ and the function $ p: [0,2 \pi] \to [0,1] $ defined by $$ \forall x \in [0,2 \pi]: \quad p(x) \stackrel{\text{df}}{=} \mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}) $$ is a Lebesgue-integrable function with mass $ 1 $, i.e., $ \displaystyle \| p \|_{1} \stackrel{\text{df}}{=} \int_{[0,2 \pi]} p ~ \mathrm{d}{\lambda} = 1 $.

The answer is ‘no’. By way of contradiction, assume that $ \Sigma $ and $ \mu $ exist with the properties listed above. Observe that the sets $$ \{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}, \quad x \in [0,2 \pi] $$ are disjoint. As $ \| p \|_{1} = 1 $, we obviously have $ p(x) > 0 $ for uncountably many $ x \in [0,2 \pi] $. By the Infinite Pigeonhole Principle, there exist (i) an $ \epsilon > 0 $ and (ii) an uncountable subset $ S $ of $ [0,2 \pi] $ such that $ p(x) > \epsilon $ for every $ x \in S $. Let $ C \subsetneq S $ be countably infinite. Then by the $ \sigma $-additivity of $ \mu $, we have \begin{align} \mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) \in C \}) & = \mu \! \left( \bigcup_{x \in C} \{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \} \right) \\ & = \sum_{x \in C} \mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}) \\ & = \sum_{x \in C} p(x) \\ & > \sum_{x \in C} \epsilon \\ & = \infty. \end{align} This contradicts the assumption that $ \mu $ is a probability measure.

This automatically renders the first question meaningless.