Finding conditional probability of "success" when success probability is random

Question

Suppose that event $A$ is a subset of event $B$.

Event $A$ happens with probability $p_A$ and event $B$ happens with probability $p_B$.

Here, we only know the distribution of $p_A$ and $p_B$.

In fact, we have $p_A\sim Beta(\alpha_1,\beta_1)$ and $p_B\sim Beta(\alpha_2,\beta_2)$ with a condition that $\alpha_1+\beta_1=\alpha_2+\beta_2$.

What I want to compute is to find a conditional probability of event $A$ given event $B$ (denote it by $p_{A|B}$).

If $p_A$ and $p_B$ were deterministic, we can simply have $$p_{A|B}=\frac{p_A}{p_B}.$$

However, how should I proceed if these $p_i$'s themselves are random variables? From the beta distribution and the assumption on the beta parameters, we have

$$f_{\mathbb p_A|\mathbb p_B=z}(p_A|p_B=z)=\frac{\Gamma(\alpha_2)\Gamma(\beta_2)}{\Gamma(\alpha_1)\Gamma(\beta_1)}\frac{p_A^{\alpha_1-1}(1-p_A)^{\beta_1-1}}{z^{\alpha_2-1}(1-z)^{\beta_2-1}},~p_A\leq z.$$

From this conditional probability, how to derive $p_{A|B}$? Can we proceed by setting $$E[p_{A|B}]=\int^1_0\int^z_0p_Af_{\mathbb p_A|\mathbb p_B=z}(p_A|p_B=z)f_{P_B}(z)dp_Adz?$$

Answer 1

If $p_A,p_B$ are both random variables here (unusual situation) then also $p_{A|B}$ must be looked at as a random variable: $p_{A\mid B}=\frac{p_A}{p_B}$ if $B\subseteq A$.

Further it is not in general true that $\mathbb E\frac{X}{Y}=\frac{\mathbb EX}{\mathbb EY}$ so it is also not guaranteed that $\mathbb Ep_{A\mid B}=\frac{\mathbb Ep_A}{\mathbb Ep_B}$

Answer 2

I doubt it is generally possible to have $P(A)$ and $P(B)$ both Beta distributions and $p_{A \mid B}=P(A \mid B)$ independent of the actual value of $B$ unless it is $0$ or $1$ or some other special case.

• The mean of $p_A =P(A)$ is $\frac{\alpha_1}{\alpha_1 + \beta_1}$ and the variance is $\frac{\alpha_1 \beta_1}{(\alpha_1 + \beta_1)^2(\alpha_1 + \beta_1+1)}$ as it has a Beta distribution

• If $P(A \mid B) = k$ then $P(A) = k P(B)$ with mean $k\frac{\alpha_2}{\alpha_2 + \beta_2}$ and variance $k^2\frac{\alpha_2 \beta_2}{(\alpha_2 + \beta_2)^2(\alpha_2 + \beta_2+1)}$

For the means and variances to match, you need $\frac{\beta_2}{\alpha_2(\alpha_2 + \beta_2+1)} = \frac{ \beta_1}{\alpha_1(\alpha_1 + \beta_1+1)}$ which is unlikely, and even then in non-trivial cases would not guarantee both followed a Beta distribution