Computing conditional probability with uniform distribution

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Suppose that parameter $\theta$ characterizes the probability that an individual takes an action. In the population, $\theta\sim U[0,1]$. Individual i has probability $\pi(\theta_i)$ of taking the action. Suppose further that

\begin{equation} \pi(\theta_i)=\begin{cases} \pi_1(\theta_i) & if & \theta_i\leq c\\ \pi_2(\theta_i) & if & else \end{cases} \end{equation}

What is the correct formula for the average probability? Is it \begin{eqnarray} \pi&=&\int \pi_i(\theta_i) di\\ &=&\int_{0}^c \pi_1(\theta_i) di +\int_{c}^1 \pi_2(\theta_i) di \end{eqnarray}

or is it

\begin{eqnarray} \pi&=&\int \pi_i(\theta_i) di\\ &=&Prob(\theta\leq c)\int_{0}^1 \pi_1(\theta_i) di +Prob(\theta\geq c)\int_{c}^1 \pi_2(\theta_i) di \end{eqnarray}

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It is the first option. You have to integrate the conditional probability with respect to the density of the random variable being conditioned on. Your second option is integrating the density separately from the conditional probability and taking the product, which is incorrect.

A minor notational issue: the variable of integration is $\theta_i$, not $i$, so the integral should be w.r.t. $d\theta_i$, not $di$.