If there are three random variables and three related thresholds, how to simplify the following expression by summation or multiply or other operators? Thank you.
$\mathbb{E}[f_1(\cdot)|H_1>\theta_1,H_2>\theta_2,H_3>\theta_3]\mathbb{P}[H_1>\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3]+\mathbb{E}[f_2(\cdot)|H_1\leq\theta_1,H_2>\theta_2,H_3>\theta_3]\mathbb{P}[H_1\leq\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3]+\mathbb{E}[f_3(\cdot)|H_1>\theta_1,H_2\leq\theta_2,H_3>\theta_3]\mathbb{P}[H_1>\theta_1]\mathbb{P}[H_2\leq\theta_2]\mathbb{P}[H_3>\theta_3]+\mathbb{E}[f_4(\cdot)|H_1>\theta_1,H_2>\theta_2,H_3\leq\theta_3]\mathbb{P}[H_1>\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3\leq\theta_3]+\mathbb{E}[f_5(\cdot)|H_1\leq\theta_1,H_2\leq\theta_2,H_3>\theta_3]\mathbb{P}[H_1\leq\theta_1]\mathbb{P}[H_2\leq\theta_2]\mathbb{P}[H_3>\theta_3]+\mathbb{E}[f_6(\cdot)|H_1\leq\theta_1,H_2>\theta_2,H_3\leq\theta_3]\mathbb{P}[H_1\leq\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3\leq\theta_3]+\mathbb{E}[f_7(\cdot)|H_1>\theta_1,H_2\leq\theta_2,H_3\leq\theta_3]\mathbb{P}[H_1>\theta_1]\mathbb{P}[H_2\leq\theta_2]\mathbb{P}[H_3\leq\theta_3]+\mathbb{E}[f_8(\cdot)|H_1\leq\theta_1,H_2\leq\theta_2,H_3\leq\theta_3]\mathbb{P}[H_1\leq\theta_1]\mathbb{P}[H_2\leq\theta_2]\mathbb{P}[H_3\leq\theta_3]$
where $f_i(\cdot)$ is the function of $H_1, H_2$ and $H_3$ and them are different.
You can start by factorization. For example for the first two terms: $\mathbb{P}[H_1>\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3]+\mathbb{P}[H_1\leq\theta_1]\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3]=(\mathbb{P}[H_1>\theta_1]+\mathbb{P}[H_1\leq\theta_1])\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3]=\mathbb{P}[H_2>\theta_2]\mathbb{P}[H_3>\theta_3].$