I am having the following expression:
$$P = \text{Pr}\biggl(h_0 h_1 < \frac{u_1}{a_1 \rho - a_2 \rho u_1}\biggr)\tag{1}$$
where $h_0,h_1$ are independent and identically distributed (i.i.d) random variables, $u_1 = 2.4, a_1 = 0.8, a_2 = 0.2$ and $\rho$ is some positive constant, say $\rho = 1$.
Let $\phi = \frac{u_1}{a_1\rho-a_2\rho u_1}$.
As $h_0$ and $h_1$ are independent, we have
$$P = \int_0^{\infty} F_{h_1}\biggl(\frac{\phi}{y}\biggr)\cdot f_{h_0}(y) \text{d}y\tag{4}$$
where $F_{h_1}(.)$ is CDF of $h_1$ and $f_{h_0}(.)$ is PDF of $h_0$.
My query is that I am not getting how eq. $(4)$ is obtained if $h_1$ and $h_0$ being independent. Any help in this regard will be highly appreciated.
You will need $H_0$ (and probably $H_1$) to be non-negative random variables.
Then, by law of total probabilities,
$$ \begin{align} \Pr\{H_0H_1 < \phi\} &= \int_0^{+\infty} \Pr\{H_0H_1 < \phi|H_0 = y\}f_{H_0}(y)dy \tag{1}\\ &= \int_0^{+\infty} \Pr\{yH_1 < \phi|H_0 = y\}f_{H_0}(y)dy \tag{2}\\ &= \int_0^{+\infty} \Pr\left\{H_1 < \frac {\phi} {y}|H_0 = y\right\}f_{H_0}(y)dy \tag{3}\\ &= \int_0^{+\infty} \Pr\left\{H_1 < \frac {\phi} {y}\right\}f_{H_0}(y)dy \tag{4}\\ &= \int_0^{+\infty} F_{H_1}\left(\frac {\phi} {y}\right)f_{H_0}(y)dy \tag{5}\\ \end{align}$$
where $(1)$ is due to the continuous version of law of total probabilities, conditional on the values of $H_1$ over its support $(0, +\infty)$; $(2)$ is basic property of conditional probabilities; $(3)$ require the support of $H_1$ to be positive so that the direction of the inequality is preserved; $(4)$ is due to the independent assumption so the conditional part can be dropped; $(5)$ is the definition of CDF of the continuous random variables $H_1$.