Joint probability in the given equation One of the equation in an article I am reading says, 
\begin{align}
& 1 - \Pr\left(\left\Vert \boldsymbol{h}_m\right\Vert^2>\tau, \  |h_s|^2> \frac{\tau \left(1 + \gamma\left\Vert \boldsymbol{h}_m\right\Vert^2\right)}{\gamma\left( \left\Vert \boldsymbol{h}_m\right\Vert^2 - \tau\right)} \right) \\
= {} & \int_0^\tau f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y)\, dy + \int_\tau^\infty f_{\left\Vert \boldsymbol{h}_{m}\right\Vert^{2}} (y) \, F_{|h_s|^2} \left( \frac{\tau(1 + \gamma y)}{\gamma(y - \tau)}\right)\, dy \tag{1}
\end{align}
where $\boldsymbol{h}_{m}$ is a column vector with elemets distributed as $\mathcal{CN}(0, \Omega)$ and $h_s$ is another random variable which is maximum of $S$ random variables each distributed as $\mathcal{CN}(0, \Omega_S)$.
What I know is 
$$1 - \Pr\left( \left\Vert \boldsymbol{h}_m\right\Vert^2 > \tau \right) = \int_0^\tau f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y)\, dy \tag{2}$$
I do not understand how the equality in $(1)$ has been obtained? Can someone please explain. Thanks!
 A: It follows from the total probability law (which is essentially the law of total expectation):
$$
\Pr(A)=\int \Pr(A\mid x)f(x)\,dx, \quad \text{ or } \quad \operatorname{E}(1_A)=\operatorname{E}(\operatorname{E}(1_A\mid X)).
$$
In your particular case
\begin{align}
1 - {}&\Pr \left(\left\Vert \boldsymbol{h}_m\right\Vert^2>\tau, \  |h_s|^2> \frac{\tau \left(1 + \gamma\left\Vert \boldsymbol{h}_m\right\Vert^2\right)}{\gamma\left( \left\Vert \boldsymbol{h}_m\right\Vert^2 - \tau\right)} \right) \\
&=1-\int_\tau^\infty\Pr\left(|h_s|^2> \frac{\tau \left(1 + \gamma y\right)}{\gamma\left( y - \tau\right)}\right) f_{\left\Vert\boldsymbol{h}_m \right\Vert^2} (y)\, dy\\
&=1-\int_\tau^\infty\left(1-F_{|h_{s}|^2}\left(\frac{\tau \left(1 + \gamma y\right)}{\gamma\left( y - \tau\right)}\right)\right) f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y)\, dy\\
&=1-\int_\tau^\infty f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y)\, dy+\int_\tau^\infty F_{|h_{s}|^{2}}\left(\dfrac{\tau \left(1 + \gamma y\right)}{\gamma\left( y - \tau\right)}\right)f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y)\, dy\\
&= \int_0^\tau f_{\left\Vert \boldsymbol{h}_{m}\right\Vert^2} (y)\, dy + \int_\tau^\infty f_{\left\Vert \boldsymbol{h}_m\right\Vert^2} (y) \, F_{|h_{s}|^2} \left( \frac{\tau(1 + \gamma y)}{\gamma(y - \tau)}\right)\, dy
\end{align}
