While reading a paper related to the random mobility model, I am stuck with the following two questions. Let us start with the following equation.
$$P(L<l) = 1 - \exp(-\lambda\pi l^2),$$ where L is the random variable and $l$ is the exact value. L denotes transition length and $L = VT$ $(V,T)$ are velocity and transition-time random varible with $v,t$ as respective values.
I understand that to find pdf of transition-time $T$ find f_T(t), we have to find CDF such that: $$P(T<t) = P(\frac{L}{V}<t) = P(L<vt) = 1-\exp(-\lambda\pi v^2t^2).$$ But since $V$ is also a random variable, how come the authors say
$$P(T<t) = P(\frac{L}{V}<t) = P(L<vt) = 1-\int_\mathcal{V}\exp(-\lambda\pi v^2t^2)dP_V(v).$$ where $\mathcal{V}$ is the range of velocity and $P_V(.)$ is the velocity distribution. (where does $dP_V(v)$ comes from)
The second question is: If the velocity is uniformly distributed on $[v_{min},v_{max}]$, I cannot understand how the pdf of transition time becomes
$$f_T(t) = \frac{g(v_{min}) - g(v_{max})}{(v_{max} - v_{min})t}, t\geq0$$ with $g(x) = xe^{-\lambda \pi t^2x^2} + \frac{1}{\sqrt{\lambda}t}Q(\sqrt{2\pi\lambda t x})$ and $Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty e^{\frac{u^2}{2}}$