You are waiting in line for a taxi. There are two people ahead of you. Taxis arrive in a Poisson process, at average rate of one every two minutes
Let T$_2$=time until 2nd taxi arrives
I'm trying to find the CDF for T$_2$, The answer I'm given is F$_T$$_2$(t)=1-(1+λt)e$^($$^-$$^λ$$^t$$^)$
Can anyone explain or give me a hint on how to do this?
Let $\tau_n$ be the interarrival times of the taxis, i.e. $\tau_n\stackrel{i.i.d.}\sim\mathsf{Exp}(\lambda)$, where we may interpret $\lambda$ as $\frac12\mathrm{min}^{-1}$ or $30\mathrm{h}^{-1}$. Then $T_2=\tau_1+\tau_2$, so the density of $T_2$ is (for $t>0$) \begin{align} f_{T_2}(t) &=f_{\tau_1}\star f_{\tau_2}(t)\\ &= \int_0^t f_{\tau_1}(s)f_{\tau_2}(t-s)\ \mathsf ds\\ &= \int_0^t \lambda e^{-\lambda s}\lambda e^{-\lambda(t-s)}\ \mathsf ds\\ &= \lambda e^{-\lambda t} \int_0^t \lambda\ \mathsf ds\\ &= (\lambda t)\lambda e^{-\lambda t}. \end{align} Integrating yields \begin{align} F_{T_2}(t) &= \int_0^t f_{T_2}(s)\ \mathsf ds\\ &= \int_0^t(\lambda s)\lambda e^{-\lambda s}\ \mathsf ds\\ &=\lambda s e^{-\lambda s}|_t^0 + \int_0^t \lambda e^{-\lambda s}\ \mathsf ds\\ &= -\lambda te^{-\lambda t} + 1 -e^{-\lambda t}\\ &= 1 - (1+\lambda t)e^{-\lambda t},\ t>0. \end{align}