Length of holding interval in renewal theory I have a Poisson process $(N_t)_{t\ge 0}$ of rate $\lambda$. This process refers to the number of renewals up to time $t$. I also define $S_{N_t}$ which refers to the length of the holding interval containing the time point $t$.
What about the disitrbution of $S_{N_t}?$ My notes say the disitrbution function is $$\mathbb P(S_{N_t}\le x)=1-(1+\lambda\min\lfloor t,x\rfloor)e^{-\lambda x}, x\ge 0$$
but where is this coming from?
 A: There might be an easier way, but you can do all the calculations by hand.
By definition, $S_{N_t}=T_{N_t+1}-T_{N_t}$. I will also denote by $\xi_n=T_{n+1}-T_n$ defined for $n\ge0$, which follows an exponential distribution of parameter $\lambda$. Also, note that
$$T_n=\sum_{k=0}^{n-1}\xi_k\sim\Gamma(n,\lambda),$$
as it is the sum of $n$ independent exponential random variables. Also by definition (or property) of the Poisson process, $T_n$ is independent of $\xi_n$.
In the calculations, I will assume $x\ge t$, although you proceed similarly if $x<t$.
\begin{align*}
\mathbb{P}(S_{N_T}\le x)&=\sum_{n\ge 0}\mathbb{P}\left(T_{N_t+1}-T_{N_t}\le x, N_t=n\right)\\
&=\sum_{n\ge 0}\mathbb{P}\left(T_{n+1}-T_{n}\le x, T_n\le t< T_{n+1}\right)\\
&=\sum_{n\ge 0}\mathbb{P}\left(\xi_n\le x, T_n\le t< T_n+\xi_n\right)\\
&=\sum_{n\ge 0}\int\int1_{u\le x}1_{v\le t<v+u}\lambda e^{-\lambda u}\frac{v^{n-1}}{(n-1)!}\lambda^ne^{-\lambda v}\,\mathrm{d}u\mathrm{d}v\\
&=\sum_{n\ge 0}\int_{v=0}^t\left(e^{-\lambda(t-v)}-e^{-\lambda x}\right)\frac{v^{n-1}}{(n-1)!}\lambda^ne^{-\lambda v}\mathrm{d}v\\
&=\sum_{n\ge 0}\left(\int_{v=0}^t\frac{v^{n-1}}{(n-1)!}\lambda^ne^{-\lambda t}\mathrm{d}v-\int_{v=0}^t\frac{v^{n-1}}{(n-1)!}\lambda^ne^{-\lambda v}e^{-\lambda x}\mathrm{d}v.\right)
\end{align*}
Now the term for $n=0$ is simply
$$\mathbb{P}(\xi_0\le x, t<\xi_0)=e^{-\lambda t}-e^{-\lambda x},$$
and thus we can apply the summation inside of the (bounded) integrals to obtain
\begin{align*}
\mathbb{P}(S_{N_T}\le x)&=e^{-\lambda t}-e^{-\lambda x}+\int_0^t\lambda e^{-\lambda t}e^{\lambda v}\mathrm{d}v-\int_0^t\lambda e^{-\lambda x}\mathrm{d}v\\
&=e^{-\lambda t}-e^{-\lambda x}+e^{-\lambda t}(e^{\lambda t}-1)-\lambda te^{-\lambda x}\\
&=1-e^{-\lambda x}(1+\lambda t),
\end{align*}
which is the required result.
