Renewal process, distribution of time between jumps I have a Poisson process, where times between jumps are random variables from exponential distribution: $T_i \sim Exp(\lambda)$. I'm considering another renewal process in which jumps occur if $T_i > d$ , where $d=const$ is some constant. After second process jumps the first one "resets" so waiting times in second one are independent. My goal is to find CDF of waiting times in second process and Laplace transform of their density.
There is how I tried to approach this problem (got stuck in one moment and I'm not sure if this anyhow close to being correct):
Let's call times between jumps in second process by $W$ then
$$ W = \sum_{i=1}^{N-1}T_i $$
(we assume that we know in advance that the waiting time will be greater than $d$ so we jump right away, not after additonal $d$ time, that's why the sum is to $N-1$), where
$$N=min\{n: T_{n} >d \}$$
So I have a sum of random amount of random variables. I've found that we can 'easly' calculate moment-generating function of $W$ as
$$ M_{W}(t) = M_{N}(M_{T}(t)) $$,
where $M_{X}$ is moment-generating function of $X$. 
Finding this would automaticly give me the answer to the second question as Laplace transform of density is equal to moment-generating function and from this I could also derive the CDF.
$M_{T}$ is easy to find as $T$ is just exponential distribution. My problem is that I'm not sure how to find $M_{N}$. Some ideas were:
$$ P(T_{n} >d | T_{1}<d, ...,T_{n-1}<d) $$,
but this doesn't give me the distribution of the index of jump only the probability that it'll happen in n-th jump (or am I wrong? It seemed correct to me at first but hen I could find use of it);
$$ N = \min(\operatorname{argmax}(T_{n}>d)) $$,
this again doesn't seem right and even if it's right then I don't know how to calculate it.
I would appreciate any kind of help, any tips or information if my approach is correct or if I should do this some other way.
 A: $N$ has discrete distribution with pmf
$$
\mathbb P(N=k) = \mathbb P(T_1\leq d,\ldots, T_{k-1}\leq d, T_k>d)=F^{k-1}(d)(1-F(d)), \ k=1,2,\ldots
$$
where $F(d)$ is CDF of exponential distribution of jumps. This is geometric distribution.
A: As @NCh pointed out, $N$ is geometrically distributed (with parameter $1-e^{-\lambda d}$). Now, for any $t>0$ we have
\begin{align}
\mathbb P(W>t) &= \sum_{n=1}^\infty \mathbb P(W>t\mid N=n)\mathbb P(N=n)\\
&= \sum_{n=1}^\infty \mathbb P\left(\sum_{i=1}^{n-1} T_i>t\right)\mathbb P(N=n)\\
&=\sum_{n=1}^\infty \sum_{k=n-1}^\infty \frac{(\lambda t)^k}{k!} e^{-\lambda t}\left(1-e^{-\lambda d}\right)^{n-1}e^{-\lambda d}\\
&= e^{-\lambda(t+d)} \sum_{n=0}^\infty \left(1-e^{-\lambda d}\right)^n \sum_{k=n}^\infty \frac{(\lambda t)^k}{k!}\\
&= e^{-\lambda(t+d)}\sum_{k=0}^\infty \frac{(\lambda t)^k}{k!}\sum_{n=0}^k \left(1-e^{-\lambda d}\right)^n\\
&= e^{-\lambda(t+d)} \sum_{k=0}^\infty \frac{(\lambda t e^{-\lambda d})^k}{k!}\\
&= e^{\lambda t(e^{-\lambda d}-d)}.
\end{align}
The CDF is $1-e^{\lambda t(e^{-\lambda d}-d)}$, so the density is obtained by differentiating: $f_W(t) = \lambda  \left(e^{-d \lambda }-d\right) \left(e^{\lambda  t \left(d-e^{-d \lambda }\right)}\right)$. We compute the moment-generating function:
\begin{align}
\mathbb E[e^{\theta W}] &= \int_0^\infty e^{\theta t}\lambda\left(e^{-d \lambda }-d\right) \left(e^{\lambda  t \left(d-e^{-d \lambda }\right)}\right)\\
&= \frac{\lambda  \left(1-d e^{d \lambda }\right)}{\lambda - e^{d \lambda } (d \lambda -\theta ) },\quad \theta < \lambda\left(d-e^{-d\lambda} \right).
\end{align}
The Laplace-Stieltjes transform would similarly be computed as
\begin{align}
\mathbb E[e^{-\theta W}] &= \int_0^\infty e^{\theta t}\lambda\left(e^{-d \lambda }-d\right) \left(e^{\lambda  t \left(d-e^{-d \lambda }\right)}\right)\\
&= \frac{\lambda  \left(1-d e^{d \lambda }\right)}{\lambda-e^{d \lambda } (d \lambda +\theta )},\quad \theta > \lambda(e^{-d\lambda}-d).
\end{align}
