# Calculating probabilities of ordinary renewal process (renewal time)

Let $(X_n : n \in \mathbb{N}_0)$ be a sequence of independent positive random variables which are identially distributed.

Define the sequence $\mathcal{S}=(S_n : n \in \mathbb{N})$ where $S_1 := X_0$ and $S_{n+1}:=S_n+X_n$. The random variable $S_n$ is called the $nth$ renewal time, whilst the time duration $X_n$ is called the $nth$ renewal interval.

Define the random variable of the number of renewals until time $t$ by $N_t:=\max\{n\in \mathbb{N} : S_n \le t \}$. Then the continuous time process $\mathcal{N} = (N_t : t \ge 0)$ is called a renewal process. $X_0$ is called the delay of $\mathcal{N}$. If $X_0$ and $X_1$ have the same distribution, then $\mathcal{N}$ is called an ordinary renewal process. Assume that $\mathbb{P}(X_1 = 0)$ and $\mathbb{E}(X_1) < \infty$.

Given an ordinary renewal process $\mathcal{N} = (N_t : t \ge 0)$, with $\mathbb{P}(X_1=1)= 1/3$ and $\mathbb{P}(X_1=2) = 2/3$, calculate the probablities $\mathbb{P}(N_1 =k)$, $\mathbb{P}(N_2 =k)$, $\mathbb{P}(N_3 =k)$ for all $k \in \mathbb{N}_0$.

What I did was I got $\mathbb{P}(N_t=k) = \mathbb{P}(S_k \le t)$. Using the fact that $S_1 := X_0$ and $S_{n+1}:=S_n+X_n$ and that $X_0$ and $X_1$ have the same distribution function, I got

$\mathbb{P}(N_1 =0) = \mathbb{P}(S_0 \le 1) = ?\\ \mathbb{P}(N_1 =1) = \mathbb{P}(S_1 \le 1) = \mathbb{P}(X_0 \le 1) = \mathbb{P}(X_1 \le 1) = 1/3 \\ \mathbb{P}(N_1 =2) = \mathbb{P}(S_2 \le 1) = \mathbb{P}(S_1 +X_1 \le 1) = \mathbb{P}(X_0 +X_1 \le 1) = ? \\ \mathbb{P}(N_1 =3) = \mathbb{P}(S_3 \le 1) = \mathbb{P}(S_2 +X_2 \le 1) + \mathbb{P}(X_0 +X_1 + X_2 \le 1) = ?$

But how would I calculate the probabilites of the others above? Is what I'm doing on the right track?

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It should be obvious $\Pr(N_1 =0) = 2/3$, $\Pr(N_1 =1) = 1/3$ and $\Pr(N_1 =k) = 0$ for $k \gt 1$. –  Henry Apr 22 '12 at 18:56
Thanks, could you explain how you got $\Pr(N_1 =0) = 2/3$? –  Richard Apr 22 '12 at 20:49
How would I also calculate $\mathbb{P}(N_2 =k)$ and $\mathbb{P}(N_3 =k)$? –  Richard Apr 22 '12 at 21:01
On your first, $\Pr(N_1=0)=Pr(S_1 \gt 1) =\Pr(X_0 \gt 1)$. Similarly $\Pr(N_2=k) = \Pr(S_{k} \le 2 \text{ and } S_{k+1} \gt 2)$ while $\Pr(N_3=k) = \Pr(S_{k} \le 3 \text{ and } S_{k+1} \gt 3)$ which you can calculate looking at possibilities for $X_0, X_1,X_2$. –  Henry Apr 22 '12 at 23:15
Add up all the elementary event that go into $N_2 = k$. For example, starting from 0 you can go 1,2 or 1 3, and $N_2$ is 2 in the first case and 1 in the second.