# renewal processes

Let $X_1, X_2,...$ be a discrete renewal process, in which $X_i$ denotes the time in between renewals with distribution: $Pr(X_i=1)=p$ and $Pr(X_i=2)=q=1-p.$ I want to show that the renewal equation $$m(n)=E(N(n))=\frac{n}{1+q}-\frac{q^2}{(1+q)^2}[1-(-q)^n]$$

This is what I know: Let $$F_k(n)=Pr(T_k\le n)$$ Then $$m(n)= \sum_{k=1}^\infty F_k(n)$$ I've noticed that $Pr(T_k \le n)= 0$ for $n\lt k$; $1-q^k$ for $k \le n \lt 2k$, and $1$, for $n \ge 2k$

which, btw im not 100% on. But now I dont know what to do with this since the summation diverges if I plug in $1-q^2$ for $F_k(n)$. (an goes to infinity). Any suggestions?

p.s. writing mathjax for the first time was a pain in the ass experience...

• You did well with mathjax :) I think your solution for $k\leq n < 2k$ is not right. I'm pretty sure it should at least depend on n and k. Also I don't see why the sum diverge since for $k>n$ you only get 0 hence your sum is finite. – wece Mar 10 '16 at 10:57
• the idea behind $1-q^k$ is: the only way $T_k \le n$ doesn't hold on (n,2n) is if $X_i=2$ is the only event that occur throughout [n,2n). I can't think of a way to generalize the idea for any n, regardless of bounds.. – ak87 Mar 10 '16 at 11:14
• For every $k$, $N(n)\geqslant k$ if and only if $X_1+\cdots+X_k\leqslant n$ hence $$E(N(n))=\sum\limits_{k\geqslant1}P(N(n)\geqslant k)=\sum\limits_{k\geqslant1}P(X_1+\cdots+X_k\leqslant n).$$ Decomposing each probability in the series along the values of $X_1$, one gets, for every $n\geqslant2$, $$E(N(n))=P(X_1\leqslant n)+p\sum\limits_{k\geqslant2}P(X_2+\cdots+X_k\leqslant n-1)+q\sum\limits_{k\geqslant2}P(X_2+\cdots+X_k\leqslant n-2),$$ that is, $$m(n)=1+pm(n-1)+qm(n-2).$$ The initial conditions are $m(0)=0$ and $m(1)=p$. Remains to solve this system of equations, can you do that? – Did Mar 10 '16 at 12:04
• yes, thank you. very instructive solution. – ak87 Mar 10 '16 at 12:08