# Renewal process: escess of life and age of the process independent.

Let $$X_1,X_2,...$$ be non-negative iid continuous random variables. Let $$W_n=X_1+\cdots+X_n$$, with $$W_0=0$$, and $$N(t)=\max\{n\geq 0:W_n\leq t\}$$. We define for this renewal process the excess of life $$\gamma_t=W_{N(t)+1}-t$$ and the age $$\delta_t=t-W_{N(t)}.$$ It is easy to prove that if $$N(t)$$ is a Poisson process then $$\gamma_t$$ and $$\delta_t$$ are independent.

I'm asked to show the converse, i.e., that if $$\gamma_t$$ and $$\delta_t$$ are independent, then $$N(t)$$ is Poisson process.

I'm trying to show that $$X_1$$ has the memoryless property, so $$X_1,X_2,...$$ will be iid exponential. I have found that given the independence condition on $$\gamma_t$$ an $$\delta_t$$, $$P(N(t+y)-N(t+x)=0)=P(N(t)-N(t-x)=0)P(N(t+y)-N(t)=0).$$ From this, if $$x\to t$$, $$P(X_1>t+y)=P(X_1>t)P(N(t+y)-N(t)=0).$$ So I just need to show that $$P(N(t+y)-N(t)=0)=P(X_1>y).$$

Is that an immediate consequence of the hypothesis of independence?

• @Benjamin N(t) is discrete. Is the maximum of a countable set. For example, if X is exponential then N Will be Poisson. So W_{N(t)} is well defined. – RLC Jul 27 at 14:50
• I think you right. the only distribution with memory-less property is exponential distribution, and an equivalent definition of poisson process could be found by using exp-dist and independence. – Benjamin Aug 31 at 5:59