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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?

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  • $\begingroup$ @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. $\endgroup$ – RLC Jul 27 at 14:50
  • $\begingroup$ 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. $\endgroup$ – Benjamin Aug 31 at 5:59

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