Aggregate arrivals from a Poisson Process

The inter-arrival time of a Poisson Process, $t$, conforms to the exponential distribution, so the probability density function for $t$ is $f(t)=λe^{−λt},~t>0$. ($λ$ is the arrival rate of the Poisson Process.)

Next we aggregate the requests according to the following pattern: from the first arrival, within the fixed-length time interval $T$, the requests in this interval are aggregated to the first arrival. In other words, the requests in this interval are removed except the first one. This procedure repeats for the rest arrivals. The following figure illustrates this aggregation pattern. My question is, what is the distribution of the inter-arrival times after the request aggregation?

Memorylessness might suggest that if the inter-arrival time in the second process is $X$ then it is simply shifted up by $T$ so has density $$g(x)=\lambda e^{−\lambda(x-T)} ,~x\gt T$$
• Clearly because the situation after the waiting time $T$ is exactly the same as at time $t=0$. Jun 30 '15 at 7:28
• @Tom-Tom Let $Y$ denote the waiting time immediately after T for the next request, because of the memorylessness, $Y$ has the exactly same distribution of the original inter-arrival distribution of the Poisson process. This is what you mean, right? Now, I can derive the distribution of $X=T+Y$ by $P(X<x)=P(T+Y<x)=P(Y<x-T)=1-e^{-\lambda (x-T)}$. This is what I thought. Jun 30 '15 at 8:15
• What if the original inter-arrival time $t$ conform to a general distribution? I have a follow-up question here: math.stackexchange.com/questions/1344277/… Would you please take a look? Jun 30 '15 at 8:23