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I need your generous urgent help!!

My Problem is ,

Assume that there is a Poisson random process which has a rate of $\lambda$ that produce samples which has an exponentially distributed inter-arrival times, t (i.e distribution function given by $\lambda e^{-\lambda t}$. Then there is another deterministic process which produces samples at a rate of 1 sample in every $T$ time interval (i.e the sample-inter arrival times of this process are not random and you get 1 sample in every $T$ time). Now assume both process start together and we measure the inter arrival times of the samples of resulting process irrespective of which process they are generated from)

I need to get the Distribution of the inter-arrival times of the resulting process?? can any one reply me how to do this!! Urgernt!!!

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In a time interval $\tau$, you know the probability of having $k$ arrivals from the Poisson process, call it $P(k,\tau)$. On the other hand, assuming the second process is independent of the first, you will have $[\tau/T]$ (round down) guaranteed jumps plus another extra jump which has probability $(T-[\tau/T])/T$, call this $Q(\tau)$. Thus the probability of seeing a total of $k$ jumps in time interval $\tau$ is:

$$P(k-[\tau/T],\tau)(1-Q(\tau))+P(k-[\tau/T]-1,\tau)Q(\tau).$$

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  • $\begingroup$ Thanks a lot Alex R for your answer!! Actually the above expression gives the probability of having k number of arrivals in time $\tau$ but how can I derive the inter-arrival times distribution from this?? $\endgroup$ – Prasad Perera Aug 9 '14 at 3:37
  • $\begingroup$ Call the interarrival time $T$, that is, the chance of seeing the first jump at time $T$. Then $P(T>t)$ is precisely the probability that there are zero jumps in time $t$. Can you finish from here? $\endgroup$ – Alex R. Aug 9 '14 at 7:03
  • $\begingroup$ if $k=0$ in $P(k-\frac{\tau}{T},\tau)$ then it can not be defined? and how can I call the inter arrival time $T$? Sorry about asking lot of questions still I cant understand the method? $\endgroup$ – Prasad Perera Aug 9 '14 at 7:55
  • $\begingroup$ It's defined and is equal to 0 when the argument is negative $\endgroup$ – Alex R. Aug 9 '14 at 16:16
  • $\begingroup$ Alex R Sorry mate I still cant work it out!! COuld you kindly show me couple of steps beyond Pls?? $\endgroup$ – Prasad Perera Aug 9 '14 at 22:41

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