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i started to study about poisson process and i having a problem with the next question:

M(t) is poisson process with with parameter -x.
Ti is the first time that M(Ti)=i.

prove that (Ti+2 - Ti) has a Erlang Distribution with parameter x,k=2.

i understand why the distribution of (Ti+1 - Ti) has Exp Distribution , but im not sure what do to when i have an arrival in my time interval. hope it was clear. thanks.

Guy
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2 Answers2

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HINT

$T_{i+2}-T_i = (T_{i+2}-T_{i+1})+(T_{i+1}-T_i)$

spaceisdarkgreen
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$T_{i+2}-T_i\stackrel{d}{=}T_2$

If $\lambda$ denotes the parameter of the Poisson process then:$$P(T_{i+2}-T_i>t)=P(T_2>t)=P(N_t\leq1)=e^{-\lambda t}\left[\frac{(\lambda t)^0}{0!}+\frac{(\lambda t)^1}{1!}\right]=e^{-\lambda t}[1+\lambda t]$$

drhab
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  • why you are allowed to say that i = 0? i mean , why looking at the first 2 arrival (T2-T0) are the same as (Ti+2-Ti)? thanks! – Guy Jun 23 '18 at 08:49
  • It can be shown that $T_i=E_1+E_2+\cdots+E_i$ where the $E_i$ are iid rv's having exponential distribution. In that context $T_{i+2}-T_i=E_{i+1}+E_{i+2}$ where $E_{i+1}+E_{i+2}$ and $E_1+E_2=T_2$ have equal distribution. – drhab Jun 23 '18 at 09:04