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Define Poisson process is a renewal process in which the interarrival intervals have exponential distribution. $S_n$ is the arrival epoch of the $n$th arrival, $N(t)$ is the number of arrivals in $(0,t]$. I don't understand some equalities in the proof below: enter image description here

In the proof above, $V_t=S_{N(t_1)+1}-t_1$, $Z_i=X_{N(t_1)+i}$. I don't understand the first and the third equalities. I tried to draw a picture for $T_i$'s: enter image description here

According to the picture:

$\bullet$ For the 1st equality, I think the event $\left\{N(t_{l+1})-N(t_l)=k_{l+1}\right\}$ should be the same as the event $\left\{T_{k_2+k_3+\cdots+k_{l+1}}\le t_{l+1}-t_1<T_{k_2+k_3+\cdots+k_{l+1}+1}\right\}$. The picture is the case $l=1$ and $l=2$.

$\bullet$ For the 3rd equality, I totally don't understand. I just can see from the picture that $T_{k}+t_1=S_{N(t_1)+k}$ for all $k$.

I really need help for the explanations of the 1st and 3rd equalities. Any help would be appreciated.

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First Inequality: You have a point. I think that line should be

$$= P(N(t_1)=k_1,\;T_{k_2}\leq t_2-t_1\lt T_{k_{2}+1},\;\ldots,\;T_{k_2+\cdots+k_{m+1}}\leq t_{m+1}-t_1\lt T_{k_2+\cdots+k_{m+1}+1}).$$

I don't know if you mean to imply that the two events you quote are equal, but they are not: $\left\{N(t_{l+1})-N(t_l)=k_{l+1}\right\}$ does not account for the number of events occurring before time $t_l$ whereas $\left\{T_{k_2+k_3+\cdots+k_{l+1}}\le t_{l+1}-t_1<T_{k_2+k_3+\cdots+k_{l+1}+1}\right\}$ does. However, the full intersection of the events either side of that first equality (excluding "$N(t_1)=k_1$") are equal, which is all that really matters.

Third Inequality: The change from "$T$" to "$S$" is due to a shifting of the time frame right by $t_1$ units, so it's like starting at time $t_1$ instead of at time $0$. So, for example, the waiting time for $k_2$ events to occur, which is $S_{k_2}$ by definition, equals $T_{k_2}$ if we consider the process having begun at time $t_1$.

The problem you found in the first equality should carry though to the third equality. The fourth equality will translate in the reverse direction and that corrects the problem so that line needs no change.

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  • $\begingroup$ Could you explain more for the point "$\left\{N(t_{l+1})-N(t_l)=k_{l+1}\right\}$ does not account for the number of events occurring before time $t_l$ whereas $\left\{T_{k_2+k_3+\cdots+k_{l+1}}\le t_{l+1}-t_1<T_{k_2+k_3+\cdots+k_{l+1}+1}\right\}$ does"? $\endgroup$ – Tien Kha Pham Jun 13 '15 at 12:33
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    $\begingroup$ @TienKhaPham $\left\{T_{k_2+k_3+\cdots+k_{l+1}}\le t_{l+1}-t_1<T_{k_2+k_3+\cdots+k_{l+1}+1}\right\}$ says that in the time interval $[t_1,t_{l+1}]$ we have exactly $k_2+k_3+\cdots+k_{l+1}$ events occur, which is the same as event $\left\{N(t_{l+1})-N(t_1)=k_2+k_3+\cdots+k_{l+1}\right\}$. But the event $\left\{N(t_{l+1})-N(t_l)=k_{l+1}\right\}$ says that in the time interval $[t_l,t_{l+1}]$ we have exactly $k_{l+1}$ events occur. $\endgroup$ – Mick A Jun 13 '15 at 13:10
  • $\begingroup$ I understood. I depend so much on the picture, which shows only small cases. Thanks a lot, Mick A. $\endgroup$ – Tien Kha Pham Jun 13 '15 at 13:54

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