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Poisson question

Note that both diagrams are refferring to the same problem. The difference is that I'm not sure if graph I'm supposed to visualize is the PDF or the CDF, so I drew them both and hope someone will correct me.

In the picture, The Xi are the poisson arrival events. So X1 is the first arrival, X2 the second etc. The parameter is lambda.

The Ti are the interarrival times of successive events. So T1 is the time between the first event and the beginning of time. T2 the time btwn the 2nd event and then first etc. The Ti have an exponential distribution because with rate lambda because we are in a poisson process.

My question is about whether I'm thinking about this correctly. Every time an event happens, is it like we are starting a new poisson process or resetting the exponential distribution? Kind of like, when an event happens, has the process or the exponential distribution renewed? If so, does it make sense to see the PDF or CDF of the Xi as the same PDF/CDF for all Xi? What I mean is, say T1 has an exponential distribution like the one drawn. Then will T2 have the same distribution (since it has the same parameter lambda) except that it will be squished/scaled? (In diagram A I am fitting the PDF of the exponential distribution between each successive event. In diagram B I am fitting the CDF of the exponential distribution between each successive event.)

Also, is diagram A or B correct? Or are they both wrong? Any other problems that you see with my thought process please let me know.

Thank you for your help and my apologies for my bad diagrams in advance.

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  • $\begingroup$ Neither the CDF nor the PDF has a vertical tangent. You shouldn't make it look as if they do. $\qquad$ $\endgroup$ – Michael Hardy Apr 30 '16 at 19:16
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    $\begingroup$ $T_2$ itself is actually not exponentially distributed; it is a sum of two iid exponentials which is not exponential. (Intuitively, as you go along the process, when $T_1$ passes, the process remembers that this has happened, and so the memoryless property of the exponential distribution becomes violated.) But $T_2-T_1$ and $T_1$ are iid. $\endgroup$ – Ian Apr 30 '16 at 19:17
  • $\begingroup$ @Ian - it depends on whether $T_2$ is the arrival time of the second event, or the interarrival time between the first and second events $\endgroup$ – Henry Apr 30 '16 at 20:31
  • $\begingroup$ To my knowledge, the standard notation is just that $T_2$ is the arrival time for the second event as Ian mentioned, even if the author meant to write the interarrival time. $\endgroup$ – Kevin Apr 30 '16 at 20:42
  • $\begingroup$ As Kevin said, that is fairly standard notation, though I now see that I misread the labeling of the graphs in the OP. $\endgroup$ – Ian Apr 30 '16 at 21:49
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As Ian said, $T_n$ for $n>1$ is actually Gamma Distributed with $k=1$. In fact, the Exponential Distribution is a special case with $k=1$. You did mention you were using $T_n$ for the interarrival times, but that isn't the standard notation - $T_n - T_{n-1}$ is the interarrival time, where $T_n$ is simply the total arrival time for $x_n$.

I think you're confusing the pdfs with the actual arrival times, also. In your diagram, if you are given the arrivals $x_1, x_2, \dots$ then the interarrival times (given these arrivals) are no longer random. They are simply the intervals that you have denoted, there is no "pdf" at all!

The pdf of the interarrival times are actually all the same: the exponential distribution. The total arrival times will have pdfs that all intersect the $y$-axis. Finally, the pdfs that you have drawn are (roughly) the conditional pdfs, given all previous arrivals - that distribution will start at the last known arrival, not the $y$-axis.

Finally as Michael mentioned, the functions do not have vertical tangents either!

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  • $\begingroup$ Thanks for your response. Your second paragraph made me realize why my diagram is incorrect. I'm gonna try again and make a new post soon and I'll add the link in the comment so you can check it out. Thanks! $\endgroup$ – Joe Apr 30 '16 at 21:27

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