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Consider a Poisson process with rate $\lambda$ in a given time interval $[0,T]$. The inter-arrival time between successive arrivals is negative exponential distributed with mean $\frac{1}{\lambda}$ such that $X_1 >0$, and $\sum_{i=1}^\text{Last} X_i < T$, where $X$ represents inter-arrival time.

What about the distribution of time between Last arrival and ending time $T$? Is it also negative exponential distributed and has a mean value of $\frac{1}{\lambda}$? Can we study time segment $[0,T]$ of Poisson process in the backward direction too? In the forward direction, time between $t=0$ and first arrival is negative exponential distributed. In the backward direction, Last arrival is the first arrival and is the time between $t=T$ and Last is also negative exponential distributed. Is there any way to justify this? or some reference?

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  • $\begingroup$ The conditional distribution of the time until the first arrival, given that the number of arrivals before time $T$ is exactly $k,$ has a scaled beta distribution. $\endgroup$ – Michael Hardy May 13 '18 at 11:08
  • $\begingroup$ You've altered the question, so that it now includes the csae where the number of arrivals during $[0,T]$ is $0.$ In that case, how will you define the time from the "last" arrival until $T\text{?} \qquad$ $\endgroup$ – Michael Hardy May 13 '18 at 11:39
  • $\begingroup$ when there is no arrival occurs between $[0,T]$, in that case, we assume that time from the "last" arrival until T is exactly $=T$ and it has a probability $e^{-\lambda T}$. $\endgroup$ – Hallian1990 May 13 '18 at 11:54
  • $\begingroup$ Do you likewise assume, for the same reason, that in that case the time until the first arrival is $T \text{ ?} \qquad$ $\endgroup$ – Michael Hardy May 13 '18 at 12:06
  • $\begingroup$ Unfortunately not. Except my assumption for last time as $T$ when you pointed case there is no arrival in $[0,T]$, all rest things are completely according to Poisson. I mean inter-arrival time strictly follow negative exponential distribution. $\endgroup$ – Hallian1990 May 13 '18 at 12:13
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Here's an answer to the question as last modified.

First suppose that on the entire real line, not just on $[0,\infty),$ we assign to each interval $(a,b)$ a random variable $N_{(a,b)}$ for which

  • $\Pr(N_{(a,b)} = n) = \dfrac{(\lambda(b-a))^n e^{-\lambda(b-a)}}{n!}$ for $n=0,1,2,3,\ldots,$ and
  • For intervals $A,B,C,\ldots$ no two of which intersect each other, $N_A,N_B,N_C,\ldots$ are independent.

Then let $X_1$ be the time of the first arrival after time $0,$ let $X_2$ be the time from then until the next arrival after that, and so on. Then we have $$ \Pr(X_n > x) = e^{-\lambda x} \text{ for } x\ge0. $$ Now consider the time $Y$ from the last arrival before time $T>0$ until time $T.$ For $x\ge0$ we have $$ \Pr(Y>x) = \Pr(\text{no arrivals during } [T-x,T]) = \frac{(\lambda x)^0 e^{-\lambda x}}{0!} = e^{-\lambda x}. $$ I.e. it has the same distribution that any one of the inter-arrival times has.

And then if you want to truncate it at $x=T,$ as suggested in comments under the question, you can modify that accordingly.

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  • $\begingroup$ These $X_n$ and $Y$ are exponentially distributed. But do these have expectation value $E[X]=\frac{1}{\lambda}$? Dont we need an additional $\lambda$ with this $e^{-\lambda x}$? $\endgroup$ – Hallian1990 May 14 '18 at 2:06
  • $\begingroup$ @Hallian1990 : The probability density function of a random variable $X$ is $x\mapsto\lambda e^{-\lambda x}$ for $x\ge0,$ (with $\lambda$ appearing as a factor of the function as a whole) if and only if $\Pr(X>x) = e^{-\lambda x}$ for $x\ge0$ (without $\lambda$ appearing as a factor of this function as a whole). $\qquad$ $\endgroup$ – Michael Hardy May 14 '18 at 2:34
  • $\begingroup$ Thanks. Moreover, I was just thinking that, is not it the case that we already assume Y to be exponentially distributed before proving it? $\endgroup$ – Hallian1990 May 14 '18 at 7:07
  • $\begingroup$ @Hallian1990 : No, I'm not assuming $Y$ is exponentially distributed; I'm proving it by recalling that the number of arrivals during a certain time has a Poisson distribution. $\endgroup$ – Michael Hardy May 14 '18 at 8:15
  • $\begingroup$ Dear @Michael. I am sorry for reaching you too much. I agree and that is correct explanation for proving it. However, I am not sure if we truncate at $x=T$, how the expressions would like like or remains same. Or I have to divide above two probabilities i.e., $P(X_n >x)$ and $P(Y >x)$ with $P(x<=T)=1-e^{-\lambda T}$ in order to normalize. Is it correct? $\endgroup$ – Hallian1990 May 14 '18 at 11:59
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What appears below answers the question as originally posted, where it was stated that the number of arrivals during $[0,T]$ is equal to a given number $k.$

Suppose the number of arrivals during $[0,T]$ is exactly $k\ge 1.$ You've called the inter-arrival times $X_i,$ so that $X_1$ is the time of the first arrival, $X_2+X_2$ is the time of the second arrival, and so on. For $t\ge 0,$ let $N_t$ be the number of arrivals before time $t,$ so you're saying $N_T=k.$

What, then, is the conditional distribution of $X_1$ given that $N_T=k$?

\begin{align} & \Pr(X_1\le t\mid N_T=k) = \Pr(N_t\ge 1 \mid N_T=k) = 1-\Pr(N_t=0\mid N_T=k) \\[10pt] = {} & 1 - \frac{\Pr(N_t=0\ \&\ N_T=k)}{\Pr(N_T=k)} = 1-\frac{\Pr(N_t=0 \ \&\ N_T -N_t =k)}{(\lambda T)^k e^{-\lambda T} /k!} \\[10pt] = {} & 1-\frac{\Pr(N_t=0)\Pr(N_T-N_t = k) }{(\lambda T)^k e^{-\lambda T} /k!} = 1 - \frac{e^{-\lambda t} \cdot (\lambda(T-t))^k e^{-\lambda(T-t)}/k!}{(\lambda T)^k e^{-\lambda T} /k!} \\[10pt] = {} & 1 - \frac{(T-t)^k}{T^k}. \end{align}

(This does not depend on $\lambda.$)

In a similar way you can show that $T - \sum_{i=1}^k X_i$ has the same conditional distribution given that $N_t=k.$

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