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Consider a Poisson process with rate $\lambda$ and let $L$ be the time of the last arrival in the interval $[0,t]$, with $L=0$ if there was no arrival.

How can I prove that t-L has exponential distribution with rat $\lambda$? I tried to prove it by the following relation \begin{equation} P(t-L>x)=P(N(x)=0) \end{equation} However it leads us to a correct answer but I think this relation can not be true. because $P(N(x))=0$ doesn't have any information about t! Actually we know that $t-L>x$ means that $N(x)=0$ but the reverse is not obvious.So all we can say is: $P(t-L>x)<=P(N(x)=0)$. The purpose of this discussion is to find $E[t-L]$ by the knowledge of distribution of $L$ or $t-L$!

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  • $\begingroup$ I seem to remember an answer that Did (a very high-reputation user on math.SE) posted on math.SE to the effect that while the time of the next arrival after any given time $t_0$ is indeed exponentially distributed, the time of the most recent arrival before $t_0$ is not exponentially distributed. But I don't seem to be able to find that answer now. $\endgroup$
    – Dilip Sarwate
    Commented Oct 18, 2015 at 22:05
  • $\begingroup$ The answer referred to in my comment above might be this one where the time elapsed since the last arrival is claimed to be of the form $\min(t,X)$ where $X$ is an exponential random variable with parameter $\lambda$ (while as everyone knows and agrees that the time till the next arrival after $t$ is exponentially distributed with parameter $\lambda$). $\endgroup$
    – Dilip Sarwate
    Commented Oct 18, 2015 at 22:22

2 Answers 2

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Let us define a random variable T=t-L

$P(T<x)$ $\\=1-P(\text{no arrival happened in the interval } [t-x, t])$ $\\= 1-P(\text{no arrival happened in the interval } [0, x])$ $\\= 1-\frac{\lambda^0}{0!} \exp{(-\lambda x)}$ $\\= 1-\exp{(-\lambda x)}$

This is the cdf of the exponential distribution.

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  • $\begingroup$ Consider the case that there is an arrival between L and t-x but there is no.arrival between t-x and t. So your first equality is not true! $\endgroup$
    – A.Soleimani
    Commented Oct 11, 2015 at 17:59
  • $\begingroup$ What do you mean there is an arrival between L and t-x? Wasn't L defined as the last arrival? $\endgroup$
    – michal
    Commented Oct 11, 2015 at 18:40
  • $\begingroup$ what you said is: $P(t-L>x)=P(no~ arrival~ in ~[t-x,t])$. we know that if $A\Rightarrow B$ then $P(A)\le P(B)$.in our discussion we know that $t-L>x\Rightarrow no~ arrival ~in~ [t-x, t]$ so $P(t-L>x)\le P(no~ arrival ~in [t-x,t])$ but $no ~arrival ~in [t-x,t]$ doesn't mean $t-L>x$ or equivalently $L<t-x$. I mean the reverse is not true. $\endgroup$
    – A.Soleimani
    Commented Oct 12, 2015 at 6:26
  • $\begingroup$ That is not true what you said. P(t-L > x) only says about what happens in the subinterval [t-x, t]. The second implication holds too. $\endgroup$
    – michal
    Commented Oct 12, 2015 at 11:01
  • $\begingroup$ x is bounded. The last period T does not follow exponential distribution, because the period between L and next arrive will be the exponential distribution, and T is always less than this period. $\endgroup$
    – Tony416
    Commented Sep 10, 2020 at 23:29
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My Professor (Dr.Pasha) has guided me to solve the problem as follows: Theorem 2.14.(Richard Durrett) If we condition on N(t), then $T_1,T_2,..,T_n$ have uniform distribution in [0,t]. thus $P(\max \{T_1,...,T_n\}<s)=\frac{s^n}{t^n}$ so $E[L|N(t)=n]=\frac{nt}{n+1}$

$E [L] = \sum_{n=0}^{\infty} E [ L \,| \, N(t) = n ] \, P \{ N(t)=n \} =\sum_{n=0}^{\infty} E [\max \{T_1,...,T_n\}\,| \, N(t) = n ] \, P \{ N(t)=n \} = \sum_{n=0}^{\infty} \frac{n t}{n+1} \, P \{ N(t)=n \} = \sum_{n=0}^{\infty} \left( 1 - \frac{1}{n+1} \right) t \, \frac{ (\lambda t)^n}{n!} e^{-\lambda t} = t e^{- \lambda t} \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!} - t e^{- \lambda t} \sum_{n=0}^\infty \frac{1}{n+1} \frac{ ( \lambda t)^n}{n!}.$\

Now after a little computation you find that

$\sum_{n=0}^\infty \frac{1}{n+1} \frac{ ( \lambda t)^n}{n!} = \frac{1}{ \lambda t} ( e^{ \lambda t} - 1 ).$\

Thus

$E [ L ] = t - t e^{- \lambda t} \frac{1}{\lambda t} ( e^{ \lambda t} - 1 ) = t - \frac{1}{\lambda } (1 - e^{ \lambda t} ).$

Therefore we indeed obtain

$E [ t - L ] = t - E [ L ] = t - \left( t - \frac{1}{\lambda } (1 - e^{ \lambda t} ) \right) = \frac{1}{\lambda } (1 - e^{ \lambda t} ).$

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  • $\begingroup$ So what was wrong with my answer? $\endgroup$
    – michal
    Commented Oct 19, 2015 at 15:35

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