Deriving the exponential distribution from the time between event in a Poisson distribution

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[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]


Let [ilmath]X\sim[/ilmath][ilmath]\text{Poi}(\lambda)[/ilmath] for some [ilmath]\lambda\in\mathbb{R}_{>0} [/ilmath]

  • Here supposed that [ilmath]X[/ilmath] models the number of events per unit time - although as with Poisson distribution - any continuum will do


  • Suppose an event happens at [ilmath]t\eq t_0[/ilmath]
    • Let [ilmath]T[/ilmath] be the random variable which is the time until the next event.
      • Let [ilmath]d\in\mathbb{R}_{>0} [/ilmath] be given, so we can investigate [ilmath]\P{T>d} [/ilmath]
      • We are interested in [ilmath]\mathbb{P}\big[\text{no events happening for time in }(t_0,t_0+d)\big]\eq\P{T>d} [/ilmath]
        • Let [ilmath]X'\sim\text{Poi}(\lambda d)[/ilmath] be used to model this interval
          • as if [ilmath]\lambda[/ilmath] events are expected to occur per unit time, then [ilmath]\lambda d[/ilmath] are expected to occur per unit [ilmath]d[/ilmath] of time
          • It is easy to see that [ilmath]\mathbb{P}\big[\text{no events happening for time in }(t_0,t_0+d)\big]\eq\mathbb{P}\big[X'\eq 0\big]\eq e^{-\lambda d} [/ilmath]
        • Thus [ilmath]\P{T>d}\eq e^{-\lambda d} [/ilmath]
          • Or: [ilmath]\P{T\le d}\eq 1-\P{T>d}\eq 1-e^{-\lambda d} [/ilmath]
    • But this is what we'd see if [ilmath]T[/ilmath] followed the exponential distribution with parameter [ilmath]\lambda d[/ilmath]
      • [ilmath]\P{T\le d}\eq 1-e^{-\lambda d} [/ilmath]

Thus we see the time between occurrences of events in a Poisson distribution is exponentially distributed, or memoryless.


Suppose instead [ilmath]t_0[/ilmath] is the start time of the process rather than the last event time, how does this change things?