# 指數分配與Poisson分配的對偶性

Update:

Poi(λ) → Exp(1/λ)

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Consider a sequence of clain arrivals over time. Let T be the time until the first arrival and for each t. Let Nt be the number of arrivals in the time interval [0,t]

Suppose that Nt~ Poi(λt) for each t

Then for t>=0

Pr(T>t)

=Pr(first claim arrives after time t)

=Pr(no claim arrives in [0,t])

=Pr (Nt=0)

=e^(- λt)

Since Nt~ Poi(λt). On the other hand, for t<0, it is obviously true that Pr(T>t)=1. Hence the survival function of T is given by

S(t)=e^(- λt) for t>=0

S(t)=1 for t<0

So the density function of T is

f(t)=λe^(- λt) for t>=0

f(t)=0 for t<0

which we recognise as the density of the exponential distribution with parameter λ

對不起，這問題不懂。但因為香港用戶不能追蹤台灣問題，所以先在這裡佔個位子。

另外，我有以下意見：

根據 wiki

http://en.wikipedia.org/wiki/Queueing_theory

"... In the Poisson probability distribution, the observer records the number of events that occur in a time interval of fixed length. In the (negative) exponential probability distribution, the observer records the length of the time interval between consecutive events. In both, the underlying physical process is memoryless."

這似乎說明這兩個分佈是用在不同的量度需要，不是有甚麼對偶性吧。

2009-02-08 13:58:45 補充：

如果在指定時間內，總事件的發生數目是Poi(λ)，那麼兩件連續事件發生之間的時間的分佈是Exp(λ)，其平均數分別為λ(事件總數)和 1/λ(時間)。這可以在很多書(如 Ross的)都有說明。

Exp(1/λ)的平均數是λ。

看不出和Poi(λ)有任何關係。

2009-02-13 19:09:50 補充：

應指出使用 memoryless property，所以証明 Interarrival time 是Exp(λ)時，只需証明 first arrival time 是Exp(λ)就足夠了。

但到底兩個要量度的東西不同，一個是"事件數"，另一個是"時間"，我不認為會有滿意答案。