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Suppose bus arrivals at a station represent a Poisson process with intensity $λ$. A passenger arrives at the station at a fixed time $t$. Define $U$ as the time she has to wait and $V$ as the time interval between the last passage of a bus and her arrival.

a) Determine the laws of $U$ and $V$ .

b) Are $U$ and $V$ independent?

c) Calculate $E(U + V)$. Why is the result surprising?

I know that $E(U)=E(V)=E(U+V)=\frac{1}{λ}$ and that $U$ and $V$ are independent but how can I prove that?

c) Since interarrival times are independent and obey the $\operatorname*{Exp}(λ)$ distribution which is $\frac{1}{λ}$

Thanks for your help in advance!!

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    Would you be surprised if in fact $E[U+V] = \dfrac2\lambda$ despite being an interarrival time? – 2017-02-04
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    For a) and b), it is true that $U$ and $V$ are independent and both have exponential distribution of rate $\lambda$. This gives $\Bbb{E}(U) = \Bbb{E}(V) = \frac{1}{\lambda}$. Then we must have $$\Bbb{E}(U+V) = \frac{2}{\lambda}.$$ It may take time to realize that $U+V$ does not have exponential distribution. Loosely speaking, this is because we know that certain fixed time $t$ lies between these two arrivals. This information serves as conditioning and consequently the distribution is affected. – 2017-02-04
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    Have a look to (http://stats.stackexchange.com/questions/122722/please-explain-the-waiting-paradox) – 2017-02-04

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