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I was wondering if there is a known characterization of an arrival process defined as follows: a "potential" arrival occurs according to a Poisson process of rate L, then a Bernoulli RV with P(1)=B determines whether the arrival officially occurs. The Bernoulli RVs are drawn iid. It would be nice if this were itself a Poisson process, but I don't think it is.

Thanks.

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It is a standard exercise to show that this is a Poisson process.

Let $X_t$ be the number of "potential" arrivals before time $t$ and let $Y_t$ be the number of actual arrivals before time $t$. Suppose $0<s<t$. Then $$ \begin{align} & {}\quad\Pr(Y_t-Y_s = y) \\[12pt] & {} = \Pr\left( \bigcup_{w=y}^\infty \Big(\left(X_t-X_s = w\right) \cap (y\text{ of the }w\text{ potential arrivals are actual arrivals})\Big) \right) \\[12pt] & {} = \sum_{w=y}^\infty \Pr((X_t-X_s=w) \cap (y\text{ of the }w\text{ potential arrivals are actual arrivals})) \\[12pt] & = \sum_{w=y}^\infty \frac{e^{-L(t-s)} (L(t-s))^w}{w!} \cdot \binom w y B^y (1-B)^{w-y} = \sum_{w=y}^\infty \frac{e^{-L(t-s)} (L(t-s))^w B^y (1-B)^{w-y}}{(w-y)!y!} \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \sum_{w=y}^\infty \frac{(L(t-s))^{w-y}(1-B)^{w-y}}{(w-y)!} \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \sum_{v=0}^\infty \frac{(L(t-s))^v (1-B)^v}{v!} \qquad\text{where } v=w-y \\[12pt] & = e^{-L(t-s)} \frac{B^y}{y!} (L(t-s))^y \cdot e^{L(t-s)(1-B)} = e^{-BL(t-s)} \frac{(BL(t-s))^y}{y!}. \end{align} $$ So you have a Poisson process with rate $BL$.

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PS: The above counts the potential arrivals that are actual arrivals. Another process counts the potential arrivals that are not actual arrivals. That is of course a Poisson process with rate $(1-B)L$. Here's another exercise: Prove that those two Poisson processes are actually independent of each other. –  Michael Hardy Aug 11 '12 at 0:12
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