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Suppose that $\{N'(t) :t \geq 0\}$ and $\{N''(t) :t \geq 0 \}$ are the counting processes of two independent homogeneous Poisson processes on the line, each with with rate $\lambda$. Fix an integer $n\geq 1$ and let $S'_n$ be the time of the $n$-th arrival for the first Poisson process. Define a new counting process $\{N(t) :t \geq 0\}$ by: $$ N(t) = \begin{cases} N''(t) &\text{for }0\leq t\leq S'_n, \\ N'(t) -N'(S_n) + N''(S_n')&\text{for } t > S'_n. \end{cases}$$

I'm trying to show that $\{N(t) :t \geq 0 \}$ is also the counting process of a homogeneous Poisson process on the line with rate $\lambda$.

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The result holds and a proof is based on the joint construction of the processes $N'$ and $N''$, as explained below.

Start with a homogenous Poisson process $N_0=\{N_0(t):t\geqslant0\}$ with rate $2\lambda$, and a sequence of i.i.d. Bernoulli $\pm1$ random variables $(B_k)_{k\geqslant1}$ such that $\mathrm P(B_k=+1)=\mathrm P(B_k=-1)=\frac12$. Assume that $(B_k)_{k\geqslant1}$ and $N_0$ are independent. Let $(T_k)_{k\geqslant1}$ denote the ordered sequence of time events in $N_0$. Then $N'$ and $N''$ may be realized as the sets of time events $\{T_k\mid k\geqslant1,B_k=-1\}$ and $\{T_k\mid k\geqslant1,B_k=+1\}$ respectively.

We now build the process $N$. Let $C_0=0$ and, for every $k\geqslant1$, $C_k=\#\{i\leqslant k\mid B_i=+1\}$. Let $(A_k)_{k\geqslant1}$ denote a sequence of $\pm1$, defined as follows: if $C_{k-1}\lt n$, then $A_k=+1$, if $C_{k-1}\geqslant n$, then $A_k=-1$. Then $N$ may be realized as the set of time events $\{T_k\mid k\geqslant1,A_kB_k=+1\}$.

Thus, it remains to check that the sequence $(A_kB_k)_{k\geqslant1}$ is:

  1. independent from $N_0$,
  2. i.i.d. with symmetric Bernoulli distribution.

Item 1. holds because the sequence $(A_kB_k)_{k\geqslant1}$ is $\sigma((B_k)_{k\geqslant1})$-measurable, and $(B_k)_{k\geqslant1}$ is independent on $N_0$. Re item 2., introduce the sigma-algebras $\mathcal B_k=\sigma((B_i)_{i\leqslant k})$. Note that $A_k$ is $\mathcal B_{k-1}$-measurable and that $B_k$ is symmetric Bernoulli and independent on $\mathcal B_{k-1}$. Hence $A_kB_k$ is symmetric Bernoulli and independent on $\mathcal B_{k-1}$ as well. This proves the desired assertion.

Ultimately, all this is based on the following simple fact, whose proof is left to the reader:

Let $A$ and $B$ denote some Bernoulli $\pm1$ random variables and $\mathcal G$ a sigma-algebra. Assume that $A$ is $\mathcal G$-measurable and that $B$ is independent on $\mathcal G$ and symmetric. Then $AB$ is a symmetric Bernoulli random variable independent on $\mathcal G$.

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