Step function of Poisson distributed number of uniformly distributed RVs is Poisson process

Question

I want to proof this:

Let $N,X_1,X_2,\dots$ be independent RVs, $N$ Poisson distributed and the $X_k\sim\text{Unif}([0,1])$. Then $$ N_t:=\sum_{k=1}^N 1_{[0,t]}(X_k)\quad (t\in [0,1]) $$

is a Poisson process (restricted to $t\in[0,1]$), i.e. $N_t(\omega)$ is an increasing and right-continuous step-function with jumps of size one and $N_t-N_s\sim\text{Pois}(\lambda(t-s))$ for some parameter $\lambda$.

I don't know how to show that the increments are Poisson distributed. I've tried this, but I don't know how to proceed or if this is even the right approach:

$$ N_t-N_s=\sum_{k=1}^N 1_{[0,t]}(X_k)-\sum_{k=1}^N 1_{[0,s]}(X_k)=\sum_{k=1}^N 1_{(s,t]}(X_k) $$ It's obvious that

$$ P(X_k\in(s,t])=t-s\ \text{ and } P(X_k\in[0,s])=s $$

So to calculate $P(N_t-N_s=n)$, $n\in\mathbb{N}$, we need to calculate the probability that exactly $n$ of the $X_k$ are in $(s,t]$ and $N-n$ of the $X_k$ are in $[0,s]$, which is

$$ P(|\{k\leq N : X_k\in (s,t]\}|=n,\ |\{k\leq N : X_k\in[0,s]\}|=N-n) \\ = \binom{N}{n}(t-s)^ns^{N-n} $$

So somehow I ended up with an binomial distribution and there is also missing the probability that $N\geq n$ and I don't know how to combine that and receive a Poisson distribution as result.

Thanks for any help!

Answer 1

In the understanding that $\binom{k}{n}=0$ if $n$ exceeds $k$ we have:$$P\left(N_{t}-N_{s}=n\mid N=k\right)=\binom{k}{n}\left(t-s\right)^{n}\left(1-t+s\right)^{k-n}$$

Note that you should take $1-t+s$ here (and not $s$).

Also be aware that $N$ is a random variable.

Working this out we find:

$\begin{aligned}P\left(N_{t}-N_{s}=n\right) & =\sum_{k=n}^{\infty}P\left(N=k\right)P\left(N_{t}-N_{s}=n\mid N=k\right)\\ & =\sum_{k=n}^{\infty}e^{-\lambda}\frac{\lambda^{k}}{k!}\binom{k}{n}\left(t-s\right)^{n}\left(1-t+s\right)^{k-n}\\ & =\sum_{k=n}^{\infty}e^{-\lambda}\frac{\lambda^{k}}{n!\left(k-n\right)!}\left(t-s\right)^{n}\left(1-t+s\right)^{k-n}\\ & =\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^{k+n}}{n!k!}\left(t-s\right)^{n}\left(1-t+s\right)^{k}\\ & =e^{-\lambda}\frac{\lambda^{n}}{n!}\left(t-s\right)^{n}\sum_{k=0}^{\infty}\frac{\lambda^{k}\left(1-t+s\right)^{k}}{k!}\\ & =e^{-\lambda\left(t-s\right)}\frac{\left[\lambda\left(t-s\right)\right]^{n}}{n!} \end{aligned} $

Answer 2

Denote by $\lambda$ for the rate of $N$. For each subset $S \subseteq [0, 1]$, define

$$ \Pi(S) := \sum_{k=1}^{N} \mathbf{1}_{S}(X_k). $$

Then the old definition $N_t$ is the same as $\Pi([0, t])$. Now we prove:

Claim. Suppose that $I_1, \cdots, I_m \subseteq [0, 1]$ are disjoint intervals. Then $\Pi(I_1), \cdots, \Pi(I_m)$ are independent and $\Pi(I_k) \sim \operatorname{Poisson}(\lambda|I_k|)$ for each $k$.

Remark. For OP's specific question, it suffices to consider $n = 1$ case. But it causes little harm to consider and prove the generality, and in fact, this kind of statement is necessary for establishing independent increment condition for Poisson process.

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1^st Proof. We examine the multivariate MGF. For real $s_1, \cdots, s_m$, define $f(\cdot) = \sum_{k=1}^{m} s_k \mathbb{1}_{I_k}(\cdot)$. Then by the towering property of conditional expectation,

$$ \mathbb{E}\Biggl[ \exp\Biggl\{ \sum_{k=1}^{m} s_k \Pi(I_k) \Biggr\} \Biggr] = \mathbb{E}\Biggl[ \prod_{j=1}^{N} e^{f(X_j)} \Biggr] = \mathbb{E}\Bigl[ \mathbb{E}\bigl[ e^{f(X_1)} \bigr]^{N} \Bigr] = \exp\bigl\{ \lambda(\mathbb{E}[e^{f(X_1)}] - 1) \bigr\}. $$

Moreover, using the fact that $I_k$'s are disjoint, we easily compute that

$$ \mathbb{E}[e^{f(X_1)}] = \int_{0}^{1} e^{f(x)} \, \mathrm{d}x = 1 + \sum_{k=1}^{m} |I_k| (e^{s_k} - 1). $$

Plugging this back, we get

$$ \mathbb{E}\Biggl[ \exp\Biggl\{ \sum_{k=1}^{m} s_k \Pi(I_k) \Biggr\} \Biggr] = \prod_{k=1}^{m} \exp\bigl\{ \lambda|I_k| (e^{s_k} - 1) \bigr\}. $$

By the uniqueness of MGF, this implies the desired claim.

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2^nd Proof. By adding some extra intervals into consideration, we may assume that $\cup_{k=1}^{m} I_k = [0, 1]$. Now for each $X_j$, define $Y_j$ by

$$ Y_j = [\text{value of index $k$ such that $X_j \in I_k$}] $$

Then note that $Y_j$'s are i.i.d. and $\mathbb{P}(Y_j = k) = |I_k|$ for each $j$ and $k$. Consequently, for each $n$, counting the number of occurrences of each symbol in the list $(Y_1, \cdots, Y_n)$ yields a multinomial distribution.

Now let $n_1, \cdots, n_m \geq 0$ be integers and write $n = \sum_{k=1}^{m} n_k$. Then

\begin{align*} \mathbb{P}\Biggl( \bigcap_{k=1}^{m} \{ \Pi(I_k) = n_k \} \Biggr) &= \mathbb{P} \Biggl( \{N = n\} \cap \bigcap_{k=1}^{m} \{ \text{the number of $k$'s in $(Y_1, \cdots, Y_n)$ is $n_k$} \} \Biggr) \\ &= \frac{\lambda^n}{n!}e^{-\lambda} \cdot \binom{n}{n_1,\cdots,n_m} |I_1|^{n_1} \cdots |I_m|^{n_m} \\ &= \prod_{k=1}^{m} \frac{(\lambda |I_k|)^{n_k}}{n_k!} e^{-\lambda |I_k|}. \end{align*}

From this, we easily check that $\Pi(I_k)$'s are independent and $\Pi(I_k) \sim \operatorname{Poisson}(\lambda |I_k))$.