Independent Bernoulli Thinning of a Poisson Process

From Poisson Arrivals to Retained Arrivals

Suppose events arrive according to a Poisson process, but only some are recorded: each arrival is kept independently with probability $p$. This model describes, for example, imperfect detection or a sampling rule applied event by event. The key question is whether the recorded events still form a Poisson process. The answer follows by combining the Poisson count in an interval with the binomial retention rule conditional on that count.

Let $(\Omega,\mathcal{F},\mathbb{P})$ support a Poisson process $(N_t)_{t\ge0}$ of rate $\lambda>0$, with arrival times $S_1,S_2,\ldots$. Thus $N_t$ counts arrivals in $(0,t]$; counts on disjoint time intervals are independent, and $N_t$ has a Poisson distribution with mean $\lambda t$. Let $(Z_k)_{k\ge1}$ be independent Bernoulli random variables, independent of the entire Poisson process, with $\mathbb{P}(Z_k=1)=p$ and $\mathbb{P}(Z_k=0)=1-p$, where $0\le p\le1$. The mark $Z_k=1$ means arrival $k$ is retained. The retained count process is

N^{(p)}_t=\sum_{k=1}^{N_t}Z_k,\qquad t\ge0,

where a sum with no terms is zero. It has time index $t\ge0$ and state space $\mathbb{Z}_{\ge0}$. One may use the filtration $\mathcal{F}_t=\sigma(N_s:s\le t;\,Z_k\mathbf{1}_{\{S_k\le t\}}:k\ge1)$, which records the original arrivals up to time $t$ and the marks of arrivals that have occurred. The count calculations below use conditioning on the original interval count and on the marks; they do not require a martingale argument.

Definition: The process $(N^{(p)}_t)_{t\ge0}$ is the Bernoulli thinning of $(N_t)_{t\ge0}$ with retention probability $p$. For an interval $I$, its retained count is the number of arrivals in $I$ whose marks equal $1$.

We use intervals of the form $(a,b]$, of length $h=b-a$. This convention assigns an arrival at a shared endpoint to only one of two adjacent intervals. A Poisson process has probability zero of an arrival at any specified time, so the choice of endpoint convention does not change the fixed-interval probabilities.

The Count in One Fixed Interval

Write $M=N_b-N_a$ for the original count in $(a,b]$, and $K$ for its retained count. If $M=m$, the $m$ marks are independent Bernoulli variables with success probability $p$, so $K$, conditional on $M=m$, has a binomial distribution. The following proposition computes the unconditional law.

Proposition: If $h=b-a\ge0$, then for every integer $k\ge0$, $$ \mathbb{P}(K=k)=e^{-p\lambda h}\frac{(p\lambda h)^k}{k!}. $$ When $p\lambda h=0$, this means $K=0$ with probability $1$. Thus the retained count in a fixed interval is Poisson with mean $p\lambda h$.

Proof. First take $p\in(0,1)$ and $h>0$, so the conditional binomial probabilities and the Poisson probabilities are defined in their usual form. For $m\ge k$, $\mathbb{P}(K=k\mid M=m)=\binom{m}{k}p^k(1-p)^{m-k}$, while $M$ is Poisson with mean $\lambda h$. Conditioning on $M$ and setting $r=m-k$ gives

\begin{aligned} \mathbb{P}(K=k) &=\sum_{m=k}^{\infty}e^{-\lambda h}\frac{(\lambda h)^m}{m!} \binom{m}{k}p^k(1-p)^{m-k}\\ &=e^{-\lambda h}\frac{(\lambda hp)^k}{k!} \sum_{r=0}^{\infty}\frac{(\lambda h(1-p))^r}{r!}\\ &=e^{-\lambda h}\frac{(\lambda hp)^k}{k!}e^{\lambda h(1-p)} =e^{-p\lambda h}\frac{(p\lambda h)^k}{k!}. \end{aligned}

The second equality uses $\binom{m}{k}/m!=1/(k!(m-k)!)$; the third uses the exponential series. If $p=0$, no arrival is retained, and if $p=1$, every arrival is retained, so the formula holds directly in both cases. If $h=0$, the interval has count zero almost surely, and the formula again gives the degenerate Poisson law of mean zero. $\square$

This calculation gives the distribution, not just the expected count. In particular, it shows how the original rate $\lambda$ and the retention probability $p$ combine: the retained mean count over length $h$ is $p\lambda h$.

Worked Example: Exactly Two Retained Arrivals

Suppose the original process has rate $\lambda=\tfrac32$, and each arrival is retained with probability $p=\tfrac23$. Find the probability of exactly two retained arrivals in an interval of length $h=2$, first using the mixture over original counts.

The original mean count is $\lambda h=\tfrac32\cdot2=3$, and the retained mean is $p\lambda h=\tfrac23\cdot3=2$. For an original count $m\ge2$, the conditional probability of retaining exactly two is $\binom{m}{2}(\tfrac23)^2(\tfrac13)^{m-2}$. Therefore

\begin{aligned} \mathbb{P}(K=2) &=\sum_{m=2}^{\infty}e^{-3}\frac{3^m}{m!} \binom{m}{2}\left(\frac23\right)^2\left(\frac13\right)^{m-2}\\ &=2e^{-3}\sum_{r=0}^{\infty}\frac{1}{r!} =2e^{-3}e =2e^{-2}. \end{aligned}

For the second equality, set $r=m-2$. Then $3^m(\tfrac13)^{m-2}=9$, $(\tfrac23)^2=\tfrac49$, and $\binom{m}{2}/m!=1/(2r!)$, so the coefficient is $9\cdot\tfrac49\cdot\tfrac12=2$. Check: the proposition gives $e^{-p\lambda h}(p\lambda h)^2/2!=e^{-2}\cdot 2^2/2=2e^{-2}$, matching the mixture calculation.

Why the Retained Process Is Poisson

A Poisson process is characterized by its initial value and its increments: it starts at zero, has independent increments, and the count in an interval of length $h$ has a Poisson distribution whose mean is the rate times $h$. The fixed-interval proposition supplies the count law. To establish the process claim, we also need to check independence across intervals.

Theorem: For $0\le p\le1$, the Bernoulli-thinned process $(N^{(p)}_t)_{t\ge0}$ is a Poisson process of rate $p\lambda$, with rate zero interpreted as the process identically equal to zero. In particular, the retained counts on any finite collection of disjoint time intervals are independent.

Proof. Let $I_1,\ldots,I_q$ be disjoint intervals, with lengths $h_1,\ldots,h_q$. Let $M_r$ be the original count and $K_r$ the retained count in $I_r$, for $1\le r\le q$. The original Poisson counts $M_1,\ldots,M_q$ are independent. Conditional on the original arrival times, each interval uses marks from a distinct set of arrival indices. Those marks are independent, so, conditional on the original arrivals, the retained counts in the intervals are independent. For any nonnegative integers $k_1,\ldots,k_q$, this gives

\begin{aligned} \mathbb{P}(K_1=k_1,\ldots,K_q=k_q) &=\mathbb{E}\left[\prod_{r=1}^q \mathbb{P}(K_r=k_r\mid M_r)\right]\\ &=\prod_{r=1}^q\mathbb{E}\left[ \mathbb{P}(K_r=k_r\mid M_r)\right]\\ &=\prod_{r=1}^q\mathbb{P}(K_r=k_r). \end{aligned}

The first equality follows from conditional independence of the marks in the disjoint intervals. In the second, each factor depends only on $M_r$, and the original counts $M_r$ are independent. The factorization proves that the retained counts are independent. By the proposition, each $K_r$ is Poisson with mean $p\lambda h_r$, a quantity depending only on the interval length. Also, $N^{(p)}_0=0$. Its paths are nondecreasing integer-valued paths, and it has finitely many jumps on every bounded time interval because it is formed by removing arrivals from the original nonexplosive Poisson process. Thus it has independent stationary increments and Poisson increments of rate $p\lambda$, as claimed. When $p=0$, all counts are zero, consistent with the same conclusion. $\square$

Worked Example: Joint Counts in Two Disjoint Intervals

Let $\lambda=2$ and $p=\tfrac14$. Consider $I_1=(0,1]$ and $I_2=(1,3]$, whose lengths are $1$ and $2$. Find the probability of exactly one retained arrival in $I_1$ and no retained arrivals in $I_2$.

The retained mean in $I_1$ is $p\lambda h_1=\tfrac14\cdot2\cdot1=\tfrac12$. Hence its probability of exactly one is $e^{-1/2}(\tfrac12)^1/1!=\tfrac12e^{-1/2}$. In $I_2$, the retained mean is $p\lambda h_2=\tfrac14\cdot2\cdot2=1$, so its probability of zero is $e^{-1}\cdot1^0/0!=e^{-1}$. The intervals are disjoint, so the theorem gives

\mathbb{P}(K_1=1,K_2=0) =\left(\frac12e^{-1/2}\right)e^{-1} =\frac12e^{-3/2}.

Check: the marginal probabilities used in the product are $\mathbb{P}(K_1=1)=\tfrac12e^{-1/2}$ and $\mathbb{P}(K_2=0)=e^{-1}$; their product is exactly $\tfrac12e^{-3/2}$, as required by independence.

Worked Example: At Least Two Retained Arrivals

Suppose $\lambda=\tfrac34$, $p=\tfrac23$, and the observation interval has length $h=4$. Find the probability of at least two retained arrivals.

The retained mean is $p\lambda h=\tfrac23\cdot\tfrac34\cdot4=2$. Thus $K$ is Poisson with mean $2$. The probabilities of zero and one retained arrivals are

\mathbb{P}(K=0)=e^{-2}\frac{2^0}{0!}=e^{-2}, \qquad \mathbb{P}(K=1)=e^{-2}\frac{2^1}{1!}=2e^{-2}.

Since the events $K=0$, $K=1$, and $K\ge2$ partition the possibilities,

\mathbb{P}(K\ge2)=1-e^{-2}-2e^{-2}=1-3e^{-2}.

Check: the complementary probability is $\mathbb{P}(K=0)+\mathbb{P}(K=1)=e^{-2}+2e^{-2}=3e^{-2}$, so the two probabilities sum to $1$.

A Common Mistake and Why Thinning Matters

A frequent error is to treat “no retained arrival” as though it meant “no original arrival.” These are different events: an interval can contain original arrivals and still have no retained ones if all their marks are zero. For example, with $\lambda=2$, $p=\tfrac12$, and $h=1$, a wrong calculation would give $\mathbb{P}(K=0)=\mathbb{P}(M=0)=e^{-2}$. The correct calculation conditions on the original count $M$: given $M=m$, all $m$ arrivals are rejected with probability $(\tfrac12)^m$. Therefore

\begin{aligned} \mathbb{P}(K=0) &=\sum_{m=0}^{\infty}\mathbb{P}(M=m)\left(\frac12\right)^m\\ &=\sum_{m=0}^{\infty}e^{-2}\frac{2^m}{m!}\left(\frac12\right)^m\\ &=e^{-2}\sum_{m=0}^{\infty}\frac{1}{m!} =e^{-2}e^{-0}e^1 =e^{-1}. \end{aligned}

The last expression is $e^{-1}$, while the incorrect expression is $e^{-2}$. The corrected value agrees with the retained Poisson mean $p\lambda h=\tfrac12\cdot2\cdot1=1$, for which $\mathbb{P}(K=0)=e^{-1}$. More generally, thinning is useful whenever an event stream is observed imperfectly or sampled randomly: the retained process has the same Poisson structure, with its rate scaled by the probability of recording an arrival.

The result concerns the distribution and independent increments of the retained process by itself. In applying it, keep the two steps separate: first determine the original Poisson mean $\lambda h$, then apply independent retention to obtain the retained mean $p\lambda h$. For counts in multiple intervals, use the fact that disjoint original counts and disjoint sets of independent marks produce independent retained counts.

Check Your Understanding

Use the conditional binomial law and the fixed-interval Poisson formula to answer these questions.

An original process has rate $\lambda=3$, and each arrival is retained with probability $p=\tfrac13$. What is the distribution of the retained count in an interval of length $2$?
For a fixed interval with original mean $\lambda h=4$ and retention probability $p=\tfrac12$, calculate the probability of zero retained arrivals by conditioning on the original count.
Why does the proof of independent retained counts require both independent original counts and independent marks?
Two disjoint intervals have lengths $1$ and $3$. If $\lambda=1$ and $p=\tfrac12$, calculate the probability that the first contains zero retained arrivals and the second contains exactly two.
State the retained process when $p=0$ and when $p=1$, and give its rate in each case.

Independent Bernoulli Thinning of a Poisson Process

What You'll Learn