Ferkans — Interactive Telecom Tutor

Beyond Constant-Rate, Unit-Jump Arrivals

The basic Poisson process counts events that all look the same and arrive at a constant rate. Real systems break both assumptions. In a wireless network, each arriving packet carries a random payload size. During rush hour, the call arrival rate at a base station increases. Two natural extensions handle these situations: the compound Poisson process (random marks on arrivals) and the non-homogeneous Poisson process (time-varying rate).

Definition:
Compound Poisson Process

Let $\{N(t)\}$ be a Poisson process with rate $\lambda$ and let $Y_1, Y_2, \ldots$ be i.i.d. random variables (called marks or jumps) independent of $\{N(t)\}$ . The process $X(t) = \sum_{k=1}^{N(t)} Y_k$ is called a compound Poisson process. Its mean and variance are: $\mathbb{E}[X(t)] = \lambda t \, \mathbb{E}[Y_1], \qquad \text{Var}(X(t)) = \lambda t \, \mathbb{E}[Y_1^2].$

The compound Poisson process has independent and stationary increments but is not a counting process (jumps can be arbitrary). It is the continuous-time analog of a random walk with random step sizes.

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Example: Aggregate Data Traffic

Packets arrive at a link according to a Poisson process with rate $\lambda = 1000$ packets/second. Each packet has a random size $Y_k$ uniformly distributed on $[100, 1500]$ bytes. Model the total data volume $X(t)$ as a compound Poisson process and find the mean and standard deviation of the data volume in a 1-second window.

Solution

Identify the compound Poisson parameters

$\lambda = 1000$ packets/s. The marks $Y_k \sim \text{Uniform}(100, 1500)$ have $\mathbb{E}[Y] = 800$ bytes and $\mathbb{E}[Y^2] = \text{Var}(Y) + (\mathbb{E}[Y])^2 = \frac{(1400)^2}{12} + 800^2 = 163{,}333 + 640{,}000 = 803{,}333$ bytes $^2$ .

Mean data volume

$\mathbb{E}[X(1)] = 1000 \times 800 = 800{,}000$ bytes $= 800$ KB.

Standard deviation

$\text{Var}(X(1)) = 1000 \times 803{,}333 = 803{,}333{,}000$ . $\text{SD}(X(1)) = \sqrt{803{,}333{,}000} \approx 28{,}343$ bytes $\approx 27.7$ KB.

Definition:
Non-Homogeneous Poisson Process

A counting process $\{N(t) : t \geq 0\}$ is a non-homogeneous (inhomogeneous) Poisson process with intensity function $\lambda(t) \geq 0$ if:

$N(0) = 0$ .
It has independent increments.
For all $0 \leq s < t$ : $N(t) - N(s) \sim \text{Poisson}\!\left(\int_s^t \lambda(u)\,du\right).$

The cumulative intensity (or mean function) is $\Lambda(t) = \int_0^t \lambda(u)\,du$ , so $\mathbb{E}[N(t)] = \Lambda(t)$ .

When $\lambda(t) = \lambda$ is constant, this reduces to the homogeneous Poisson process. The non-homogeneous version models time-varying arrival rates such as diurnal traffic patterns in cellular networks.

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Theorem: Thinning and Superposition of Poisson Processes

(Thinning.) Let $\{N(t)\}$ be a Poisson process with rate $\lambda$ . If each arrival is independently retained with probability $p$ and deleted with probability $1-p$ , the retained arrivals form a Poisson process with rate $p\lambda$ , independent of the deleted process (rate $(1-p)\lambda$ ).

(Superposition.) If $\{N_1(t)\}, \ldots, \{N_m(t)\}$ are independent Poisson processes with rates $\lambda_1, \ldots, \lambda_m$ , then the merged process $N(t) = N_1(t) + \cdots + N_m(t)$ is Poisson with rate $\lambda_1 + \cdots + \lambda_m$ .

Thinning is like coin-flipping each arrival: heads keeps it, tails removes it. Since independent coin flips on independent arrivals preserve independence, the sub-processes remain Poisson. Superposition works because independent increments and orderliness are preserved under merging.

Proof

Proof of thinning (sketch)

Let $N_A(t)$ count retained arrivals in $[0,t]$ and $N_B(t)$ the deleted ones. For disjoint intervals, $N_A$ inherits independent increments from $N$ . For any interval of length $h$ : $\mathbb{P}(N_A(h)=1) = \mathbb{P}(N(h)=1)\cdot p + o(h) = p\lambda h + o(h)$ and $\mathbb{P}(N_A(h) \geq 2) = o(h)$ . So $\{N_A(t)\}$ satisfies the Poisson process axioms with rate $p\lambda$ . Independence of $N_A$ and $N_B$ follows by conditioning on $N$ and noting the binomial splitting is independent across disjoint intervals. $\blacksquare$

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Why This Matters: Spatial Poisson Process for Base Station Locations

In stochastic geometry models for cellular networks, the locations of base stations in $\mathbb{R}^2$ are modeled as a homogeneous Poisson point process (PPP) with spatial intensity $\lambda_{\text{BS}}$ (base stations per km $^2$ ). This is the two-dimensional extension of the temporal Poisson process studied in this section.

Superposition explains why the PPP is such a good fit: when many independent operators deploy base stations according to their own (possibly non-Poisson) point processes, the merged set of all base stations converges to a PPP by the superposition property.

Thinning models heterogeneous networks (HetNets): from the PPP of all base stations, independently classify each as macro, micro, or pico cell. Each tier forms an independent PPP with a fraction of the total intensity.

The key results — coverage probability, rate distribution, handover rate — all build on the Poisson properties of this chapter. See the FSP/Telecom chapters on stochastic geometry for the full development.

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Common Mistake: Inter-Arrival Times of a Non-Homogeneous Poisson Process

Mistake:

Assuming that the inter-arrival times of a non-homogeneous Poisson process are exponential. Students often write $T_k \sim \text{Exp}(\lambda(t))$ , which is not well-defined since $\lambda(t)$ changes over time.

Correction:

For a non-homogeneous Poisson process, the inter-arrival times are not exponential (unless $\lambda(t) = \lambda$ is constant). The survival function of the first arrival time is $\mathbb{P}(T_1 > t) = e^{-\Lambda(t)}$ , which is exponential only when $\Lambda(t) = \lambda t$ . The correct characterization uses the cumulative intensity $\Lambda(t) = \int_0^t \lambda(u)\,du$ .

Common Mistake: Variance of a Compound Poisson Process

Mistake:

Computing $\text{Var}(X(t)) = \lambda t \cdot \text{Var}(Y)$ . This is wrong because it ignores the randomness in the number of terms.

Correction:

By the law of total variance: $\text{Var}(X(t)) = \mathbb{E}[N(t)]\text{Var}(Y) + \text{Var}(N(t))(\mathbb{E}[Y])^2 = \lambda t(\text{Var}(Y) + (\mathbb{E}[Y])^2) = \lambda t \, \mathbb{E}[Y^2]$ . The correct formula uses the second moment of $Y$ , not just the variance.

Quick Check

Claims arrive at an insurance company as a Poisson process with rate $\lambda = 2$ /day. Each claim amount is $\text{Exp}(\mu)$ with mean $\mu^{-1} = 500$ euros. What is the expected total payout in a 30-day month?

$30{,}000$ euros

$60{,}000$ euros

$1{,}000$ euros

$15{,}000$ euros

Correction:

30{,}000

euros

$\mathbb{E}[X(30)] = \lambda \cdot 30 \cdot \mathbb{E}[Y] = 2 \times 30 \times 500 = 30{,}000$ .

Compound Poisson process

The process $X(t) = \sum_{k=1}^{N(t)} Y_k$ where $N(t)$ is a Poisson process and $Y_k$ are i.i.d. marks independent of $N(t)$ . Models random-sized arrivals.

Related: Poisson process

Non-homogeneous Poisson process

A Poisson process with time-varying intensity $\lambda(t)$ . The count in $(s,t]$ is Poisson with mean $\int_s^t \lambda(u)\,du$ . Models time-varying arrival rates.

Related: Poisson process

Compound and Non-Homogeneous Poisson Processes

Beyond Constant-Rate, Unit-Jump Arrivals

Definition: Compound Poisson Process

Example: Aggregate Data Traffic

Identify the compound Poisson parameters

Mean data volume

Standard deviation

Definition: Non-Homogeneous Poisson Process

Theorem: Thinning and Superposition of Poisson Processes

Proof of thinning (sketch)

Why This Matters: Spatial Poisson Process for Base Station Locations

Common Mistake: Inter-Arrival Times of a Non-Homogeneous Poisson Process

Common Mistake: Variance of a Compound Poisson Process

Quick Check

Compound Poisson process

Non-homogeneous Poisson process

Definition:
Compound Poisson Process

Definition:
Non-Homogeneous Poisson Process