Poisson Distribution

Context: FIT1058_MOC, FIT2086_MOC · counts independent events in an interval · pmf $e^{-\mu }{\mu }^k/k!$ · large-$n$ small-$np$ approximation to the binomial · the ${\mathbb{Z}}^{+}$ member of the parametric zoo

Quick Revision

• 🎯 Objective: count of rare independent events in an interval ➔ $\mathrm{Pr}(X=k)=e^{-\mu }{\mu }^k/k!$ ($\mu \equiv \lambda$, the rate).
• 📦 Core Components: one parameter $\mu$ ➔ $E=\mathrm{Var}=\mu$ ➔ unbounded support ➔ closed under addition and interval-splitting.
• ⚡ Key Constraint: approximates $\mathrm{Bin}(n,p)$ for large $n$, small $np$ ($\mu =np$); the four modelling conditions must actually hold.

📝 Core

1. The Distribution

• Definition ➔ $\mathrm{Pr}(X=k)=\frac{e^{-\mu }{\mu }^k}{k!}$, $k\in {\mathbb{N}}_0$, $\mu >0$; FIT2086 writes $X\sim Pois(\lambda )$ with $\lambda$ called the rate.
• One parameter ➔ $E(X)=\mu$, $\mathrm{Var}(X)=\mu$, $\sigma =\sqrt{\mu }$ ➔ the family where the variance grows with the mean.

2. When It Arises

• Count of independent occurrences ➔ calls per hour, visits, radioactive emissions — sample space ${\mathbb{Z}}^{+}=\{0,1,2,\dots \}$.
• Interval-scaled ➔ $\mu$ scales with interval length.
• Unbounded ➔ any nonnegative integer (unlike the binomial’s cap at $n$).
• Four appropriateness conditions (FIT2086) ➔ (i) one event’s occurrence does not affect the probability of a second — events are independent; (ii) the rate is constant across intervals, never higher in some than others; (iii) two events cannot occur at the same instant; (iv) the probability of an event in a small interval is proportional to that interval’s length. Violating (i) or (ii) is what makes real count data spread wider than $V=\lambda$ allows.

3. Binomial Approximation

• Condition ➔ large $n$, small $np$ ⟹ $\mathrm{Bin}(n,p)\approx \mathrm{Poisson}(np)$.
• Benefit ➔ trades a two-parameter formula for one.

4. Additivity and Thinning (FIT2086)

• Sums stay Poisson ➔ $X_1\sim Poi({\lambda }_1)$, $X_2\sim Poi({\lambda }_2)$ ⟹ $X_1+X_2\sim Poi({\lambda }_1+{\lambda }_2)$ — rates add.
• Decomposition ➔ any $X\sim Poi(\lambda )$ splits as $X={\sum }_{i=1}^n{X}_i$ with $X_i\sim Poi({\lambda }_i)$ and ${\sum }_i{\lambda }_i=\lambda$.
• Interval rescaling (thinning) ➔ if $X_T\sim Poi(\lambda )$ counts events in a period $T$, then over $T/k$: $X_{T/k}\sim Poi(\lambda /k)$ ➔ the rate is per unit time ➔ always rescale $\lambda$ to the interval the question asks about before evaluating any probability.

Key identities:

${\sum }_{k=0}^{\infty }\frac{e^{-\mu }{\mu }^k}{k!}=e^{-\mu }{\sum }_{k=0}^{\infty }\frac{{\mu }^k}{k!}=e^{-\mu }{e}^{\mu }=1$

⚖️ Core Decision Matrix

Aspect	Poisson	vs Binomial
parameter	$\mu$	$n,p$
$E$, $\mathrm{Var}$	both $\mu$	$np$, $np(1-p)$
support	unbounded	$\{0,\dots ,n\}$
relation	$\approx \mathrm{Bin}$ large $n$	$\mu =np$
closure under $+$	rates add: $Poi({\lambda }_1+{\lambda }_2)$	trials add (same $\theta$): $Bin(\theta ,n_1+n_2)$

When It Flips: Poisson suits "could be any count" of rare independent events; it is the unbounded limit of the binomial. The $e^x$-series normalisation mirrors the geometric-series argument for other pmfs. $V=E$ is the family fingerprint — sample variance far above the sample mean falsifies the constant-rate assumption.

📊 Exam Execution Trace

Applied Exercise 1

Problem: Poisson $\mu =2$ — find $\mathrm{Pr}(X=0)$, $\mathrm{Pr}(X=1)$, $E$, $\mathrm{Var}$. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}\mathrm{Pr}(X=0)& =e^{-2}\approx 0.135,\mathrm{Pr}(X=1)=2e^{-2}\approx 0.271\\ E(X)& =\mu =2,\mathrm{Var}(X)=\mu =2\end{array}$$

Final Extracted Output: $\approx 0.135$, $\approx 0.271$; mean and variance both 2.

Applied Exercise 2 — rescaling the interval (FIT2086)

Problem: calls arrive at $X_T\sim Poi(12)$ per hour. Find $P(\text{no call in a given }15\text{ minutes})$.

$$\begin{array}{rl}k& =4(\text{an hour splits into four }15\text{-minute periods})\Rightarrow X_{T/4}\sim Poi(12/4)=Poi(3)\\ P(X_{T/4}=0)& =\frac{3^0{e}^{-3}}{0!}=e^{-3}\approx 0.0498\end{array}$$

Final Extracted Output: $\approx 0.05$. Key move: rescale $\lambda$ to the requested interval first; using $\lambda =12$ would answer a different question entirely.

Applied Exercise 3 — hospital heart attacks, week ➔ day (Studio 2)

Problem: a hospital sees on average $6$ heart-attack patients per week, rate independent of the day. Find (i) $P(\le 2$ in a week$)$; (ii) $P(\text{exactly }1$ on a given day$)$; (iii) $P(\ge 1$ on a given day$)$.

$$\begin{array}{rl}\text{(i)}& P(X\le 2\mid \lambda =6)=\text{ppois(2, 6)}\approx 0.0620\\ \text{(ii)}& X={\sum }_{i=1}^7{X}_i\sim Poi\left({\sum }_{i=1}^7{\lambda }_i\right)\text{ with all }{\lambda }_i\text{ equal}\Rightarrow {\lambda }_i=\frac{6}{7}\\ & P(X_i=1\mid \lambda =\frac{6}{7})=\frac{(6/7{)}^1{e}^{-6/7}}{1!}\approx 0.3637\text{dpois(1, 6/7)}\\ \text{(iii)}& P(X_i\ge 1)=1-P(X_i=0)=1-\frac{(6/7{)}^0{e}^{-6/7}}{0!}=1-e^{-6/7}\approx 0.5756\end{array}$$

Final Extracted Output: $0.0620$; $0.3637$; $0.5756$. Key move: the identical-days assumption is what forces ${\lambda }_1=\cdots ={\lambda }_7$, so additivity run backwards gives ${\lambda }_i=\lambda /7$. And “at least one” always collapses to $1-e^{-\lambda }$ because $P(X=0)=e^{-\lambda }$ — no summation needed.

⚙️ In R

The pois suffix (details ➔ R Simulation and Random Sampling)

dpois(0, lambda = 3)                  # P(X = 0)  = 0.0498
ppois(2, lambda = 3)                  # P(X <= 2) = 0.4232
ppois(2, lambda = 3, lower.tail = FALSE)  # P(X >= 3)
qpois(0.95, lambda = 3)               # quantile
rpois(10, lambda = 3)                 # 10 random counts
 
# Studio 2 — the pmf by hand vs the built-in (identical results)
4^1 * exp(-4) / factorial(1)          # 0.07326  =  dpois(1, 4)
4^0*exp(-4)/factorial(0) + 4^1*exp(-4)/factorial(1)   # P(X < 2) = 0.09158
dpois(0, 4) + dpois(1, 4)             # same, and equals ppois(1, 4)
1 - ppois(5, 4)                       # P(X > 5) = 0.2149

💡 Common Mistake: lambda must already match the interval in the question ➔ do the $\lambda /k$ thinning by hand before the call. Also exp() and factorial() reproduce the pmf exactly — useful for checking that you translated $P(X<2)$ into $P(X\le 1)$ and not $P(X\le 2)$.

⚠️ Common Mistakes

• 💡 $\mu$ is mean AND variance ➔ the single parameter does double duty; the binomial→Poisson swap needs large $n$ with small $np$.
• 💡 Forgetting to rescale the rate ➔ $\lambda$ is defined per interval; a question about a half/quarter period needs $Poi(\lambda /k)$.
• 💡 Applying Poisson to clustered events ➔ if one event makes another more likely (contagion, queues, bursts) the independence and constant-rate conditions fail, and the $V=\lambda$ prediction breaks.
• 💡 Strict vs weak inequality on a discrete support ➔ $P(X<2)=P(X\le 1)=$ ppois(1, λ), not ppois(2, λ); unlike the continuous case the boundary value carries real mass.
• 💡 Poisson for a binary or continuous variable ➔ “land vs water” is binary (binomial) and “average weight” is continuous (normal); Poisson needs an unbounded count. Selection drill ➔ Parametric Probability Distributions.

🧠 Active Recall

State the Poisson pmf, show it sums to 1, and give mean and variance.

• Hint: $e^x$ series.

Answer

• Short answer: $\mathrm{Pr}(X=k)=e^{-\mu }{\mu }^k/k!$; $\sum =e^{-\mu }{e}^{\mu }=1$; $E=\mathrm{Var}=\mu$.
• Why: Power series ➔ ${\sum }_k{\mu }^k/k!=e^{\mu }$.

When does the Poisson distribution arise, and how does it relate to the binomial?

• Hint: Rare independent counts.

Answer

• Short answer: Counts of independent occurrences in a fixed interval; $\mathrm{Bin}(n,p)\approx \mathrm{Poisson}(np)$ for large $n$, small $np$.
• Why: Unbounded limit ➔ one parameter replaces two.

Which four conditions must hold for a Poisson model, and what happens to the rate if the interval is shortened by a factor $k$?

• Hint: Independence, constancy, no ties, proportionality.

Answer

• Short answer: events independent; constant rate across intervals; no two events at the same instant; probability in a small interval proportional to its length. Shortening the interval to $T/k$ gives $X_{T/k}\sim Poi(\lambda /k)$.
• Why: The rate is per unit time ➔ proportionality (condition iv) plus additivity means the $k$ sub-intervals of $T$ each carry rate $\lambda /k$ and sum back to $Poi(\lambda )$; breaking independence or constancy leaves the data spread wider than $V=\lambda$ permits, falsifying the model.