Geometric Distribution

Context: FIT1058_MOC · the number of trials until the first success · pmf $(1-p{)}^{k-1}p$ · memoryless · drives the Coupon Collector’s Problem

Quick Revision

• 🎯 Objective: trials until the first success ➔ $\mathrm{Pr}(X=k)=(1-p{)}^{k-1}p$.
• 📦 Core Components: $E=1/p$ ➔ $\mathrm{Var}=(1-p)/p^2$ ➔ memoryless.
• ⚡ Key Constraint: memorylessness refutes the “law of averages”; the unique $\mathbb{N}$-valued such distribution.

📝 Core

1. The Distribution

• Definition ➔ $X\sim \mathrm{Geom}(p)$ = trials up to and including the first success.
• pmf ➔ $\mathrm{Pr}(X=k)=(1-p{)}^{k-1}p$, $k\in \mathbb{N}$.
• Moments ➔ $E(X)=1/p$, $\mathrm{Var}(X)=(1-p)/p^2$.

2. Memorylessness

• Property ➔ given $X\ge t$, $X-t$ is again $\mathrm{Geom}(p)$ — the wait restarts.
• Unique ➔ only $\mathbb{N}$-valued memoryless distribution.
• Refutes ➔ the gambler’s “law of averages”.

Key identities:

$\mathrm{Pr}(X=k)=(1-p{)}^{k-1}p,{\sum }_{k\ge 1}(1-p{)}^{k-1}p=\frac{p}{1-(1-p)}=1$

⚖️ Core Decision Matrix

Aspect	Geometric	vs Binomial
counts	trials to first success	successes in fixed $n$
support	unbounded $\mathbb{N}$	$\{0,\dots ,n\}$
$E$	$1/p$	$np$
memoryless	yes	no

When It Flips: smaller $p$ ⟹ longer wait ($E=1/p$). Models loop iterations to a stopping condition, time-to-failure; a sum of geometric waits gives the Coupon Collector's Problem.

📊 Exam Execution Trace

Manual Execution Trace

Fair coin to first Heads, $\mathrm{Geom}(\frac{1}{2})$:

Step / State	$k$	$(1-p{)}^{k-1}$	$\mathrm{Pr}(X=k)$
0 (Init)	—	—	—
1	1	1	$\frac{1}{2}$
2	2	$\frac{1}{2}$	$\frac{1}{4}$
3	3	$\frac{1}{4}$	$\frac{1}{8}$

⚠️ Common Mistakes

• 💡 Law of averages is a fallacy ➔ a run of failures does not make success “due”; each trial keeps probability $p$ (memorylessness formalises this).

🧠 Active Recall

Derive the geometric pmf and show it sums to 1.

• Hint: $k-1$ failures then success.

Answer

• Short answer: $\mathrm{Pr}(X=k)=(1-p{)}^{k-1}p$; sums to $\frac{p}{1-(1-p)}=1$.
• Why: Geometric series ➔ first term $p$, ratio $1-p$; $E=1/p$, $\mathrm{Var}=(1-p)/p^2$.

What is the memoryless property, and why does it refute the "law of averages"?

• Hint: Wait restarts.

Answer

• Short answer: Given $X\ge t$, $X-t\sim \mathrm{Geom}(p)$ — past failures don’t change the future.
• Why: Independence ➔ each trial keeps $p$; success is never “overdue”.