Mathematical Statistics

🎯 Learning Objectives

By the end of this lecture, you will be able to:

Apply the multiplicative law \(P(A \cap B) = P(A)P(B \mid A)\), in either order
Extend it to a chain of three or more dependent stages
Apply the additive law \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), and say what the subtraction repairs
Use the complement \(P(A) = 1 - P(\bar{A})\) to convert an “at least one” event into an easy one
Choose between the two laws by reading whether the event is an intersection or a union

🗺️ Where We Are

Wackerly §2.8

Wednesday: \(P(A \mid B) = P(A \cap B)/P(B)\). We used it to read information out of a table.

Today we read it the other way. Multiply both sides by \(P(B)\) and the definition stops being a measurement and becomes a construction: a way to build the probability of an intersection out of pieces.

Two laws, one for \(\cap\) and one for \(\cup\). Between them they handle every compound event in this chapter.

📐 Theorem 2.5: The Multiplicative Law

Theorem 2.5

For any two events \(A\) and \(B\), \[P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)\] and if \(A\) and \(B\) are independent, \[P(A \cap B) = P(A)P(B)\]

The proof is Definition 2.9 with the denominator moved across. There is no new content — only a new use.

🔑 Two Forms, One of Which You Can Evaluate

\[P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)\]

The two expressions are always equal, but a problem rarely hands you both. It gives you one marginal and the conditional that goes with it.

So the first question is which conditioning you were given, not which formula you prefer. Choosing the other form leaves you with two unknowns and one equation.

📊 Worked Example: Two Loans From a Pool

A securitisation pool holds 12 mortgages, of which 4 are in arrears. Two are drawn without replacement for a file review. What is the probability that both are in arrears?

\[P(A_1 \cap A_2) = P(A_1)\,P(A_2 \mid A_1) = \frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = 0.0909\]

Note what the law did: it turned a question about a pair into two questions about a single draw, and the second one was easy because the first had already been answered.

⛓️ The Extension to Three

Three events

\[P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)\]

Each factor conditions on everything already assumed. Drawing a third mortgage from the same pool: \[P(A_1 \cap A_2 \cap A_3) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = 0.0182\]

The pattern continues for any number of stages, and it is how sampling without replacement is handled throughout the course.

🤝 The Independent Case

Azerbaijani and Georgian trade receivables are insured under separate policies, and a claim on one is taken to be independent of a claim on the other. With \(P(A) = 0.07\) and \(P(G) = 0.05\), \[P(A \cap G) = 0.07 \times 0.05 = 0.0035\]

Independence is what licenses the shortcut — it is not the default. When a problem does not say “independent”, the conditional does not disappear, and assuming it does is the most common way to arrive confidently at the wrong number.

📐 Theorem 2.6: The Additive Law

Theorem 2.6

For any two events \(A\) and \(B\), \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] and if \(A\) and \(B\) are mutually exclusive, \[P(A \cup B) = P(A) + P(B)\]

The second line is Axiom 3 itself. The first is what Axiom 3 becomes when the events overlap.

➖ What the Subtraction Repairs

Write \(A \cup B\) as a union of three disjoint pieces: \(A \cap \bar{B}\), \(\bar{A} \cap B\), and \(A \cap B\). Axiom 3 applies to those, and adding them gives the law.

Adding \(P(A)\) and \(P(B)\) counts the overlap twice — once inside \(A\), once inside \(B\). Subtracting \(P(A \cap B)\) removes one of the two copies.

So the subtraction is not a correction for an error. It is bookkeeping, and it disappears exactly when there is nothing to double count.

🏦 Worked Example: A Customer Relationship

Of a bank’s retail customers, 62% hold a current account (\(C\)), 45% hold a card (\(K\)), and 31% hold both.

\[P(C \cup K) = 0.62 + 0.45 - 0.31 = 0.76\]

Drop the subtraction and you get 1.07 — a probability above 1, which Axiom 2 forbids. An answer above 1 is almost always a forgotten intersection.

And \(P(\bar{C} \cap \bar{K}) = 1 - 0.76 = 0.24\) hold neither: 24% of the customer base carries no product at all.

📐 Theorem 2.7: The Complement

Theorem 2.7

\[P(A) = 1 - P(\bar{A})\]

\(A\) and \(\bar{A}\) are mutually exclusive and their union is \(S\), so \(P(A) + P(\bar{A}) = P(S) = 1\) by Axiom 2 and Axiom 3.

Three lines of proof, and it is the most useful result in the chapter.

💡 “At Least One” Is an Intersection in Disguise

A bank guarantees 8 SME loans. Each is called with probability 0.06, independently. What is the probability that at least one is called?

As a union, this is eight events overlapping in every combination — 255 terms of inclusion and exclusion.

As a complement, it is one line. “At least one is called” fails exactly when none is called, and none is an intersection: \[P(\text{at least one}) = 1 - (1 - 0.06)^8 = 1 - 0.6096 = 0.3904\]

💻 At Least One, Across Portfolio Sizes

Code

p_call <- 0.06
n      <- c(1, 4, 8, 12, 20, 40)

data.frame(
  guarantees   = n,
  none_called  = round((1 - p_call)^n, 4),
  at_least_one = round(1 - (1 - p_call)^n, 4)
)

  guarantees none_called at_least_one
1          1      0.9400       0.0600
2          4      0.7807       0.2193
3          8      0.6096       0.3904
4         12      0.4759       0.5241
5         20      0.2901       0.7099
6         40      0.0842       0.9158

A 6% chance on one guarantee is better than even odds by twelve. Small independent risks accumulate, and the complement is what makes that visible in one line.

🔬 Interactive: The Accumulation

Code

viewof p_each = {
  const input = Inputs.range([0.01, 0.30], {value: 0.06, step: 0.01,
    label: "Probability each guarantee is called:"});
  ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove']
    .forEach(e => input.addEventListener(e, ev => ev.stopPropagation()));
  return input;
}

Code

Plot.plot({
  width: 1150,
  height: 320,
  marginLeft: 78,
  marginBottom: 58,
  style: {fontSize: "18px"},
  x: {label: "Number of guarantees", domain: [1, 40]},
  y: {label: "P(at least one called)", domain: [0, 1], tickFormat: ".1f"},
  marks: [
    Plot.ruleY([0.5], {stroke: "#cbb8a9", strokeDasharray: "4 4"}),
    Plot.line(curve, {x: "n", y: "p", stroke: "#14130f", strokeWidth: 2}),
    Plot.dot(half ? [half] : [], {x: "n", y: "p", r: 8, fill: "#8b2635"}),
    Plot.ruleY([0])
  ]
})

🧠 Think-Pair-Share

A borrower must clear two hurdles to draw on a credit line: an affordability test (\(T\)) and a collateral valuation (\(V\)).

\(P(T) = 0.80\), and among those who pass the affordability test, 70% also pass the valuation.

Four minutes, in pairs:

What is \(P(T \cap V)\)?
If additionally \(P(V) = 0.62\), what is \(P(T \cup V)\)?
What is the probability the borrower fails both?

✅ Think-Pair-Share: Solution

The 70% is a conditional, \(P(V \mid T) = 0.70\). Theorem 2.5: \[P(T \cap V) = 0.80 \times 0.70 = 0.56\]
Theorem 2.6: \(P(T \cup V) = 0.80 + 0.62 - 0.56 = 0.86\).

“Fails both” is \(\bar{T} \cap \bar{V}\), which by De Morgan is \(\overline{T \cup V}\). Theorem 2.7: \[P(\bar{T} \cap \bar{V}) = 1 - 0.86 = 0.14\]

✅ Think-Pair-Share: The Habit

Part 3 is where the three results become one tool: De Morgan turned an intersection of complements into the complement of a union, and the union was already computed.

Before reaching for a formula, write the event in set notation. “Neither”, “both”, “at least one”, “only one” each have exactly one reading, and the reading picks the law.

It is the same discipline as Wednesday’s order matters or not — the mathematics is easy once the event is named correctly.

🧭 Which Law? A Reading Guide

The event says	In set notation	Reach for
“both”, “and”, “all of”	\(A \cap B\)	Theorem 2.5
“either”, “or”, “at least one of two”	\(A \cup B\)	Theorem 2.6
“at least one of many”	\(\overline{\bar{A_1} \cap \cdots \cap \bar{A_n}}\)	2.7, then 2.5
“neither”, “none”	\(\bar{A} \cap \bar{B}\)	De Morgan, then 2.7
“exactly one”	\((A \cap \bar{B}) \cup (\bar{A} \cap B)\)	disjoint, so Axiom 3

The last row is worth remembering: those two pieces never overlap, so no subtraction is needed.

📝 Quiz #1: Which Law?

\(P(A) = 0.5\), \(P(B) = 0.4\), \(P(A \cap B) = 0.2\). What is \(P(A \cup B)\)?

\(0.7\)
\(0.9\)
\(0.2\)
\(1.1\)

📝 Quiz #2: The Conditional Is Not Optional

\(P(A) = 0.6\) and \(P(B \mid A) = 0.25\). What is \(P(A \cap B)\)?

\(0.15\)
\(0.85\)
\(0.25\)
Cannot be found without \(P(B)\)

📝 Quiz #3: At Least One

Three independent suppliers each deliver late with probability 0.2. What is the probability that at least one is late?

\(1 - 0.8^3 = 0.488\)
\(3 \times 0.2 = 0.6\)
\(0.2^3 = 0.008\)
\(1 - 0.2^3 = 0.992\)

📝 Quiz #4: Neither

\(P(A \cup B) = 0.75\). What is \(P(\bar{A} \cap \bar{B})\)?

\(0.25\)
\(0.75\)
\(0.5625\)
Cannot be found from \(P(A \cup B)\) alone

📋 Key Formulas

	Statement
Theorem 2.5	\(P(A \cap B) = P(A)P(B \mid A) = P(B)P(A \mid B)\)
independent case	\(P(A \cap B) = P(A)P(B)\)
Theorem 2.5, three events	\(P(A \cap B \cap C) = P(A)P(B \mid A)P(C \mid A \cap B)\)
Theorem 2.6	\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
mutually exclusive case	\(P(A \cup B) = P(A) + P(B)\)
Theorem 2.7	\(P(A) = 1 - P(\bar{A})\)

📋 Summary

The multiplicative law is Definition 2.9 rearranged — it builds an intersection from a marginal and a conditional
Use the form whose conditioning the problem actually gave you
The additive law subtracts the intersection because adding the marginals counts it twice
The complement turns “at least one” from a 255-term union into a single line
De Morgan converts between “neither” and “not either”, and the laws do the rest
Name the event in set notation first; the law follows from the name

📚 Practice Problems

Wackerly, 7th edition

Exercises at the end of §2.8 — start with 2.85 – 2.91, then 2.95 and 2.97
Return to Wednesday’s bond table and verify Theorem 2.5 on it directly, computing \(P(S \cap D)\) both ways round

Week 3, Problem Set 2 opens today and closes 3 October at 23:59 on WeBWorK, covering §2.8. The TA-led tutorial runs this week, before both deadlines.

Next class: the event-composition method — how to decide, for a compound event with no obvious decomposition, which events to name in the first place.

🙏 Thank You

Dr. Samir Orujov

📧 sorujov@ada.edu.az 🏢 Building D, Room D325 🕓 Office hours: Wednesday, 16:00 – 18:00

Slides and readings: sorujov.net/teaching

❓ Questions

Given only \(P(A)\) and \(P(B)\), what are the largest and smallest values \(P(A \cap B)\) can take?
The additive law subtracts one term for two events. How many terms does it need for three?
Mutually exclusive events add, independent events multiply. Which of the two is the stronger assumption to make without evidence?