Mathematical Statistics

Conditional Probability and the Independence of Events

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2026-09-20

🎯 Learning Objectives

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

• State Definition 2.9 and compute \(P(A \mid B)\) from a table or from stated probabilities

• Explain the division by \(P(B)\) as a rescaling of the sample space

• Distinguish \(P(A \mid B)\) from \(P(B \mid A)\), and say why they differ

• Test a pair of events for independence using Definition 2.10

• Argue why two mutually exclusive events of positive probability are necessarily dependent

🗺️ Where We Are

Wackerly §2.7

Saturday we counted. Every probability was \(n_a/N\), and every sample point carried the same weight \(1/N\).

That weighting was an assumption about ignorance: we knew nothing that separated one point from another.

Today: information arrives. A bond is known to be speculative-grade; a borrower is known to have missed a payment. What does the arrival of information do to a probability?

💡 The Motivating Question

Of 1,000 corporate bond issues tracked over five years, 80 defaulted. Unconditionally, \[P(D) = \frac{80}{1000} = 0.08\]

Now you are told the issue was speculative grade. Is 0.08 still your answer?

Obviously not — and “obviously” is exactly what needs a definition. The intuition is that we are no longer working in a sample space of 1,000 issues.

📖 Definition 2.9

Definition 2.9

The conditional probability of an event \(A\), given that an event \(B\) has occurred, is \[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\] provided \(P(B) > 0\).

The proviso is not a technicality. Conditioning on something that cannot happen is not a harder problem; it is not a problem.

⚖️ Why Divide by \(P(B)\)?

Learning that \(B\) occurred does two things.

It discards every sample point outside \(B\). The ones left are exactly those in \(A \cap B\) — which is why the numerator is \(P(A \cap B)\) and not \(P(A)\).

It rescales what remains. The surviving points carry total weight \(P(B)\), not 1, so dividing by \(P(B)\) restores Axiom 2: \(P(B \mid B) = 1\).

Conditioning is not a new kind of probability. It is the same probability measure on a smaller sample space.

📊 The Bond Book

1,000 issues, classified by rating at issue and by whether they defaulted within five years.

	Defaulted (\(D\))	Did not default	Total
Investment grade (\(I\))	12	688	700
Speculative grade (\(S\))	68	232	300
Total	80	920	1000

One issue is drawn at random, every issue equally likely. Then \[P(D \mid S) = \frac{P(D \cap S)}{P(S)} = \frac{68/1000}{300/1000} = \frac{68}{300} = 0.2267\]

🔍 The Counts Were Enough

\[P(D \mid S) = \frac{68/1000}{300/1000} = \frac{68}{300}\]

The 1000s cancel. With equiprobable points, a conditional probability is just the row read on its own.

That is a computational convenience, not the definition. It fails the moment the sample points stop being equally likely — and Definition 2.9 does not.

↔︎️ \(P(A \mid B)\) Is Not \(P(B \mid A)\)

\[P(D \mid S) = \frac{68}{300} = 0.2267 \qquad P(S \mid D) = \frac{68}{80} = 0.8500\]

Same numerator, different denominator, and they answer different questions.

• \(P(D \mid S)\): of the speculative issues, how many defaulted?
• \(P(S \mid D)\): of the defaults, how many were speculative?

Reading one as the other is the most expensive single error in applied probability. We will meet it again on 3 October under its own name.

💻 Both Conditionals in R

bonds <- matrix(c(12, 688, 68, 232), nrow = 2, byrow = TRUE,
                dimnames = list(c("Investment", "Speculative"),
                                c("Default", "No default")))

data.frame(
  quantity = c("P(D)", "P(D | S)", "P(D | I)", "P(S | D)"),
  value    = round(c(sum(bonds[, 1]) / sum(bonds),
                     bonds["Speculative", "Default"] / sum(bonds["Speculative", ]),
                     bonds["Investment",  "Default"] / sum(bonds["Investment",  ]),
                     bonds["Speculative", "Default"] / sum(bonds[, "Default"])), 4)
)

  quantity  value
1     P(D) 0.0800
2 P(D | S) 0.2267
3 P(D | I) 0.0171
4 P(S | D) 0.8500

\(P(D \mid S) = 0.2267\) against \(P(D \mid I) = 0.0171\). Rating carries information about default, and that comparison is the whole of the next definition.

📖 Definition 2.10

Definition 2.10

Two events \(A\) and \(B\) are independent if any one of \[P(A \mid B) = P(A), \qquad P(B \mid A) = P(B), \qquad P(A \cap B) = P(A)P(B)\] holds. If none holds, \(A\) and \(B\) are dependent.

Any one — the three are equivalent whenever both conditioning probabilities are defined. The third is the one to use, because it needs no proviso and is symmetric in \(A\) and \(B\).

🧮 Testing the Bond Book

\[P(S \cap D) = \frac{68}{1000} = 0.068 \qquad P(S)P(D) = 0.300 \times 0.080 = 0.024\]

\(0.068 \neq 0.024\), so rating class and default are dependent — and the direction is informative: defaults among speculative issues run at nearly three times the rate the product rule would predict.

Equivalently, \(P(D \mid S) = 0.2267 \neq 0.08 = P(D)\). Conditioning moved the probability, and that movement is the dependence.

🎚️ Independence Is a Statement About a Pair

Independence is not a property of \(A\), nor of \(B\), nor of the experiment. It is a property of the pair.

A classification with three or more categories can be independent of an event in one category and dependent in another. Verifying one pair tells you nothing about the rest.

Nor is it a physical claim about causation. Independence is an arithmetic relation among three numbers, and nothing more.

⚠️ Independence Is Not Mutual Exclusivity

These two are confused more often than any other pair of ideas in Chapter 2, and they are close to opposites.

• Mutually exclusive: \(A \cap B = \varnothing\). If one happens, the other cannot.
• Independent: \(P(A \cap B) = P(A)P(B)\). If one happens, the other is no less likely than before.

One says the events are maximally informative about each other. The other says they carry no information at all.

📐 The Argument in One Line

A small result worth stating

If \(P(A) > 0\), \(P(B) > 0\) and \(A \cap B = \varnothing\), then \(A\) and \(B\) are dependent.

\[P(A \cap B) = P(\varnothing) = 0, \qquad P(A)P(B) > 0\] so the product rule fails. Equivalently \(P(A \mid B) = 0 \neq P(A)\): learning that \(B\) occurred has driven \(A\)’s probability to zero, which is as much information as one event can carry about another.

🏛️ A Concrete Case

A ministry awards a single contract. Let \(A\) be “firm A wins”, \(B\) be “firm B wins”, with \(P(A) = 0.45\) and \(P(B) = 0.35\).

One contract, so \(A \cap B = \varnothing\) and \(P(A \cap B) = 0\). But \(P(A)P(B) = 0.1575\).

They are dependent — strongly so. Told that firm B won, you know firm A did not. Two firms competing for the same contract are about as far from independent as events get.

🔬 Interactive: How Far From Independence?

viewof spec_def = {
  const input = Inputs.range([6, 140], {value: 68, step: 1,
    label: "Speculative issues that defaulted:"});
  ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove']
    .forEach(e => input.addEventListener(e, ev => ev.stopPropagation()));
  return input;
}

tot = 1000
nS = 300
nI = 700
invDef = (s) => Math.max(0, 150 - s)
pD = (s) => (s + invDef(s)) / tot
bars = {
  const s = spec_def;
  return [
    {what: "P(D | S)", p: s / nS},
    {what: "P(D | I)", p: invDef(s) / nI},
    {what: "P(D)", p: pD(s)}
  ];
}

md`With **${spec_def}** speculative defaults: P(D | S) = **${(spec_def/nS).toFixed(4)}**,
P(D) = **${pD(spec_def).toFixed(4)}**. The two agree exactly when the bars level out —
and that is independence.`

Plot.plot({
  width: 1150,
  height: 300,
  marginLeft: 120,
  marginBottom: 55,
  style: {fontSize: "18px"},
  x: {label: "Probability", domain: [0, 0.5]},
  y: {label: null, domain: ["P(D | S)", "P(D | I)", "P(D)"]},
  marks: [
    Plot.barX(bars, {y: "what", x: "p", fill: "#8b2635"}),
    Plot.ruleX([pD(spec_def)], {stroke: "#14130f", strokeDasharray: "4 4"}),
    Plot.ruleX([0])
  ]
})

🧠 Think-Pair-Share

A bank’s currency desk records, for each trading day, whether the euro book made a profit (\(E\)) and whether the dollar book made a profit (\(U\)).

Over 500 days: \(E\) on 300, \(U\) on 250, both on 150.

Three minutes, in pairs:

1. Compute \(P(U \mid E)\) and \(P(U)\).

2. Are \(E\) and \(U\) independent?

3. Are they mutually exclusive? Could they be both?

✅ Think-Pair-Share: Solution

1. \(P(U \mid E) = \dfrac{150/500}{300/500} = \dfrac{150}{300} = 0.50\), and \(P(U) = \dfrac{250}{500} = 0.50\).

2. They are equal, so by Definition 2.10 the two books are independent. The product check agrees: \(P(E \cap U) = 0.30 = 0.60 \times 0.50\).

3. Not mutually exclusive — they happened together on 150 days. And they could not be both: mutually exclusive events of positive probability are always dependent.

🔄 Conditioning in Sequence

A placement of 12 mortgage loans contains 4 already in arrears. Two are drawn for review, without replacement.

\[P(A_1) = \frac{4}{12} = 0.3333\]

The second draw is made from what is left, so its probability is conditional by construction: \[P(A_2 \mid A_1) = \frac{3}{11} = 0.2727, \qquad P(A_2 \mid \bar{A_1}) = \frac{4}{11} = 0.3636\]

The two differ, so \(A_1\) and \(A_2\) are dependent. Sampling without replacement always produces dependence; sampling with replacement never does.

📝 Quiz #1: Reading the Definition

\(P(A \cap B) = 0.12\) and \(P(B) = 0.40\). What is \(P(A \mid B)\)?

• \(0.30\)
• \(0.048\)
• \(0.52\)
• \(3.33\)

📝 Quiz #2: Which Way Round?

Of 200 flagged transactions, 30 were genuinely fraudulent; the bank processed 5,000 transactions in all. Which quantity is \(30/200\)?

• \(P(\text{fraud} \mid \text{flagged})\)
• \(P(\text{flagged} \mid \text{fraud})\)
• \(P(\text{fraud} \cap \text{flagged})\)
• \(P(\text{flagged})\)

📝 Quiz #3: Exclusive or Independent?

\(A\) and \(B\) are mutually exclusive, with \(P(A) = 0.3\) and \(P(B) = 0.5\). Are they independent?

• No — \(P(A \cap B) = 0\) but \(P(A)P(B) = 0.15\)
• Yes — mutually exclusive events are always independent
• Yes, because \(P(A) + P(B) < 1\)
• There is not enough information to say

📝 Quiz #4: What Independence Buys

\(A\) and \(B\) are independent with \(P(A) = 0.6\) and \(P(B) = 0.25\). What is \(P(A \mid B)\)?

• \(0.60\)
• \(0.15\)
• \(0.25\)
• \(2.40\)

📋 Key Formulas

	Statement	Requires
Definition 2.9	\(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\)	\(P(B) > 0\)
Definition 2.10	\(P(A \cap B) = P(A)P(B)\)	nothing
equivalently	\(P(A \mid B) = P(A)\)	\(P(B) > 0\)
mutually exclusive	\(P(A \cap B) = 0\)	nothing

The last two rows are compatible only when one of the events has probability zero.

📋 Summary

• Conditioning discards the points outside \(B\) and rescales what is left by \(P(B)\)

• \(P(A \mid B)\) and \(P(B \mid A)\) share a numerator and answer different questions

• Independence means the conditioning changes nothing: \(P(A \mid B) = P(A)\)

• Independence is a property of a pair, verified pair by pair

• Mutually exclusive events of positive probability are dependent, never independent

• Sampling without replacement is the everyday source of dependence

📚 Practice Problems

Wackerly, 7th edition

• Exercises at the end of §2.7 — start with 2.69 – 2.75, then 2.79 and 2.81

• Redo the bond table with the roles of the two classifications swapped, and check that the dependence verdict is unchanged

Week 3, Problem Set 1 opens today and closes 3 October at 23:59 on WeBWorK, covering §2.7. Twenty problems; unlimited attempts before the deadline.

Next class: the multiplicative and additive laws — Definition 2.9 rearranged into a tool, and the rule for the probability of a union.

🙏 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

• If \(A \subset B\), what are \(P(A \mid B)\) and \(P(B \mid A)\), without computing either?

• Can an event be independent of itself? Which events, exactly?

• The bond table’s dependence was read as “rating carries information about default”. What would have to be true of the table for the phrase to be reversed?