Chapter 04

Medium

Independence of Events

00 · Symbol Glossary

$A\perp B$A independent of B

Shorthand for $\mathbb{P}(A\cap B)=\mathbb{P}(A)\,\mathbb{P}(B)$ . It is a statement about the probability measure, not about the sets. Some texts draw the same relation with a double-stemmed symbol.

$\mathbb{P}(A\cap B)$Joint probability of A and B

Probability that both events occur. Independence is the claim that this number equals the product of the two individual probabilities.

$A^c$A complement

The event that $A$ does not occur, with $\mathbb{P}(A^c)=1-\mathbb{P}(A)$ . Independence of $A$ and $B$ passes automatically to complements.

$A\perp B\mid C$A independent of B given C

Conditional independence: inside the world where $C$ occurred, $A$ and $B$ factor. Formally $\mathbb{P}(A\cap B\mid C)=\mathbb{P}(A\mid C)\,\mathbb{P}(B\mid C)$ when $\mathbb{P}(C)>0$ . This can hold in every regime and still fail for the unconditional probabilities.

$A_1,\ldots,A_n$A family of events

Several events considered together — for example, "bond $i$ defaults this year" for each bond in a portfolio. Independence of a family is stronger than independence of each pair.

$p_i$p sub i — marginal default probability

$p_i=\mathbb{P}(\text{name } i \text{ defaults})$ , the probability for one issuer on its own. Multiplying the $p_i$ is valid only under mutual independence, and even then it is one point inside a wider range the marginals allow.

01 · When Information Does Not Help

Chapter 03 showed that learning $B$ generally changes the probability of $A$ . Sometimes it does not: knowing that $B$ occurred leaves $\mathbb{P}(A)$ exactly where it was. That special case is independence.

Independence is the assumption behind "i.i.d. shocks", "idiosyncratic risk diversifies away", and every probability computed by multiplying marginals. It is the most useful simplification in probability and one of the most dangerous in finance, because the events that matter most — defaults, margin calls, liquidity runs, VaR breaches — tend to arrive together.

The test is always arithmetic. A story about causes ("the two desks do not talk") is not a proof, and a story about overlap ("the events share an outcome") is not a disproof. Sections 03 and 04 separate those confusions.

02 · The Product Rule

Definition — Independence of Two Events

Events $A$ and $B$ are independent, written $A\perp B$ , if

\mathbb{P}(A\cap B)=\mathbb{P}(A)\,\mathbb{P}(B)

Otherwise they are dependent.

Plain language: the probability of the overlap is exactly what you would get by multiplying the two separate probabilities. No extra mass sits on the overlap, and none has been pulled away from it.

The definition is symmetric in $A$ and $B$ , and it makes sense even when one event has probability zero. The intuitive version — "learning $B$ does not change the chance of $A$ " — is equivalent whenever the conditional probability exists.

Step-by-step — Product rule and the conditional form

Assume

\mathbb{P}(B)>0

, so

\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)

is defined (Chapter 03).

Product implies conditional. If

\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)

, divide by

\mathbb{P}(B)

\mathbb{P}(A\mid B)=\mathbb{P}(A)

Conditional implies product. If

\mathbb{P}(A\mid B)=\mathbb{P}(A)

, the multiplication rule gives

\mathbb{P}(A\cap B)=\mathbb{P}(A\mid B)\mathbb{P}(B)=\mathbb{P}(A)\mathbb{P}(B)

Symmetry. If also

\mathbb{P}(A)>0

, the same argument shows

\mathbb{P}(B\mid A)=\mathbb{P}(B)

. Information flows neither way.

Chain rule collapses. Under independence the Chapter 03 chain rule loses every condition:

\mathbb{P}(A_1\cap\cdots\cap A_n)=\mathbb{P}(A_1)\cdots\mathbb{P}(A_n)

, but only when the whole family is mutually independent (Section 06), not merely when each consecutive pair is.

The product form is taken as the definition because it needs no positivity condition. An event with $\mathbb{P}(B)=0$ satisfies $\mathbb{P}(A\cap B)=0=\mathbb{P}(A)\cdot 0$ , so it is independent of every event; so is an event with $\mathbb{P}(B)=1$ . Such events carry no information.

Example — Independence you cannot see in the picture

Roll a fair die. Let $A$ = "even" $=\{2,4,6\}$ and $B=\{1,2\}$ . The sets overlap in one outcome and look unrelated to parity. Check the numbers: $\mathbb{P}(A)=1/2$ , $\mathbb{P}(B)=1/3$ , and

\mathbb{P}(A\cap B)=\mathbb{P}(\{2\})=\frac16=\frac12\cdot\frac13=\mathbb{P}(A)\,\mathbb{P}(B)

So $A\perp B$ : within $B$ , exactly half the outcomes are even, matching the whole space. Replace $B$ with $\{1,2,3\}$ and $\mathbb{P}(A\mid B)$ becomes $1/3$ ; independence is gone. Whether events are independent is decided by the arithmetic, not by whether they "seem related".

Independence also survives taking complements. If $A$ and $B$ are independent, then so are $A$ and $B^c$ , $A^c$ and $B$ , and $A^c$ and $B^c$ (Exercise 4.1). Knowing that $B$ did not happen is just as uninformative about $A$ as knowing that it did. That fact is what lets a "none of them default" probability become a product of survival probabilities.

03 · Independent Is Not Disjoint

Beginners often merge "independent" and "mutually exclusive" because both sound like "having nothing to do with each other". They are close to opposites.

❌ Treating disjoint events as independent

Take one issuer and let $A$ = "first default occurs in year 1" and $B$ = "first default occurs in year 2", each with positive probability. An analyst calls them unrelated and multiplies: $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ .

Why it breaks: a first default can happen in only one year, so $A\cap B=\emptyset$ and $\mathbb{P}(A\cap B)=0$ , while $\mathbb{P}(A)\mathbb{P}(B)>0$ . The events are maximally dependent: learning $A$ drives $\mathbb{P}(B\mid A)$ to $0$ .

Consequence: disjoint events with positive probability are never independent. Mixing up the two notions leads to adding probabilities that should be multiplied, or multiplying probabilities whose product is structurally zero. Disjointness is a property of the sets; independence is a property of $\mathbb{P}$ .

Two further extremes belong with the same warning. An event is independent of itself only when its probability is $0$ or $1$ (Exercise 4.2): anything you can still be uncertain about tells you everything about itself. And $A$ is never independent of $A^c$ when $0<\mathbb{P}(A)<1$ , because those two events are disjoint.

04 · Independence Depends on the Measure

Change the probabilities on the same sample space and independence can appear or disappear. The sets $A$ and $B$ stay the same; only $\mathbb{P}$ changes.

Example — Same sets, two measures

On $\Omega=\{1,2,3,4\}$ let $A=\{1,2\}$ and $B=\{1,3\}$ . Under the uniform measure, $\mathbb{P}(A\cap B)=\mathbb{P}(\{1\})=1/4=(1/2)(1/2)$ : independent.

Now put $\mathbb{P}(\{1\})=1/2$ and $\mathbb{P}(\{2\})=\mathbb{P}(\{3\})=\mathbb{P}(\{4\})=1/6$ . Then $\mathbb{P}(A)=\mathbb{P}(B)=2/3$ , so $\mathbb{P}(A)\mathbb{P}(B)=4/9$ , but $\mathbb{P}(A\cap B)=1/2$ . Dependent, with $A$ and $B$ now occurring together more often than independence would allow.

The finance version is the one that matters. A conditional probability $\mathbb{P}(\cdot\mid\text{regime})$ is a different measure from $\mathbb{P}$ (Chapter 03), so events can be independent under one and dependent under the other.

Example — Default clustering from a common regime

Two firms' defaults depend on the economy. In a good year (probability $0.9$ ) each defaults with probability $0.01$ ; in a bad year (probability $0.1$ ) each defaults with probability $0.10$ . Within a regime the two defaults are independent. Let $A$ , $B$ be "firm 1 defaults" and "firm 2 defaults".

By total probability, $\mathbb{P}(A)=\mathbb{P}(B)=0.9\cdot 0.01+0.1\cdot 0.10=0.019$ . Applying the product rule inside each regime and then total probability,

\mathbb{P}(A\cap B)=0.9\cdot(0.01)^2+0.1\cdot(0.10)^2=0.00009+0.001=0.00109

whereas $\mathbb{P}(A)\mathbb{P}(B)=0.019^2=0.000361$ . The joint default probability is about three times the independent value, and $\mathbb{P}(B\mid A)=0.00109/0.019\approx 0.057$ — learning that firm 1 defaulted triples the chance for firm 2.

No direct link connects the two firms. The dependence comes entirely from both reacting to an unobserved common state: a default by firm 1 is evidence (via Bayes) that the year is bad, and a bad year is bad for firm 2. Independence given the regime does not imply independence overall. The next section names that distinction.

Dependence does not require a hidden regime. One event can change the odds of the next directly.

Example — A VaR breach that raises the next day's odds

Let $B_1$ be "the book breaches its one-day VaR on Monday" and $B_2$ the same event on Tuesday, with $\mathbb{P}(B_1)=0.1$ . After a breach the desk de-risks badly and $\mathbb{P}(B_2\mid B_1)=0.4$ ; after a clean day, $\mathbb{P}(B_2\mid B_1^c)=0.1$ .

\mathbb{P}(B_2)=0.4\cdot 0.1+0.1\cdot 0.9=0.13, \qquad \mathbb{P}(B_1\cap B_2)=0.4\cdot 0.1=0.04

The product of the marginals is $0.1\cdot 0.13=0.013$ . The joint is just over three times larger, and $\mathbb{P}(B_2\mid B_1)=0.4\neq 0.13=\mathbb{P}(B_2)$ . A limit that treats successive days as independent understates the chance of a two-day breach cluster even though each day's marginal was computed honestly.

05 · Conditional Independence

The regime example and the "second test" update in Chapter 03 use the same idea: independence inside a smaller world.

Definition — Conditional Independence

Let $\mathbb{P}(C)>0$ . Events $A$ and $B$ are conditionally independent given $C$ , written $A\perp B\mid C$ , if

\mathbb{P}(A\cap B\mid C)=\mathbb{P}(A\mid C)\,\mathbb{P}(B\mid C)

If also $\mathbb{P}(B\cap C)>0$ , this is equivalent to $\mathbb{P}(A\mid B\cap C)=\mathbb{P}(A\mid C)$ : once you know $C$ , learning $B$ does not move $A$ any further.

Plain language: $C$ has already extracted the shared information. Whatever is left of $A$ and of $B$ does not talk to each other.

If $R_1,\ldots,R_m$ partition the outcomes (the regimes) and $A\perp B\mid R_i$ for every $i$ with $\mathbb{P}(R_i)>0$ , the law of total probability gives

\mathbb{P}(A\cap B)=\sum_{i=1}^m \mathbb{P}(A\mid R_i)\,\mathbb{P}(B\mid R_i)\,\mathbb{P}(R_i)

The right-hand side equals $\mathbb{P}(A)\mathbb{P}(B)$ only in special cases — for instance when the conditional probabilities do not actually depend on $i$ . The Section 04 arithmetic is this formula with two regimes: $0.9\cdot(0.01)^2+0.1\cdot(0.10)^2$ .

What Chapter 03 was assuming

The odds form in Chapter 03 said that a second test multiplies the likelihood ratio again when the two tests "err independently given the bank's true state." That phrase is $A\perp B\mid F$ and $A\perp B\mid F^c$ : conditional independence given fragile and given sound. It is not a claim that the two flags are independent before you know the bank's state. They are not — both flags are evidence of the same state, exactly as the two defaults are evidence of the same regime.

Conditional independence does not imply ordinary independence, and ordinary independence does not survive arbitrary conditioning. The second failure is easy to miss because the filter looks like "we only kept the interesting cases."

❌ Estimating dependence from the cases you bothered to open

Two alert systems fire independently, each with probability $1/5$ . A review team opens a file only when at least one alert fires, then measures how often the alerts arrive together inside those files.

Why it breaks: unconditionally $\mathbb{P}(A\cap B)=(1/5)(1/5)=1/25$ and $\mathbb{P}(A\cup B)=9/25$ , so the alerts are independent. Inside the opened files,

\mathbb{P}(A\cap B\mid A\cup B)=\frac{1/25}{9/25}=\frac19, \qquad \mathbb{P}(A\mid A\cup B)\,\mathbb{P}(B\mid A\cup B)=\left(\frac{5}{9}\right)^2=\frac{25}{81}

and $1/9\neq 25/81$ . Given that a file was opened, learning that alert $B$ fired drops the chance of alert $A$ from $5/9$ down to $\mathbb{P}(A\mid B)=1/5$ , because $B$ already explains why the file exists.

Consequence: the filtered sample invents a negative relationship. The same distortion hits any study that keeps only days with a breach, only names that were downgraded, or only months in which "something happened," and then reads the dependence off that slice. Conditioning on a union of the events under study destroys independence that was really there.

06 · Several Events: Pairwise Versus Mutual

With three or more events, there is more than one thing "independent" could mean.

Definition — Mutual Independence

Events $A_1,\ldots,A_n$ are mutually independent if for every choice of distinct indices $i_1,\ldots,i_k$ with $k\ge 2$ ,

\mathbb{P}(A_{i_1}\cap A_{i_2}\cap\cdots\cap A_{i_k})=\mathbb{P}(A_{i_1})\,\mathbb{P}(A_{i_2})\cdots\mathbb{P}(A_{i_k})

They are pairwise independent if this holds just for every pair ( $k=2$ ). A sequence of events is independent when every finite subcollection is mutually independent.

For $n$ events, mutual independence imposes $2^n-n-1$ product conditions; pairwise independence imposes only $n(n-1)/2$ of them. For $n\ge 3$ the extra conditions are genuine constraints.

Step-by-step — Pairwise but not mutually independent

Setup. Flip two fair coins:

\Omega=\{HH,HT,TH,TT\}

, uniform. Let

A

= "first coin heads"

=\{HH,HT\}

B

= "second coin heads"

=\{HH,TH\}

C

= "coins match"

=\{HH,TT\}

. Each has probability

1/2

Pairs.

A\cap B=\{HH\}

A\cap C=\{HH\}

B\cap C=\{HH\}

. Each has probability

1/4=(1/2)(1/2)

, so every pair is independent.

Triple.

A\cap B\cap C=\{HH\}

has probability

1/4

, but

\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)=1/8

Interpretation. Any one event tells you nothing about any other one. But

A

and

B

together determine

C

completely:

\mathbb{P}(C\mid A\cap B)=1\neq 1/2

The failure can also go the other way: the single triple equation $\mathbb{P}(A\cap B\cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)$ can hold while some pair is dependent (Exercise 4.4). Mutual independence needs every subcollection to factor. In this library, "independent" for a family of events always means mutually independent unless "pairwise" is said explicitly; the same convention carries over to the i.i.d. random variables of Chapter 05.

❌ Treating a pairwise check as a license to multiply the whole book

A weekly risk report checks that no pair of trading days has dependent VaR-breach flags, then writes the probability of a clean week as $0.99^5$ .

Why it breaks: "no breach on any of the five days" is one intersection of five complements. Pairwise independence controls two-day overlaps only. The coin construction above is the template at $n=3$ : every pair factors, and the triple does not, so a product over all three events is already wrong.

Consequence: the clean-week number, the "all five names survive" number, and any digital that pays only when every leg finishes in the money are statements about the full intersection. They need mutual independence. A correlation matrix of pairs cannot certify them.

07 · What the Marginals Still Allow

Independence is one joint probability, not the only legal one. Chapter 02 already bounded the overlap using nothing but the axioms.

Proof — Fréchet bounds for two events

Monotonicity gives $\mathbb{P}(A\cap B)\le\mathbb{P}(A)$ and $\mathbb{P}(A\cap B)\le\mathbb{P}(B)$ , hence $\mathbb{P}(A\cap B)\le\min(\mathbb{P}(A),\mathbb{P}(B))$ .

For the other end, inclusion–exclusion and $\mathbb{P}(A\cup B)\le 1$ rearrange to $\mathbb{P}(A\cap B)\ge\mathbb{P}(A)+\mathbb{P}(B)-1$ . Combined with non-negativity,

\max\!\big(0,\ \mathbb{P}(A)+\mathbb{P}(B)-1\big)\ \le\ \mathbb{P}(A\cap B)\ \le\ \min\!\big(\mathbb{P}(A),\mathbb{P}(B)\big)

Both ends are attainable. The upper end is the rarer event sitting entirely inside the more likely one. The lower end is an empty overlap when $\mathbb{P}(A)+\mathbb{P}(B)\le 1$ ; when the probabilities sum to more than $1$ , an empty overlap would break normalization, and the bound $\mathbb{P}(A)+\mathbb{P}(B)-1$ is achieved by making the overlap exactly that large and no larger. For the rare names below, the sum is far under $1$ , so the legal joint default probability really does start at $0$ .

Example — Two names, marginals 4% and 3%

Let $\mathbb{P}(A)=0.04$ and $\mathbb{P}(B)=0.03$ . The joint default probability must lie in

\max(0,\ 0.04+0.03-1)\ \le\ \mathbb{P}(A\cap B)\ \le\ \min(0.04,\ 0.03)

so $0\le\mathbb{P}(A\cap B)\le 0.03$ . Independence picks the single point $0.04\cdot 0.03=0.0012$ , twenty-five times smaller than the upper bound.

The probability of at least one default is $\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$ , so it runs the other way: from $0.04$ (joint at its maximum — the $3\%$ name defaults only when the $4\%$ name does) up to $0.07$ (joint at $0$ — the defaults never coincide). Independence gives $0.0688$ , close to the disjoint end and far from the pile-up end.

Read the two ends as books, not as algebra. A contract that loses only when both names default is most expensive at the upper bound. A contract that loses when either name defaults is most expensive at the lower bound. Multiplying marginals prices neither contract at its worst legal value; it prices a model in the interior. For rare names the interior point is near the bottom of the joint-default range, which is why "assume independence" is a quiet way to cheapen senior protection. Chapter 10 builds dependence between random variables directly; the bounds here need no model beyond the marginals.

08 · Multiplying Default Probabilities

Under mutual independence, joint probabilities of a whole portfolio follow from the marginals alone:

\mathbb{P}(\text{all } n \text{ names default})=\prod_{i=1}^n p_i, \qquad \mathbb{P}(\text{at least one default})=1-\prod_{i=1}^n(1-p_i)

The second formula applies the product rule to the complements $A_i^c$ , which are mutually independent whenever the $A_i$ are (the $n=2$ case is Exercise 4.1; the general case is the same argument applied to every subcollection). Both formulas are exact only under mutual independence. Pairwise independence is not enough, since the joint event involves all $n$ names at once. And even mutual independence is a point inside the Fréchet range, not a bound.

Example — Five bonds, two models with the same marginals

Five bonds each have a one-year default probability of $2\%$ .

Independent model. $\mathbb{P}(\text{all five default})=0.02^5=3.2\times 10^{-9}$ and $\mathbb{P}(\text{at least one})=1-0.98^5\approx 0.096$ .

Regime model. In a good year (probability $0.9$ ) each bond defaults with probability $0.01$ ; in a bad year (probability $0.1$ ) with probability $0.11$ ; defaults are conditionally independent within each regime. The marginal is still $0.9\cdot 0.01+0.1\cdot 0.11=0.02$ . But

\mathbb{P}(\text{all five default})=0.9\cdot(0.01)^5+0.1\cdot(0.11)^5\approx 1.6\times 10^{-6}

about $500$ times the independent answer, while $\mathbb{P}(\text{at least one})=1-\left(0.9\cdot 0.99^5+0.1\cdot 0.89^5\right)\approx 0.088$ is slightly lower.

What multiplying marginals assumes

The two models agree on every single-name number and disagree by orders of magnitude on the joint tail. Positive dependence moves probability toward the extremes: more years with no defaults, and more years with many. A senior tranche that loses only when most names default is priced almost entirely off that joint tail, so a desk that multiplies marginal $p_i$ is asserting mutual independence exactly where it is least defensible. The same product, with $0.99$ in place of $0.98$ , is the "probability of at least one $1\%$ VaR breach in ten days equals $1-0.99^{10}\approx 0.096$ " calculation — exact if the ten days are mutually independent, and not otherwise.

Practice — Spot the hidden assumption

For each statement, decide whether it relies on independence, and of which kind (pairwise, mutual, or conditional): (a) "the chance of at least one $1\%$ VaR breach in ten trading days is $1-0.99^{10}$ "; (b) "the chance of three breaches in a row is $10^{-6}$ "; (c) "the chance that both counterparties fail is the product of their default probabilities"; (d) "a stock is equally likely to rise on Monday whether or not it rose on Friday"; (e) "given the regime, multiplying the two default probabilities is legitimate, and we then average over regimes."

09 · Exercises

EXERCISE 4.1

Write $A$ as the disjoint union of $A\cap B$ and $A\cap B^c$ , so $\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$ .

$\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)=\mathbb{P}(A)-\mathbb{P}(A)\mathbb{P}(B)=\mathbb{P}(A)(1-\mathbb{P}(B))=\mathbb{P}(A)\mathbb{P}(B^c)$ , so $A$ and $B^c$ are independent. Swapping roles gives $A^c$ and $B$ independent, and applying the result once more to the pair $A^c$ , $B$ gives $A^c$ and $B^c$ independent. The survival product $1-\prod(1-p_i)$ in Section 08 is this fact applied to every name at once, once mutual independence of the $A_i$ is known.

Prove that if $A$ and $B$ are independent, then $A$ and $B^c$ are independent. Deduce that $A^c$ and $B^c$ are independent.

EXERCISE 4.2

Independence of $A$ with itself means $\mathbb{P}(A\cap A)=\mathbb{P}(A)^2$ . For $A$ and $A^c$ , compare $\mathbb{P}(A\cap A^c)$ with $\mathbb{P}(A)\mathbb{P}(A^c)$ .

Since $A\cap A=A$ , self-independence reads $\mathbb{P}(A)=\mathbb{P}(A)^2$ , i.e. $\mathbb{P}(A)(1-\mathbb{P}(A))=0$ , so $\mathbb{P}(A)$ is $0$ or $1$ . Conversely, Section 02 already noted that such events are independent of every event, hence of themselves. For the complement, $\mathbb{P}(A\cap A^c)=0$ while $\mathbb{P}(A)\mathbb{P}(A^c)=\mathbb{P}(A)(1-\mathbb{P}(A))$ , which is strictly positive exactly when $0<\mathbb{P}(A)<1$ . So a nontrivial event is never independent of itself or of its complement — both are extreme cases of Section 03's disjoint-versus-independent point.

Show that an event $A$ is independent of itself if and only if $\mathbb{P}(A)\in\{0,1\}$ . Then show that if $0<\mathbb{P}(A)<1$ , the events $A$ and $A^c$ are dependent.

EXERCISE 4.3

The defaults are conditionally independent given the year. Use total probability over the two regimes for both $\mathbb{P}(A)$ and $\mathbb{P}(A\cap B)$ , applying the product rule only inside each regime. For the last part, use Bayes from Chapter 03.

$\mathbb{P}(A)=\mathbb{P}(B)=0.95\cdot 0.02+0.05\cdot 0.20=0.019+0.010=0.029$ . Because of conditional independence inside each regime, $\mathbb{P}(A\cap B)=0.95\cdot 0.0004+0.05\cdot 0.04=0.00038+0.002=0.00238$ , while $\mathbb{P}(A)\mathbb{P}(B)=0.029^2=0.000841$ . The events are unconditionally dependent, and the joint default probability is about $2.8$ times the independent value. $\mathbb{P}(B\mid A)=0.00238/0.029\approx 0.082$ , up from $0.029$ . By Bayes, $\mathbb{P}(\text{stress}\mid A)=0.05\cdot 0.20/0.029\approx 0.345$ , against a prior of $0.05$ : one default makes stress roughly seven times as likely, which is what raises the second firm's default probability.

Two firms each default with probability $0.02$ in a calm year (probability $0.95$ ) and $0.20$ in a stress year (probability $0.05$ ), conditionally independently within each regime. Let $A$ and $B$ be the two default events. Compute $\mathbb{P}(A)$ , $\mathbb{P}(A\cap B)$ , and $\mathbb{P}(B\mid A)$ , decide whether $A$ and $B$ are independent, and find $\mathbb{P}(\text{stress}\mid A)$ to explain the mechanism.

EXERCISE 4.4

There are eight atomic patterns for which of $A$ , $B$ , $C$ occur. Add the atoms that sit inside each event, and compare each pair-intersection with the product of its marginals. One triple equality is not mutual independence.

Count scenarios inside each event. $A$ occurs in the "all three" scenario, the "only $A$ and $C$ " scenario, and the two "only $A$ " scenarios: $1+1+2=4$ , so $\mathbb{P}(A)=4/8=1/2$ . The same count gives $\mathbb{P}(B)=\mathbb{P}(C)=1/2$ . The triple intersection is the single "all three" scenario, probability $1/8$ , which equals $(1/2)^3$ . But $A\cap B$ is only that same scenario ("only $A$ and $B$ " never occurs), so $\mathbb{P}(A\cap B)=1/8$ , while $\mathbb{P}(A)\mathbb{P}(B)=1/4$ . The single three-event product holds and the pair $A,B$ is dependent: mutual independence fails. The other pairs happen to factor, $\mathbb{P}(A\cap C)=\mathbb{P}(B\cap C)=2/8=1/4$ . One well-behaved pair, or one well-behaved triple, never certifies the rest.

Eight equally likely market scenarios are labeled by which of three events occur, with weights: all three of $A,B,C$ occur in $1$ scenario; only $A$ and $C$ in $1$ ; only $A$ in $2$ ; only $B$ and $C$ in $1$ ; only $B$ in $2$ ; only $C$ in $1$ ; and the patterns "only $A$ and $B$ " and "none" never occur. Show that $\mathbb{P}(A\cap B\cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)$ but that $A,B,C$ are not mutually independent.

EXERCISE 4.5

Unconditionally the alerts are independent, so the joint is the product and the union comes from inclusion–exclusion. Divide by $\mathbb{P}(A\cup B)$ for the filtered probabilities. Compare $\mathbb{P}(A\mid B)$ with $\mathbb{P}(A\mid A\cup B)$ .

$\mathbb{P}(A\cap B)=0.01$ and $\mathbb{P}(A\cup B)=0.1+0.1-0.01=0.19$ , so unconditionally $A\perp B$ . Inside opened files, $\mathbb{P}(A\cap B\mid A\cup B)=0.01/0.19=1/19\approx 0.0526$ and $\mathbb{P}(A\mid A\cup B)=0.1/0.19=10/19\approx 0.526$ , whose square is $100/361\approx 0.277$ . These are not equal, so the alerts are not conditionally independent given $A\cup B$ .

Given a file was opened, learning that $B$ fired moves $\mathbb{P}(A\mid \cdot)$ from $10/19$ down to $\mathbb{P}(A\mid B)=0.1$ . The review sample will report a negative association that the two systems do not have. This is the Section 05 selection effect at a different firing rate; the direction does not depend on the rate being $1/5$ .

Two surveillance alerts fire independently, each with probability $0.1$ . A review opens a case only when at least one alert fires. Show that the alerts are independent, then show they are not conditionally independent given that a case was opened. By how much does observing alert $B$ inside an opened case change the probability of alert $A$ ?

10 · Chapter Summary

Concept	Meaning
Independence	$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ ; equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$ when $\mathbb{P}(B)>0$
Complements	independence of $A,B$ passes to $A^c$ and $B^c$
Disjoint vs independent	disjoint events with positive probability are always dependent
Conditional independence	product rule inside $\mathbb{P}(\cdot\mid C)$ ; given every regime, this still need not give unconditional independence
Selection	conditioning on "at least one of them happened" can create dependence
Mutual vs pairwise	mutual needs every subcollection to factor; one triple equality, or every pair, is not enough
Fréchet bounds	given only the marginals, the joint lies between $\max(0,p+q-1)$ and $\min(p,q)$
Multiplying default probabilities	the independence point inside that range, valid for the full portfolio only under mutual independence

Next: Chapter 05 — Discrete Random Variables, which attaches numbers to outcomes so that probability can describe distributions, and extends independence from events to random variables.