Statistics Graduate Level Quiz: Sigma Algebras And Probability Axioms

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Sigma Algebras And Probability AxiomsQuestion 1 of 10

Let $\Omega$ be uncountable, and define $\mathcal C$ to be the collection of all subsets of $\Omega$ that are either countable or have countable complement. For $E\in\mathcal C$ , define $P(E)=0$ if $E$ is countable and $P(E)=1$ if $E^c$ is countable. Which conclusion is correct?

\mathcal C

is a sigma-algebra, and

P

is a countably additive probability measure on it.

\mathcal C

is a sigma-algebra, but

P

fails countable additivity on disjoint countable sets.

\mathcal C

is only an algebra because a countable union of co-countable sets need not belong to it.

\mathcal C

is not an algebra because the complement of a countable set is outside the collection.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Sigma Algebras And Probability Axioms

Practice Sigma Algebras And Probability Axioms in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sigma Algebras And Probability Axioms, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\mathcal C$ is a sigma-algebra, and $P$ is a countably additive probability measure on it. (correct answer)
$\mathcal C$ is a sigma-algebra, but $P$ fails countable additivity on disjoint countable sets.
$\mathcal C$ is only an algebra because a countable union of co-countable sets need not belong to it.
$\mathcal C$ is not an algebra because the complement of a countable set is outside the collection.

Explanation: When you encounter a question like this, you're being tested on two things simultaneously: whether a construction forms a valid sigma-algebra, and whether a set function defined on it is a legitimate probability measure. Work through each property methodically. The collection

\mathcal{C}

— all subsets of

\Omega

that are countable or have countable complement — is called the countable-cocountable sigma-algebra, a classical example worth knowing cold. To verify it's a sigma-algebra:

\emptyset

is countable, so

\emptyset \in \mathcal{C}

. If

E

is countable, then

E^c

has countable complement (namely

E

), so

E^c \in \mathcal{C}

; and if

E^c

is countable, then

E^c \in \mathcal{C}

trivially. For countable unions, if every

E_n

is countable, their union is countable; if even one

E_n

has countable complement, then

(\bigcup E_n)^c = \bigcap E_n^c \subseteq E_n^c

, which is countable. So

\mathcal{C}

is closed under countable unions — it's a full sigma-algebra. Now check

P

: it assigns 0 to countable sets and 1 to cocountable sets. For countable additivity on disjoint sets

\{E_n\}

: at most one

E_n

can be cocountable (since disjoint cocountable sets would require

\Omega

to be countable, contradicting our assumption). So the sum of measures equals either

0

(all countable) or

1

(exactly one cocountable), matching

P(\bigcup E_n)

. This confirms A is correct. Choice B is wrong because the additivity argument above shows no failure occurs — disjoint countable sets sum to 0, matching the countable union. Choice C incorrectly claims cocountable sets break closure under unions; the argument above shows the opposite. Choice D is simply false — the complement of a countable set is cocountable, which belongs to

\mathcal{C}

by definition. Study tip: Memorize the countable-cocountable sigma-algebra as a canonical example — it appears frequently on graduate probability exams to test whether you can verify sigma-algebra axioms and countable additivity from first principles rather than by intuition.

Question 2

Suppose $Q$ is a finitely additive set function on $\mathcal P(\mathbb N)$ satisfying $Q(\mathbb N)=1$ and $Q(\{n\})=0$ for every $n\in\mathbb N$ . Which sequence directly demonstrates that $Q$ cannot satisfy continuity from below?

$E_k=\{2,4,\ldots,2k\}$ , since $E_k\uparrow 2\mathbb N$ while every $Q(E_k)=0$ .
$E_k=\{k,k+1,\ldots\}$ , since $E_k\downarrow\varnothing$ while every $Q(E_k)=1$ .
$E_k=\{1,\ldots,k\}$ , since $E_k\uparrow\mathbb N$ while every $Q(E_k)=0$ . (correct answer)
$E_k=\mathbb N$ for every $k$ , since $E_k\uparrow\mathbb N$ while every $Q(E_k)=1$ .

Explanation: Whenever you see a question about continuity from below, recall the precise definition: if

E_k \uparrow E

(i.e.,

E_1 \subseteq E_2 \subseteq \cdots

and

\bigcup_k E_k = E

), then continuity from below requires

Q(E_k) \to Q(E)

. A failure occurs when the limit of the sets exists and is identifiable, yet

\lim_k Q(E_k) \neq Q(E)

. Choice C gives

E_k = \{1, \ldots, k\}

. These sets are clearly increasing, and

\bigcup_k E_k = \mathbb{N}

, so

E_k \uparrow \mathbb{N}

. By finite additivity and the assumption

Q(\{n\}) = 0

, each finite set has measure zero:

Q(E_k) = \sum_{n=1}^k Q(\{n\}) = 0

. But

Q(\mathbb{N}) = 1

. So

\lim_k Q(E_k) = 0 \neq 1 = Q(\mathbb{N})

, which is a direct, clean violation of continuity from below. This is the correct answer. Choice A also has

Q(E_k) = 0

for each

k

, and

E_k \uparrow 2\mathbb{N}

(the even naturals). However, we don't know

Q(2\mathbb{N})

; it could be 0, making no contradiction. The violation isn't guaranteed without additional information. Choice B describes

E_k = \{k, k+1, \ldots\}

, which is decreasing (

E_k \downarrow \varnothing

), so this tests continuity from above, not below — a different property entirely. Choice D is trivially useless:

E_k = \mathbb{N}

for all

k

gives

Q(E_k) = 1 \to 1 = Q(\mathbb{N})

, which actually satisfies continuity from below rather than violating it. As a strategy, always verify three things: the sets are genuinely increasing (for "from below"), the union is explicitly known, and the measure of that union contradicts the limit.

Question 3

On $\Omega=\{1,2,3,4,5,6\}$ , define a random variable by $X(1)=X(2)=0$ , $X(3)=1$ , and $X(4)=X(5)=X(6)=2$ . Which event belongs to the sigma-algebra $\sigma(X)$ ?

$\{1,3,4,5,6\}$ , because it contains outcomes from all three values of $X$ .
$\{1,2,4,5,6\}$ , because it is a union of complete fibers of $X$ . (correct answer)
$\{2,3,4,5,6\}$ , because its complement is contained in one fiber of $X$ .
$\{1,2,3,4,5\}$ , because it omits only one outcome from the largest fiber.

Explanation: When you encounter a question about

\sigma(X)

, the generated sigma-algebra of a random variable, the key principle is this:

\sigma(X)

consists precisely of all unions of complete fibers (preimages) of

X

, along with the empty set. A fiber is the full set of outcomes that map to a single value. Here, the three fibers are

X^{-1}(0) = \{1,2\}

X^{-1}(1) = \{3\}

, and

X^{-1}(2) = \{4,5,6\}

. Every element of

\sigma(X)

must be a union of some subcollection of these fibers — you cannot split a fiber apart. Choice B,

\{1,2,4,5,6\}

, equals

\{1,2\} \cup \{4,5,6\} = X^{-1}(0) \cup X^{-1}(2)

, which is a perfect union of complete fibers. This makes it a valid element of

\sigma(X)

, confirming B is correct. Choice A,

\{1,3,4,5,6\}

, splits the fiber

\{1,2\}

by including 1 but not 2. This violates the completeness requirement. Choice C,

\{2,3,4,5,6\}

, similarly splits the fiber

\{1,2\}

by including 2 but not 1. The reasoning given — about complements — is irrelevant; complements of fiber-unions are themselves fiber-unions, but

\{2,3,4,5,6\}

is not. Choice D,

\{1,2,3,4,5\}

, splits the fiber

\{4,5,6\}

by omitting 6, which is equally invalid. The study tip: always identify the fibers of

X

first, then check whether the candidate set is exactly a union of whole fibers. Any set that "cuts through" a fiber cannot belong to

\sigma(X)

Question 4

Let $P$ and $Q$ be probability measures on $(\Omega,\mathcal F)$ . Suppose $\mathcal P$ is a pi-system satisfying $\sigma(\mathcal P)=\mathcal F$ and $P(E)=Q(E)$ for every $E\in\mathcal P$ . Which conclusion is justified?

$P=Q$ need not hold beyond $\mathcal P$ , because a pi-system is closed only under finite intersections and cannot generate full sigma-algebra agreements.
$P=Q$ only on the algebra generated by $\mathcal P$ , unless $\mathcal P$ is itself a sigma-algebra.
$P=Q$ on $\mathcal F$ , because the agreement class is automatically a sigma-algebra under finite additivity of measures.
$P=Q$ on $\mathcal F$ , because the agreement class is a lambda-system that contains $\mathcal P$ and the pi-lambda theorem applies. (correct answer)

Explanation: Whenever you see a question about probability measures agreeing on a generating class, your instinct should be to reach for the Dynkin pi-lambda theorem: if two measures agree on a pi-system, and that pi-system generates the full sigma-algebra, then the measures agree everywhere. The key to D is constructing the agreement class

\mathcal{L} = \{A \in \mathcal{F} : P(A) = Q(A)\}

and showing it is a lambda-system (also called a Dynkin system). You verify:

\Omega \in \mathcal{L}

since both are probability measures; if

A \subset B

with

A,B \in \mathcal{L}

, then

P(B \setminus A) = P(B) - P(A) = Q(B) - Q(A) = Q(B \setminus A)

, so

B \setminus A \in \mathcal{L}

; and countable disjoint unions are preserved by sigma-additivity. So

\mathcal{L}

is a lambda-system containing the pi-system

\mathcal{P}

. By the pi-lambda theorem,

\mathcal{L} \supseteq \sigma(\mathcal{P}) = \mathcal{F}

, so

P = Q

on all of

\mathcal{F}

. A is wrong because it confuses closure properties of

\mathcal{P}

with what can be concluded from agreement on

\mathcal{P}

. The pi-lambda theorem is precisely what bridges this gap. B is wrong because agreement extends beyond the generated algebra — sigma-additivity, not just finite additivity, is what lifts agreement to the full sigma-algebra. C is wrong in its reasoning: finite additivity alone is insufficient; the critical structure is the lambda-system, not a sigma-algebra. As a study tip, memorize the three lambda-system axioms cold — they appear constantly in uniqueness-of-measures arguments, and the exam will test whether you can verify them quickly.

Question 5

Let $(\Omega,\mathcal F,P)$ be a probability space. Suppose $N\in\mathcal F$ , $P(N)=0$ , and $V\subseteq N$ but $V\notin\mathcal F$ . Which statement correctly describes the completion of this probability space?

The completion contains $V$ , and the extended probability of $V$ is necessarily $0$ . (correct answer)
The completion contains $V$ , but its probability may be assigned any value in $[0,1]$ .
The completion excludes $V$ unless $V$ can be written as a countable union of measurable sets.
The completion leaves $V$ nonmeasurable because completion adds null sets but not their subsets.

Explanation: Whenever you see a question about completing a probability space, anchor yourself to the formal definition: the completion of

(\Omega, \mathcal{F}, P)

is the smallest

\sigma

-algebra

\overline{\mathcal{F}}

containing

\mathcal{F}

along with every subset of every

P

-null set in

\mathcal{F}

. The extended measure assigns probability

0

to all such subsets. In this problem,

N \in \mathcal{F}

with

P(N) = 0

, and

V \subseteq N

. By the very construction of the completion, every subset of a null set is added to

\overline{\mathcal{F}}

and assigned measure zero. So

V \in \overline{\mathcal{F}}

and

\overline{P}(V) = 0

. Answer A is correct. B is wrong because the extended probability is not a free choice — it is forced to be

0

. Assigning any other value would violate monotonicity of measure: since

V \subseteq N

and

P(N) = 0

, we must have

\overline{P}(V) \leq \overline{P}(N) = 0

, so

\overline{P}(V) = 0

. C is wrong because it conflates completion with a different construction (like the Borel

\sigma

-algebra built from countable unions). Completion specifically targets subsets of null sets — no countable union condition is required. D reflects a common misconception: that completion only adds the null sets themselves, not their subsets. In fact, adding subsets of null sets is the entire point of completion. Without this step, the space would not be complete in the measure-theoretic sense. Study tip: Memorize the two-part definition of a complete measure space — a measure is complete if every subset of a null set is measurable (and has measure zero). Exam questions often test whether you know what gets added and what probability it receives.

Question 6

A sequence of events $A_1,A_2,\ldots$ satisfies $P(A_n)=0.40$ for every $n$ . No independence or monotonicity assumptions are made. What is the strongest universal conclusion about $P(\limsup_n A_n)$ ?

It must equal $0.40$ because every event in the sequence has the same probability.
It is at least $0.40$ , and this lower bound is attainable without independence. (correct answer)
It is at most $0.40$ because membership infinitely often is more restrictive.
It may be arbitrarily close to $0$ because the events can move across the sample space.

Explanation: When you see a question about

P(\limsup_n A_n)

, you should immediately think about the Fatou's Lemma for probabilities and what

\limsup

structurally means: the event

\limsup_n A_n = \{A_n \text{ i.o.}\}

is the set of outcomes that belong to infinitely many

A_n

. The key tool here is the reverse Fatou lemma for probabilities:

P(\limsup_n A_n) \geq \limsup_n P(A_n)

. Since every

P(A_n) = 0.40

, we get

\limsup_n P(A_n) = 0.40

, so

P(\limsup_n A_n) \geq 0.40

. This bound is tight and attainable: if the events are all identical,

A_n = A

for all

n

, then

\limsup_n A_n = A

and

P(\limsup_n A_n) = 0.40

exactly — no independence needed. A is wrong because equal probabilities do not force the

\limsup

to equal 0.40; the probability could be anywhere from 0.40 up to 1. C inverts the logic. Being in infinitely many events is less restrictive than being in a single specific one, so the

\limsup

probability is bounded below, not above. Membership infinitely often is not more restrictive — it is a weaker requirement spread over the whole sequence. D is the most tempting distractor. Yes, events can "move" across the sample space, potentially avoiding any fixed outcome — but the reverse Fatou inequality prevents the probability from falling below 0.40 regardless. The floor is guaranteed by measure theory. Your study tip: memorize both directions of Fatou —

P(\liminf) \leq \liminf P

and

P(\limsup) \geq \limsup P

— and always test whether the equality case (identical events) achieves the bound.

Question 7

Let $(\Omega,\mathcal F)$ be a measurable space and let $D$ be an arbitrary subset of $\Omega$ , not necessarily an element of $\mathcal F$ . Define $\mathcal F_D=\{A\cap D:A\in\mathcal F\}$ . Which statement is correct?

$\mathcal F_D$ equals $\{A\in\mathcal F:A\subseteq D\}$ whether or not $D$ is measurable.
$\mathcal F_D$ is a sigma-algebra on $\Omega$ , although its largest event may be smaller than $\Omega$ .
$\mathcal F_D$ is a sigma-algebra on $D$ only when the subset $D$ belongs to $\mathcal F$ .
$\mathcal F_D$ is a sigma-algebra on $D$ , and it contains $D$ even when $D\notin\mathcal F$ . (correct answer)

Explanation: When a question asks you to restrict a sigma-algebra to a subset, your first instinct should be to check what space the resulting collection lives on — that single observation resolves most of the confusion here. The collection

\mathcal{F}_D = \{A \cap D : A \in \mathcal{F}\}

consists of subsets of

D

, so the natural question is whether it forms a sigma-algebra on $D$ . You can verify the three axioms directly. First,

\Omega \in \mathcal{F}

, so

\Omega \cap D = D \in \mathcal{F}_D

, giving you the "whole space"

D

itself — no assumption about

D \in \mathcal{F}

needed. Second, if

A \cap D \in \mathcal{F}_D

, its complement relative to

D

D \setminus (A \cap D) = A^c \cap D

, and since

A^c \in \mathcal{F}

, this complement is back in

\mathcal{F}_D

. Third, countable unions of sets of the form

A_n \cap D

equal

(\bigcup_n A_n) \cap D

, which is again in

\mathcal{F}_D

since

\mathcal{F}

is closed under countable unions. This confirms D is correct. Choice A confuses

\mathcal{F}_D

with the trace sigma-algebra restricted to measurable subsets of

D

; these two collections coincide only when

D \in \mathcal{F}

. Choice B fails because

\mathcal{F}_D

is not a sigma-algebra on

\Omega

— it lives on

D

, and

\Omega \in \mathcal{F}_D

only if

D = \Omega

. Choice C imposes an unnecessary condition; the construction works for any

D \subseteq \Omega

, measurable or not. As a study habit, whenever you see a "trace" or "restriction" construction, immediately identify the underlying space — sigma-algebra axioms are always relative to that space, not to the original

\Omega

Question 8

Let $\Omega=\{1,2,3,4,5,6\}$ , and let $\mathcal F$ be the sigma-algebra generated by $A=\{1,2,3\}$ and $B=\{3,4\}$ . Which statement about $\mathcal F$ is correct?

$\mathcal F$ has $16$ elements and contains $\{1,2,4\}$ . (correct answer)
$\mathcal F$ has $16$ elements and contains $\{1,4\}$ .
$\mathcal F$ has $8$ elements and contains $\{2,3,4\}$ .
$\mathcal F$ has $32$ elements and contains $\{1,2,5\}$ .

Explanation: When you see a question about generated sigma-algebras, your job is to systematically build the smallest collection of sets closed under complementation and countable unions that contains the generating sets. Start by identifying the "atoms" — the finest partition created by intersecting the generators and their complements. With

A = \{1,2,3\}

and

B = \{3,4\}

, compute:

A \cap B = \{3\}

A \cap B^c = \{1,2\}

A^c \cap B = \{4\}

, and

A^c \cap B^c = \{5,6\}

. These four disjoint atoms partition

\Omega

. A sigma-algebra generated by

n

atoms contains exactly

2^n

elements — here,

2^4 = 16

sets, formed by taking all possible unions of these atoms. Every set in

\mathcal{F}

is a union of some subset of

\{\{3\},\{1,2\},\{4\},\{5,6\}\}

. Checking answer A:

\{1,2,4\} = \{1,2\} \cup \{4\}

, which is a valid union of atoms, so

\{1,2,4\} \in \mathcal{F}

. Answer A is correct. Answer B claims

\{1,4\} \in \mathcal{F}

, but

\{1,4\}

splits the atom

\{1,2\}

, so it cannot appear in

\mathcal{F}

— this is the key trap. Answer C claims only 8 elements, which would require only 3 atoms. With 4 atoms, you get

2^4 = 16

, not

2^3 = 8

. Answer D's count of 32 would require 5 atoms, which doesn't match our construction. Study tip: Always find the atoms first. If you can express a candidate set as a union of atoms, it's in the sigma-algebra; if it splits any atom, it's not.

Question 9

Let $A_1\supseteq A_2\supseteq\cdots$ be events with $P(A_n)\to0.30$ . Suppose an event $B$ satisfies $B\subseteq A_n$ for every $n$ and $P(B)=0.30$ . What is necessarily true?

$\bigcap_{n=1}^{\infty}A_n=B$ exactly, because their probabilities have the same limit.
$P\!\left(\bigcap_{n=1}^{\infty}A_n\right)=0.30$ and $P\!\left((\bigcap_n A_n)\setminus B\right)=0$ . (correct answer)
$P\!\left(\bigcap_{n=1}^{\infty}A_n\right)=0$ because infinitely many restrictions are imposed.
$P\!\left(\bigcap_{n=1}^{\infty}A_n\right)$ may exceed $0.30$ because only finite additivity applies.

Explanation: When you see a decreasing sequence of events with a probability limit, your first instinct should be the continuity of probability: for any decreasing sequence

A_1 \supseteq A_2 \supseteq \cdots

, we have

P\!\left(\bigcap_{n=1}^\infty A_n\right) = \lim_{n\to\infty} P(A_n)

. This is a theorem that follows directly from countable additivity — not just finite additivity — and it pins down the intersection's probability exactly as

0.30

. Now, since

B \subseteq A_n

for every

n

, it follows that

B \subseteq \bigcap_n A_n

. Both sets have probability

0.30

, so

P\!\left(\bigcap_n A_n\right) - P(B) = 0

, which means

P\!\left(\left(\bigcap_n A_n\right) \setminus B\right) = 0

. This is exactly what B asserts — and notice it carefully says "probability zero difference," not set equality. That distinction matters. A is wrong because equal probabilities do not imply equal sets. There could be extra measure-zero "junk" in

\bigcap_n A_n

beyond

B

, so you cannot conclude set equality from probability alone. C is wrong because it confuses this scenario with the case

P(A_n) \to 0

. Continuity of probability gives

0.30

, not

0

. More restrictions shrink the set, but the limit is controlled by the probability limit. D is wrong because countable additivity — which is a standard axiom of probability — guarantees continuity for monotone sequences. Finite additivity alone would be insufficient, but we have the full axiom. Your takeaway: memorize continuity of probability for monotone sequences, and remember that equal probabilities never force set equality — only a probability-zero symmetric difference.

Question 10

For three events $A$ , $B$ , and $C$ , suppose the union bound is attained exactly: $P(A\cup B\cup C)=P(A)+P(B)+P(C)$ . Which condition is both necessary and sufficient for this equality?

The triple intersection has probability zero, so $P(A\cap B\cap C)=0$ .
The events are mutually independent and at least one of them has probability zero.
Every pairwise intersection has probability zero, although the sets need not be literally disjoint. (correct answer)
The events are pairwise disjoint as sets, with no shared outcomes including null outcomes.

Explanation: Whenever you see a question about the union bound (inclusion-exclusion), anchor yourself to the full formula:

P(A\cup B\cup C) = P(A)+P(B)+P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)

For the union bound to be attained exactly, the subtracted and added correction terms must cancel to zero:

-P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C) = 0

Since all pairwise intersections contain the triple intersection, we have

P(A\cap B\cap C) \leq \min\{P(A\cap B), P(A\cap C), P(B\cap C)\}

. If every pairwise intersection has probability zero, then the triple intersection is automatically zero too, and the entire correction vanishes — confirming C is sufficient. Conversely, if any pairwise intersection had positive probability, the correction would be nonzero, making C necessary as well. A is wrong because zeroing only the triple intersection leaves the pairwise terms untouched; the correction need not vanish. This is a classic trap — students confuse the triple term being small with the whole correction being zero. B is wrong because independence plus a zero-probability event is far more restrictive than needed, and independence alone doesn't guarantee pairwise intersections have zero probability unless probabilities are zero. D is wrong because literal set-disjointness (no shared outcomes, even null ones) is sufficient but stronger than necessary. Probability-zero overlaps are harmless — events can share null sets and still satisfy the equality. Your study tip: always return to the full inclusion-exclusion formula. The union bound is tight when all pairwise probabilities vanish, not just the triple intersection.