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✓ 9 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Modes of Convergence for Random Variables — Examples

1 · Prerequisites

2 · Summary

These constructions delimit the implication diagram and make the subsequence and uniform-integrability mechanisms concrete.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Convergence in probability need not be almost sure

Statement refuted

Convergence in probability need not imply almost-sure convergence.

Facts & Assumptions

Given: Independent events $E_{n}$ with $P (E_{n}) = 1 / (n + 1)$ , and $X_{n} = 1_{E_{n}}$ .

[L1]

Pairwise independent events with divergent probability sum occur infinitely often almost surely (Second Borel-Cantelli lemma under pairwise independence).

[L2]

Convergence in probability is fixed-threshold tail convergence (Convergence in probability).

Counterexample

technique · direct

1.1L2

For $0 < ε < 1$ , $P (∣ X_{n} ∣ > ε) = 1 / (n + 1) \to 0$ ; for $ε \geq 1$ it is zero. Hence $X_{n} \to 0$ in probability by [L2].

2.1step 1.1L1∎

But $\sum_{n} P (E_{n}) = \infty$ , so [L1] gives $E_{n}$ infinitely often almost surely. On that event $X_{n}$ has infinitely many values $1$ , and therefore cannot converge to $0$ .

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Almost-sure convergence need not imply $L^{p}$ convergence

Statement refuted

Almost-sure convergence need not imply $L^{p}$ convergence, even for $1 \leq p < \infty$ .

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ , $1 \leq p < \infty$ , and $X_{n} = (n + 1)^{1 / p} 1_{(0, 1 / (n + 1))}$ for $n \in N$ .

[L1]

Almost-sure convergence is pointwise convergence outside a null set (Almost-sure convergence of real random variables).

[L2]

$L^{p}$ convergence requires the $L^{p}$ norm of the difference to vanish ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1L1

For every $x \in (0, 1)$ , eventually $1 / (n + 1) < x$ , so $X_{n} (x) = 0$ . Thus $X_{n} \to 0$ everywhere on $(0, 1)$ , hence almost surely by [L1].

2.1step 1.1L2∎

Yet $E ∣ X_{n} ∣^{p} = \int_{0}^{1 / (n + 1)} (n + 1) d x = 1$ , so [step 1.1, L2] $∥ X_{n} ∥_{p} = 1$ for every $n$ . By [L2], there is no $L^{p}$ convergence to zero.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

$L^{p}$ convergence need not imply almost-sure convergence

Statement refuted

$L^{p}$ convergence need not imply almost-sure convergence, for any fixed $1 \leq p < \infty$ .

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ and, for $m \geq 0$ , the $2^{m}$ dyadic intervals $I_{m, j} = [j 2^{- m}, (j + 1) 2^{- m})$ , listed block by block; let $X_{n}$ be their indicators in that order.

[L1]

$L^{p}$ convergence is vanishing of the $p$ th moment of the difference ( $L^{p}$ convergence for random variables).

[L2]

Almost-sure convergence requires pointwise convergence off a null set (Almost-sure convergence of real random variables).

Counterexample

technique · direct

1.1L1

In the block of level $m$ , every $X_{n}$ has $E ∣ X_{n} ∣^{p} = 2^{- m}$ . As the block level tends to infinity with $n$ , $∥ X_{n} ∥_{p} = 2^{- m / p} \to 0$ ; hence $X_{n} \to 0$ in $L^{p}$ by [L1].

2.1step 1.1L2∎

Every non-dyadic $x \in (0, 1)$ belongs to exactly one interval in each level- $m$ block, but misses all the other intervals in that block. Thus $X_{n} (x)$ equals both $1$ and $0$ infinitely often. The exceptional dyadic endpoints are null, so [L2] rules out almost-sure convergence.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Convergence in distribution need not be convergence in probability

Statement refuted

Convergence in distribution need not imply convergence in probability.

Facts & Assumptions

Given: A random variable $X$ with $P (X = 1) = P (X = - 1) = 1 / 2$ , and $X_{n} = - X$ .

[L1]

Distributional convergence is convergence of the corresponding CDFs at continuity points (Convergence in distribution for real random variables).

[L2]

Probability convergence makes every positive error probability vanish (Convergence in probability).

Counterexample

technique · direct

1.1L1

The symmetric two-point law of $- X$ equals that of $X$ , so $F_{X_{n}} = F_{X}$ for every $n$ . Therefore $X_{n} \Rightarrow X$ by [L1].

2.1step 1.1L2∎

But $∣ X_{n} - X ∣ = 2$ almost surely, so $P (∣ X_{n} - X ∣ > 1) = 1$ for every $n$ . By [L2], $X_{n}$ does not converge to $X$ in probability.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Convergence in probability need not imply $L^{p}$ convergence

Statement refuted

Convergence in probability need not imply $L^{p}$ convergence, for $1 \leq p < \infty$ .

Facts & Assumptions

Given: $1 \leq p < \infty$ , events $E_{n}$ with $P (E_{n}) = 1 / (n + 1)$ , and $X_{n} = (n + 1)^{1 / p} 1_{E_{n}}$ for $n \in N$ .

[L1]

Convergence in probability is fixed-threshold tail convergence (Convergence in probability).

[L2]

$L^{p}$ convergence requires the $p$ th moment of the difference to tend to zero ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1L1

For fixed $ε > 0$ , $X_{n}$ is nonzero only on $E_{n}$ , hence [L1] $P (∣ X_{n} ∣ > ε) \leq 1 / (n + 1) \to 0$ . Thus $X_{n} \to 0$ in probability by [L1].

2.1step 1.1L2∎

Nevertheless, [step 1.1, L2] $E ∣ X_{n} ∣^{p} = (n + 1) P (E_{n}) = 1$ for all $n$ . By [L2], $X_{n}$ does not converge to zero in $L^{p}$ .

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

$L^{p}$ convergence need not imply $L^{q}$ convergence when $p < q$

Statement refuted

For $1 \leq p < q < \infty$ , $L^{p}$ convergence need not imply $L^{q}$ convergence.

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ , $1 \leq p < q < \infty$ , and $X_{n} = (n + 1)^{1 / q} 1_{(0, 1 / (n + 1))}$ for $n \in N$ .

[L1]

$L^{r}$ convergence is vanishing of $E ∣ X_{n} - X ∣^{r}$ for the relevant exponent $r$ ( $L^{p}$ convergence for random variables).

Counterexample

technique · direct

1.1L1

Direct calculation gives [L1] $E ∣ X_{n} ∣^{p} = (n + 1)^{p / q - 1} \to 0$ , since $p / q - 1 < 0$ . Thus $X_{n} \to 0$ in $L^{p}$ by [L1].

2.1step 1.1L1∎

But $E ∣ X_{n} ∣^{q} = (n + 1) / (n + 1) = 1$ for every $n$ , so the [step 1.1, L1] $L^{q}$ norm never tends to zero. Hence [L1] rules out $L^{q}$ convergence.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Almost-sure convergence does not imply convergence of expectations

Statement refuted

Almost-sure convergence alone need not imply convergence of expectations.

Facts & Assumptions

Given: Lebesgue probability space $(0, 1)$ and $X_{n} = (n + 1) 1_{(0, 1 / (n + 1))}$ for $n \in N$ .

[L1]

Almost-sure convergence is pointwise convergence off a null set (Almost-sure convergence of real random variables).

[L2]

Expectation is integration against the probability measure (Expectation of a nonnegative or integrable random variable).

Counterexample

technique · direct

1.1L1

For every $x \in (0, 1)$ , eventually $x > 1 / (n + 1)$ , so $X_{n} (x) = 0$ . [L1] Hence $X_{n} \to 0$ almost surely by [L1].

2.1step 1.1L2∎

Yet [L2] gives [step 1.1, L2] $E X_{n} = \int_{0}^{1 / (n + 1)} (n + 1) d x = 1$ for every $n$ , whereas $E 0 = 0$ . Thus the expectations do not converge.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-09-07Open item page →

A probability-convergent sequence with a prescribed fast almost-sure subsequence

Example

Suppose $X_{n} \to X$ in probability. Given any sequences $a_{k} > 0$ with $a_{k} ↓ 0$ and $b_{k} > 0$ with $\sum_{k} b_{k} < \infty$ , one can choose increasing $n_{k}$ so that $P (∣ X_{n_{k}} - X ∣ > a_{k}) < b_{k}$ and $X_{n_{k}} \to X$ almost surely.

Facts & Assumptions

Given: $X_{n} \to X$ in probability, $a_{k} > 0$ with $a_{k} ↓ 0$ , and $b_{k} > 0$ with $\sum_{k} b_{k} < \infty$ .

[L1]

Probability convergence supplies a later index for every positive threshold and positive bound (Convergence in probability).

[L2]

The least-index construction with a summable error schedule yields an almost-surely convergent subsequence (An almost-surely convergent subsequence from convergence in probability).

Verification

technique · constructive

1.1L1construct

After $n_{k - 1}$ , choose $n_{k}$ least with $P (∣ X_{n_{k}} - X ∣ > a_{k}) < b_{k}$ ; [L1] makes every choice possible.

2.1step 1.1L2discharge-construct∎

The proof of [L2] uses only that the displayed probabilities are summable and that $a_{k} \to 0$ . Thus it applies to these $n_{k}$ and gives almost-sure convergence.

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Uniform integrability repairs the expectation limit

Example

Let $X_{n} \to X$ in probability and suppose $∣ X_{n} ∣ \leq M$ almost surely for one finite constant $M$ and every $n$ . Then $X_{n} \to X$ in $L^{1}$ and $E X_{n} \to E X$ .

Facts & Assumptions

Given: $X_{n} \to X$ in probability and $∣ X_{n} ∣ \leq M$ almost surely for all $n$ .

[L1]

Uniform integrability is the uniform decay of tail integrals (A uniformly integrable family).

[L2]

Under probability convergence, uniform integrability is equivalent to $L^{1}$ convergence (Uniform integrability characterizes $L^{1}$ convergence under probability convergence).

Verification

technique · direct

1.1L1

If $R > M$ , then $∣ X_{n} ∣ 1_{∣ X_{n} ∣ > R} = 0$ almost surely for every $n$ . Hence the family $(X_{n})$ is uniformly integrable by [L1].

2.1step 1.1L2∎

By [L2], $X_{n} \to X$ in $L^{1}$ . Therefore $∣ E X_{n} - E X ∣ \leq E ∣ X_{n} - X ∣ \to 0$ , proving convergence of expectations.

Modes of Convergence for Random Variables — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Convergence in probability need not be almost sure

Statement refuted

Facts & Assumptions

Counterexample

Almost-sure convergence need not imply LpLp convergence

Statement refuted

Facts & Assumptions

Counterexample

LpLp convergence need not imply almost-sure convergence

Statement refuted

Facts & Assumptions

Counterexample

Convergence in distribution need not be convergence in probability

Statement refuted

Facts & Assumptions

Counterexample

Convergence in probability need not imply LpLp convergence

Statement refuted

Facts & Assumptions

Counterexample

LpLp convergence need not imply LqLq convergence when p<qp<q

Statement refuted

Facts & Assumptions

Counterexample

Almost-sure convergence does not imply convergence of expectations

Statement refuted

Facts & Assumptions

Counterexample

A probability-convergent sequence with a prescribed fast almost-sure subsequence

Example

Facts & Assumptions

Verification

Uniform integrability repairs the expectation limit

Example

Facts & Assumptions

Verification

Sources

Almost-sure convergence need not imply $L^{p}$ convergence

$L^{p}$ convergence need not imply almost-sure convergence

Convergence in probability need not imply $L^{p}$ convergence

$L^{p}$ convergence need not imply $L^{q}$ convergence when $p < q$