Convergence in Lp - Statistics Graduate Level

1. If $X_n \to X$ in $L^p$ on any measure space, which is forced?
   Answer: Convergence in measure By Markov's inequality, the measure of the set where |$X_n$ - X| > epsilon is at most ||$X_n$ - X||_p^p / epsilon^p, which goes to 0, so convergence in measure is forced. The tempting wrong answer is almost sure convergence: $L^p$ convergence only guarantees a subsequence converging almost surely, not the whole sequence. $L^r$ for r<p also fails on infinite measure spaces because $L^p$ does not imply $L^r$.
2. Let $F_1\subseteq F_2\subseteq\cdots$ be increasing sigma-fields, let $F_\infty$ be the sigma-field generated by their union, and let $X\in L^2$. Define $M_n=E[X\mid F_n]$. Which statement is correct?
   Answer: $M_n\to E[X\mid F_\infty]$ in $L^2$, because these conditional expectations are nested orthogonal projections. When you see a question involving conditional expectations built from a filtration, think immediately about Hilbert space geometry. Conditional expectation $E[X \mid \mathcal{F}_n]$ is the orthogonal projection of $X$ onto the closed subspace $L^2(\mathcal{F}_n)$. As $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \cdots$, these subspaces are nested and increasing, so the projections converge to the projection onto their limiting subspace — which is precisely $L^2(\mathcal{F}_\infty)$, where $\mathcal{F}_\infty = \sigma\!\left(\bigcup_n \mathcal{F}_n\right)$. This is the Lévy Upward Theorem: $M_n \to E[X \mid \mathcal{F}_\infty]$ both almost surely and in $L^2$ for any $X \in L^2$. The nested projection structure guarantees $\|M_n - E[X \mid \mathcal{F}_\infty]\|_2 \to 0$, confirming D is correct. Choice A is wrong because conditioning on more information doesn't average it away — that logic applies to decreasing sigma-fields (the Lévy Downward Theorem), where you'd converge toward $E[X \mid \mathcal{F}_\infty]$ for the intersection. Increasing filtrations move toward $X$, not toward $E[X]$. Choice B nearly has the right spirit but overclaims: $M_n \to X$ in $L^2$ only if $X \in L^2(\mathcal{F}_\infty)$, i.e., $X$ is already measurable with respect to the limiting sigma-field. If $X$ has components outside $\mathcal{F}_\infty$, those are irretrievably lost. Choice C invents a false condition. $L^2$ convergence holds generally for any increasing filtration — no "eventually constant" requirement exists. Study tip: Always pair the direction (increasing vs. decreasing filtration) with the correct limit ($E[X \mid \mathcal{F}_\infty]$ upward, $E[X \mid \mathcal{F}_\infty]$ for the intersection downward). The orthogonal projection interpretation makes both theorems intuitive and memorable.
3. On a finite measure space, convergence in measure plus which condition is equivalent to $L^1$ convergence?
   Answer: Uniform integrability Convergence in measure ensures the sequence is close to the limit except on small sets; uniform integrability ensures no mass escapes to infinity on those small sets. Together they imply L1 convergence, and if L1 convergence holds, both conditions follow. $L^p$ boundedness for p>1 is tempting because on a finite measure space it implies uniform integrability, but it is only sufficient: L1 convergence does not require any uniform $L^p$ bound.
4. Let $1\le p<\infty$. Assume that $\{X_n\}$ is Cauchy in $L^p$ and that $X_n\to X$ in probability for some random variable $X$. What must be true?
   Answer: The sequence converges to $X$ in $L^p$, by completeness of $L^p$ and uniqueness of limits in probability. When you see a question combining an $L^p$-Cauchy sequence with convergence in probability, you should immediately think about two foundational theorems: the **completeness of $L^p$ and the uniqueness of limits in probability**. Here's the chain of reasoning. Since $\{X_n\}$ is Cauchy in $L^p$ and $L^p$ is a complete metric space (for $1 \le p < \infty$), the sequence must converge in $L^p$ to some limit, call it $Y$. Now, convergence in $L^p$ implies convergence in probability (by Markov's inequality). So $X_n \to Y$ in probability. But we're also told $X_n \to X$ in probability. Since limits in probability are unique up to almost-sure equality, we conclude $X = Y$ a.s. Therefore $X_n \to X$ in $L^p$, confirming A is correct. B is wrong because $L^p$-Cauchy sequences do not generally converge pointwise or almost surely — convergence in $L^p$ is a statement about norms, not trajectories. Classic counterexamples (the "typewriter sequence") show $L^p$ convergence without almost-sure convergence. C is wrong because it contradicts the uniqueness of limits in probability. Two $L^p$ limits that differ on a positive-probability event would define different probability limits, which is impossible. D is wrong because domination by an integrable function is sufficient for applying the Dominated Convergence Theorem, but it is not necessary here. Completeness of $L^p$ alone guarantees the conclusion without any domination condition. Study tip: Memorize this chain — Cauchy in $L^p$ $\Rightarrow$ $L^p$ limit exists (completeness) $\Rightarrow$ probability limit exists $\Rightarrow$ both probability limits must agree a.s. This three-step argument appears frequently on graduate probability exams.
5. On a probability space, $X_n\to X$ in $L^4$. Which statement is false?
   Answer: It also converges in $L^6$ On a probability space, L4 convergence implies Lq convergence for every q < 4 by Lyapunov's inequality, so L2 follows and L6 is not guaranteed. It also implies convergence in probability, and convergence in probability always gives a subsequence that converges almost surely. So the false statement is the L6 claim; the tempting mistake is thinking the almost-sure subsequence result is false, but it follows from convergence in probability.
6. Fix $p\ge 1$. Suppose $X_n \to X$ in probability and the family $\{|X_n|^p:n\ge 1\}$ is uniformly integrable. Which conclusion follows?
   Answer: $X_n \to X$ in $L^p$, because uniform integrability upgrades the probabilistic convergence of the powered errors. When you see uniform integrability (UI) paired with convergence in probability, you should immediately think about the Vitali Convergence Theorem: convergence in probability plus uniform integrability of $\{|X_n|^p\}$ is precisely the condition that upgrades to $L^p$ convergence. The key mechanism is that UI controls the "tail mass" of $|X_n|^p$, preventing probability mass from escaping to infinity and spoiling the integral. More precisely, since $X_n \to X$ in probability implies $|X_n - X|^p \to 0$ in probability, and UI ensures the tails of $|X_n|^p$ are uniformly small, we can exchange the limit and expectation to get $\mathbb{E}[|X_n - X|^p] \to 0$. This makes A correct. B is wrong because convergence in probability does not imply almost sure convergence along the entire sequence — you can only extract an a.s.-convergent subsequence. UI does nothing to fix this gap; it controls integrals, not sample-path behavior. C is wrong because UI of $\{|X_n|^p\}$ does not grant uniform boundedness of higher moments. In fact, the $r$-th moments ($r > p$) may be infinite. UI only controls the $p$-th moment integrals, so $L^r$ convergence for $r > p$ is not guaranteed. D is wrong because UI is an $L^1$-type integrability condition on tails — it says nothing about essential supremum ($L^\infty$) control. The $X_n$ can still be unbounded a.s. Study tip: Memorize the Vitali theorem as a clean package: in probability + UI of $|X_n|^p$ $\Rightarrow$ $L^p$ convergence. This trio appears repeatedly on graduate probability exams.
7. Suppose $X_n\to X$ almost surely. A researcher wants to conclude that $X_n\to X$ in $L^2$. Which additional condition is sufficient as stated?
   Answer: There exists an integrable random variable $Y$ such that $|X_n-X|^2\le Y$ almost surely for all $n$, permitting an application of the dominated convergence theorem. When bridging almost sure convergence to $L^2$ convergence, your go-to tool is the Dominated Convergence Theorem (DCT). Almost sure convergence tells you pointwise behavior; to upgrade to $L^2$, you need to control the integrals of $|X_n - X|^2$, which requires an integrable dominating function. Choice B is correct because if $|X_n - X|^2 \le Y$ a.s. for all $n$ with $E[Y] < \infty$, then since $X_n \to X$ a.s. implies $|X_n - X|^2 \to 0$ a.s., the DCT directly gives $E[|X_n - X|^2] \to 0$, which is exactly $L^2$ convergence. Choice A is backwards and false. Jensen's inequality says $E[|X_n - X|]^2 \le E[|X_n - X|^2]$, so $L^1$ convergence of differences is weaker than $L^2$, not stronger. You cannot conclude $L^2$ from $L^1$. Choice C is a common trap. A uniform bound $\sup_n E[X_n^2] < \infty$ does not imply uniform integrability of $|X_n - X|^2$, nor does it alone yield $L^2$ convergence — you need tightness conditions beyond just bounded second moments. Choice D fails because controlling tail probabilities at a single threshold gives you convergence in probability, not $L^2$ convergence. Moment convergence requires controlling the size of deviations, not merely their probability. Study tip: Whenever almost sure convergence appears and you need $L^p$ convergence, immediately ask: "Can I find an integrable dominating function?" That's your DCT checklist.
8. Let $U$ be uniformly distributed on $[0,1]$, and define $X_n=n^{1/2}1_{\{U\le 1/n\}}$. Which statement correctly describes the convergence of $X_n$?
   Answer: $X_n \to 0$ in $L^p$ exactly when $1\le p<2$; it also converges almost surely to zero. When you encounter a "spike sequence" like this one, your instinct should be to compute $\|X_n\|_p$ directly rather than rely on intuition about small probabilities alone — those two factors (height and probability) interact in a way that depends critically on $p$. Here $X_n = n^{1/2} \cdot \mathbf{1}_{\{U \le 1/n\}}$, so: $E[|X_n|^p] = (n^{1/2})^p \cdot \frac{1}{n} = n^{p/2 - 1}$ This expression tends to zero if and only if $p/2 - 1 < 0$, i.e., $p < 2$. So $X_n \to 0$ in $L^p$ for all $1 \le p < 2$, and it fails to converge in $L^2$ (where the norm stays at 1 for all $n$) and in any higher $L^p$. For almost sure convergence: for any fixed $\omega$, $X_n(\omega) \ne 0$ only when $U(\omega) \le 1/n$. Since $\sum_n P(U \le 1/n) = \sum_n 1/n = \infty$, Borel–Cantelli doesn't give a.s. convergence directly — but these events are nested (decreasing), so $P(X_n \ne 0 \text{ i.o.}) = \lim P(U \le 1/n) = 0$, confirming $X_n \to 0$ a.s. This confirms B is correct. A is wrong because it assumes small probability alone guarantees $L^p$ convergence for all $p$ — it ignores how the spike height grows. C reverses the truth: $X_n$ converges in $L^1$ but not in $L^2$. D is wrong on two counts: a.s. convergence holds (correct), but $L^p$ convergence for $p < 2$ implies convergence in probability, so the probability convergence claim is false. Your strategy: always compute $n^{p/2-1}$ for spike sequences — the threshold $p = 2$ is a classic exam trap.
9. Let $1 \le p < q < \infty$, and suppose the underlying probability space has total mass one. If $X_n \to X$ in $L^q$, which conclusion is necessarily valid without any additional assumptions?
   Answer: $X_n \to X$ in $L^p$, because the $L^p$ distance is bounded by the $L^q$ distance on a probability space. Whenever you see a question mixing $L^p$ spaces on a probability space (total mass one), your first instinct should be the norm comparison inequality: for $1 \le p < q < \infty$ and a probability measure, Jensen's inequality gives $\|Y\|_p \le \|Y\|_q$. This is the heart of this problem. Applying this to $Y = X_n - X$, you get $\|X_n - X\|_p \le \|X_n - X\|_q$. So if $\|X_n - X\|_q \to 0$, it immediately follows that $\|X_n - X\|_p \to 0$. The $L^q$ norm controls the $L^p$ norm from above when $p < q$ and the measure is a probability measure — this is why A is correct. B is wrong because $L^q$ convergence implies only convergence in probability (and hence a subsequence converges a.s.), but not a.s. convergence of the full sequence. A classic counterexample is the "typewriter sequence" of indicator functions, which converges in every $L^p$ but nowhere almost surely. C reverses the direction of the inequality. Being bounded in $L^q$ tells you nothing about finiteness of $L^r$ norms for $r > q$; higher moments can be infinite even if lower ones are finite. D is false because $L^\infty$ (essential supremum) is the largest $L^p$ space — it is not controlled by finite $L^q$ norms. The essential supremum can be infinite even when all finite-order moments exist. Study tip: On probability spaces, the norm inequality $\|Y\|_p \le \|Y\|_q$ for $p < q$ flows downward — stronger (higher) norms control weaker (lower) norms. Keep this direction firmly in mind; many distractors exploit reversing it.
10. Suppose $X_n\to X$ in $L^1$ and $\sup_n\|X_n\|_3<\infty$. Assume also that $X\in L^3$. Which conclusion is justified?
    Answer: $X_n\to X$ in $L^2$, by interpolating between the vanishing $L^1$ distance and the bounded $L^3$ distance. When you see a problem combining $L^p$ convergence with uniform boundedness in a higher norm, your instinct should be interpolation inequalities. The key tool here is the $L^p$ interpolation inequality: for $1 \leq p \leq q \leq r$, we have $\|f\|_q \leq \|f\|_p^{\lambda} \|f\|_r^{1-\lambda}$ where $\lambda \in [0,1]$ satisfies $\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r}$. Applying this to $X_n - X$ with $p=1$, $q=2$, $r=3$, you get $\lambda = 1/2$ and therefore $\|X_n - X\|_2 \leq \|X_n - X\|_1^{1/2} \cdot \|X_n - X\|_3^{1/2}.$ Since $X_n \to X$ in $L^1$, the first factor vanishes. Since $\sup_n \|X_n\|_3 < \infty$ and $X \in L^3$, the quantity $\|X_n - X\|_3$ is uniformly bounded. A product of something going to zero and something staying bounded goes to zero — confirming D is correct. A is wrong because it inverts the actual relationship: bounded higher norms absolutely do constrain intermediate-norm convergence through interpolation. B overclaims; $L^1$ convergence plus bounded $L^3$ norm is not enough to force $L^3$ convergence — you'd need tightness or uniform integrability of $|X_n|^3$. C is wrong because $L^\infty$ control requires far more than moment bounds; no finite moment condition, however large, implies essential supremum control. A good study habit: whenever you see "$L^p$ convergence + bounded $L^r$ norm," immediately write down the interpolation inequality. It almost always bridges to an intermediate $L^q$ conclusion.
11. If $X_n$ is Cauchy in $L^p$ and $X_n\to X$ a.e., then
    Answer: Converges to X in $L^p$ Because $L^p$ is complete, a Cauchy sequence has an $L^p$ limit Y. The a.e. limit X must equal Y almost everywhere, so X is in $L^p$ and $X_n$ converges to X in $L^p$. The tempting wrong view is that a.e. convergence alone doesn't force $L^p$ convergence; but here the Cauchy condition supplies the needed convergence.
12. On the measure space consisting of the real line with Lebesgue measure, define $f_n(x)=n^{-1}1_{[0,n]}(x)$. Which statement is correct?
    Answer: $f_n\to 0$ in $L^2$ but not in $L^1$, illustrating that finite-measure norm inclusions can fail here. When working with $L^p$ convergence on infinite measure spaces like $(\mathbb{R}, \mathcal{B}, \lambda)$, you cannot rely on intuition built from finite-measure settings. The key is to compute the norms directly rather than reason qualitatively about "height" or "support size" alone. For $f_n(x) = n^{-1}\mathbf{1}_{[0,n]}(x)$, compute each norm explicitly. The $L^1$ norm is $\|f_n\|_1 = \int_0^n n^{-1}\,dx = n^{-1} \cdot n = 1,$ which stays constant at 1 for all $n$, so $f_n \not\to 0$ in $L^1$. The $L^2$ norm is $\|f_n\|_2 = \left(\int_0^n n^{-2}\,dx\right)^{1/2} = \left(n^{-2} \cdot n\right)^{1/2} = n^{-1/2} \to 0,$ confirming $f_n \to 0$ in $L^2$. This makes C correct. The intuition here is that on infinite measure spaces, $L^1 \not\subset L^2$ and $L^2 \not\subset L^1$. This sequence exploits exactly that gap — the shrinking height decays fast enough for $L^2$ (which penalizes large values more heavily) but not for $L^1$, because the expanding support perfectly compensates the shrinking height. A is wrong because it incorrectly concludes $L^1$ convergence from pointwise decay — pointwise convergence to zero never guarantees $L^1$ convergence without uniform integrability. B gets the $L^1$ conclusion backward — the constant $L^1$ norm means it fails in $L^1$, not converges. D is wrong because the $L^2$ norm calculation clearly shows convergence to zero. Your strategy: always compute norms explicitly before concluding anything about $L^p$ convergence, especially on infinite measure spaces where norm inclusions from the finite setting no longer hold.
13. Suppose $X_n\to X$ in $L^2$ and $Y_n\to Y$ in $L^2$. Which assertion about the products is necessarily correct?
    Answer: $X_nY_n\to XY$ in $L^1$, by decomposing the product error and applying the Cauchy–Schwarz inequality. When combining two convergent sequences through multiplication, you need to think carefully about which mode of convergence is preserved and why — this question tests exactly that interplay between $L^2$ convergence and product stability. The key technique is the algebraic decomposition: write $X_nY_n - XY = (X_n - X)Y_n + X(Y_n - Y)$. Now apply the Cauchy–Schwarz inequality to each term. For the first: $\|{(X_n-X)Y_n}\|_1 \leq \|X_n - X\|_2\|Y_n\|_2$. Since $X_n \to X$ in $L^2$, the first factor vanishes; since $Y_n \to Y$ in $L^2$, the sequence $(Y_n)$ is bounded in $L^2$. The second term $\|X(Y_n - Y)\|_1 \leq \|X\|_2\|Y_n - Y\|_2 \to 0$ similarly. Both terms go to zero, so $X_nY_n \to XY$ in $L^1$ — confirming C is correct. A is wrong because $L^2$ convergence implies only convergence in probability (and thus a.s. along a subsequence), not almost sure convergence along the entire sequence. Claiming a.s. convergence for the full sequence is an unjustified leap. B fails because multiplication does not generally preserve $L^2$ convergence. The product $X_nY_n$ need not even be in $L^2$ without additional assumptions like uniform boundedness, and the bound you'd need requires $L^4$ control, not $L^2$. D is false because $L^2$ boundedness says nothing about essential supremum ($L^\infty$) control; the sequences could be unbounded pointwise. Your takeaway: whenever you see a product of two $L^2$-convergent sequences, reach immediately for the decomposition $X_nY_n - XY = (X_n-X)Y_n + X(Y_n-Y)$ paired with Cauchy–Schwarz — it's the canonical tool for dropping to $L^1$.
14. On [0,1], let $X_n=\sqrt{n}$ on $[0,1/n]$, else $0$. Then
    Answer: Converges in $L^1$, not $L^2$ For each x>0, once n>1/x you get $X_n$(x)=0, so $X_n$ converges to 0 a.e. The L1 norm is sqrt(n)*(1/n)=1/sqrt(n), which tends to 0, so it converges in L1. The L2 norm is n*(1/n)=1, so it does not converge in L2. The tempting wrong answer is 'converges a.e. but not in L1'; the spike at 0 grows, but that single point has measure zero and the area under the spike still vanishes.
15. Suppose $X_n\to X$ in $L^p$ for some finite $p\ge 1$. Which statement about other modes of convergence is guaranteed?
    Answer: The sequence converges in probability, and at least one subsequence converges almost surely to $X$. When working with modes of convergence in probability theory, you need to know the precise relationships between $L^p$ convergence, convergence in probability, and almost sure convergence — a classic hierarchy tested at the graduate level. $L^p$ convergence (meaning $E[|X_n - X|^p] \to 0$) implies convergence in probability by Markov's inequality: for any $\varepsilon > 0$, $P(|X_n - X| > \varepsilon) \leq \varepsilon^{-p} E[|X_n - X|^p] \to 0$. However, $L^p$ convergence does not generally imply almost sure convergence for the full sequence. What it does guarantee is that some subsequence converges almost surely — this follows from the fact that convergence in probability implies existence of an a.s.-convergent subsequence (a standard theorem). This makes B correct: the sequence converges in probability, and at least one subsequence converges almost surely to $X$. A is wrong on two counts: $L^p$ convergence does not guarantee almost sure convergence of the entire sequence, and there is certainly no universal deterministic convergence rate across all subsequences. C is false because $L^p$ says nothing about the essential supremum ($L^\infty$) norm, and convergence in $L^p$ for finite $p$ can fail in $L^\infty$ even after removing finitely many terms. D is directly contradicted by the Markov's inequality argument above — $L^p$ convergence always implies convergence in probability. As a study tip, memorize the strict hierarchy: $L^p \Rightarrow$ in probability $\Rightarrow$ subsequence a.s., with none of these implications reversing in general. Questions will often test whether you conflate the full-sequence and subsequence results.

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