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2023-Q2 – Equivalences and Counterexamples in Modes of Convergence

2023 Probability Prelim Exam (PDF)

Problem Statement (verbatim from prelim)

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Let ${X_k}$, $k = 1,2,\dots$ be a sequence of random variables. In this problem all convergences are as $k \to \infty$.

(a) Prove that $X_k \to 0$ in probability if and only if $X_k \to 0$ in distribution.

(b)
(i) Prove that if $X_k \to 0$ a.s. then
\(P\!\left(\bigcup_{m=k}^{\infty}\{|X_m| > \varepsilon\}\right)\to 0 \quad \text{for each } \varepsilon>0.\)

(ii) Show how to conclude from (i) that convergence a.s. implies convergence in probability.

(c) Assume that ${X_k}$ are independent,
\(P(|X_k| > 1)=\frac{1}{k}, \qquad P(|X_k| \le 1/\sqrt{k}) \to 1.\)

Show that $X_k \to 0$ in probability, but $X_k \to 0$ a.s. is false.

Solution

(a) Convergence in probability iff convergence in distribution

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Claim.

If $X_k \to 0$ in probability, then $X_k \to 0$ in distribution, and conversely.

Proof.

($\Rightarrow$) Assume $X_k \to 0$ in probability.
Let $F_k$ denote the cdf of $X_k$, and let $F$ be the cdf of the constant $0$.
We must show that $F_k(x) \to F(x)$ for every continuity point of $F$.

Since $F(x) = \mathbf{1}{x \ge 0}$, the only point of discontinuity is $x = 0$.
Thus we only check $x < 0$ and $x > 0$.

  • Case 1: $x < 0$.
    Pick $\varepsilon = |x| > 0$. Then \(F_k(x) = P(X_k \le x) \le P(|X_k|>\varepsilon).\) Since $X_k \to 0$ in probability, the RHS goes to $0$.
    Also, $F(x)=0$, so $F_k(x)\to 0$.

  • Case 2: $x > 0$.
    Pick $\varepsilon = x$. Then \(P(X_k \le x) \ge 1 - P(|X_k| > x).\) The probability $P(|X_k| > x) \to 0$, so $F_k(x) \to 1 = F(x)$.

Thus $F_k(x)\to F(x)$ for all continuity points.

($\Leftarrow$)
To show: convergence in distribution to $0$ implies convergence in probability to $0$.

For any $\varepsilon>0$, consider \(P(|X_k|>\varepsilon) = P(X_k<-\varepsilon) + P(X_k>\varepsilon).\) Let $x_1=-\varepsilon$ and $x_2=\varepsilon$. These are continuity points of $F$, so
\(F_k(x_1)\to 0, \qquad F_k(x_2)\to 1.\) Hence \(P(|X_k|>\varepsilon) = F_k(-\varepsilon)+ (1-F_k(\varepsilon)) \to 0+0 = 0.\)

Conclusion.

$X_k \to 0$ in probability if and only if $X_k \to 0$ in distribution.

(b)(i) If $X_k\to0$ a.s., then tail events vanish

Claim.

If $X_k \to 0$ a.s., then for each $\varepsilon > 0$,
\(P\left(\bigcup_{m=k}^{\infty} \{|X_m| > \varepsilon\}\right) \longrightarrow 0.\)

Proof.

Fix $\varepsilon>0$ and define the tail events \(A_k = \bigcup_{m=k}^{\infty}\{|X_m|>\varepsilon\}.\)

As $k$ increases, each $A_k$ loses elements, so ${A_k}$ is a decreasing sequence: \(A_{k+1}\subseteq A_k.\)

Since $X_k\to0$ a.s., for almost every $\omega$ there exists some $K(\omega)$ such that
$|X_m(\omega)|\le \varepsilon$ for all $m\ge K(\omega)$.
Thus for almost all $\omega$, \(\omega\notin \bigcap_{k=1}^\infty A_k.\)

Hence
\(P\!\left(\bigcap_{k=1}^\infty A_k \right)=0.\)

Using continuity from above for decreasing events: \(P(A_k)\downarrow P\!\left(\bigcap_{k=1}^\infty A_k\right)=0.\)

Conclusion.

The probability that the sequence ever exceeds $\varepsilon$ after time $k$ tends to $0$.

(b)(ii) Almost sure convergence implies convergence in probability

Claim.

If $X_k\to0$ a.s., then $X_k\to0$ in probability.

Proof.

From part (i), for any $\varepsilon>0$, \(P\left(\bigcup_{m=k}^{\infty}|X_m|>\varepsilon\right)\to0.\) But \(\{|X_k|>\varepsilon\} \subseteq \bigcup_{m=k}^{\infty}|X_m|>\varepsilon,\) so \(P(|X_k|>\varepsilon) \le P\left(\bigcup_{m=k}^{\infty}|X_m|>\varepsilon\right)\to0.\)

Conclusion.

$X_k \to 0$ in probability.

(c) Convergence in probability but not almost surely

Claim.

Under the assumptions

  • independence,
  • $P( X_k >1)=1/k$,
  • $P( X_k \le 1/\sqrt{k})\to 1$,

we have $X_k \to 0$ in probability but not almost surely.

Proof.

Step 1: Convergence in probability

For any fixed $\varepsilon>0$:

  • If $\varepsilon \le 1$, then \(P(|X_k|>\varepsilon)\le P(|X_k|>1)=\frac1k\to 0.\)

  • If $\varepsilon > 1$, then eventually $|X_k|\le 1/\sqrt{k}<\varepsilon$ with probability $\to 1$, so again
    \(P(|X_k|>\varepsilon)\to0.\)

Thus $X_k \to 0$ in probability.

Step 2: Almost sure convergence fails

Consider the events \(A_k=\{|X_k|>1\}.\) We are told \(P(A_k)=\frac1k.\) Since the $A_k$ are independent, apply the Second Borel–Cantelli Lemma: \(\sum_{k=1}^{\infty} P(A_k) = \sum_{k=1}^{\infty}\frac1k = \infty, \qquad\Rightarrow\qquad P(A_k \text{ i.o.})=1.\)

Thus with probability 1, infinitely many $k$ satisfy $|X_k|>1$, which makes \(X_k \not\to 0 \quad a.s.\)

Conclusion.

We have
\(X_k \to 0 \text{ in probability}, \qquad X_k \not\to 0 \text{ a.s.}\)

Key Takeaways

  • Distribution vs. Probability Convergence:
    When the limit is a constant, convergence in distribution and convergence in probability are equivalent.

  • Monotone Tail Events:
    The sets
    $A_k=\cup_{m=k}^{\infty}{|X_m|>\varepsilon}$
    form a decreasing sequence. Continuity from above is a powerful tool for a.s. arguments.

  • a.s. ⇒ in Probability:
    A standard technique: relate single-time deviations to tail deviations, then squeeze using monotonicity.

  • Using Tail Bounds to Show Convergence in Probability:
    Split the analysis into “regular part” ($|X_k|\le 1$) and “tail part” ($|X_k|>1$).
    Evaluate each separately.

  • Borel–Cantelli II:
    Independence + $\sum P(A_k)=\infty$ ⇒ events occur infinitely often a.s.
    This is the standard way to build counterexamples where convergence in probability holds but a.s. fails.

  • General Strategy:
    If asked whether $X_k\to0$ a.s., examine
    $P(|X_k|>\varepsilon)$ for fixed $\varepsilon$,
    then ask whether large deviations occur infinitely often.

If you’d like, I can now generate 2023-Q3, Q4, etc., in the same format.

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