Probability Theory Solution – Problem 34: prove that\\

Question

Let the random variables $X_n$ be independent and identically distributed with the exponential distribution of parameter 1. Given a positive real number $\alpha$ , prove that\\ $P(X_n > \alpha \ln n \text{ i.o.}) = \begin{cases} 1, & 0 < \alpha \le 1, \\ 0, & \alpha > 1. \end{cases}$

Step-by-step solution

Step 1. Compute the probability of each event $P(A_n)$ . Define the events. Let $A_n = \{X_n > \alpha \ln n\}$ . The quantity under investigation is $P(\limsup_{n \rightarrow \infty} A_n) = P(A_n \text{ i.o.})$ . Use the exponential distribution formula to compute the probability. Since $X_n$ follows the exponential distribution with parameter 1, its probability density function is $f(x) = e^{-x} \quad (x \ge 0)$ , and its cumulative distribution function is $F(x) = 1 - e^{-x}$ . The tail probability formula is: $P(X_n > x) = e^{-x}$ Substituting $x = \alpha \ln n$ (noting that $\alpha > 0$ and $n \ge 1$ by hypothesis, so $\alpha \ln n \ge 0$ ): $P(A_n) = P(X_n > \alpha \ln n) = e^{-\alpha \ln n}$ Simplify the expression. By the logarithmic identity $e^{\ln(b^c)} = b^c$ : $P(A_n) = (e^{\ln n})^{-\alpha} = n^{-\alpha} = \frac{1}{n^{\alpha}}$

Step 2. The case $\alpha > 1$ . Examine the convergence of the probability series. Consider the series $\sum_{n=1}^{\infty} P(A_n)$ : $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ This is a $p$ -series with $p = \alpha$ . By a classical result in calculus, when $\alpha > 1$ the series converges, i.e., $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} < \infty$ Apply the first Borel–Cantelli lemma: for any sequence of events $A_n$ , if $\sum_{n=1}^{\infty} P(A_n) < \infty$ , then $P(A_n \text{ i.o.}) = 0$ . Therefore, when $\alpha > 1$ : $P(X_n > \alpha \ln n \text{ i.o.}) = 0$

Step 3. The case $0 < \alpha \le 1$ . Examine the convergence of the probability series. Again consider the series $\sum_{n=1}^{\infty} P(A_n)$ : $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ For the $p$ -series, when $0 < \alpha \le 1$ the series diverges, i.e., $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} = \infty$ Verify the independence condition. The problem explicitly states that the sequence of random variables $\{X_n\}$ is independent and identically distributed. Therefore, the sequence of events $A_n = \{X_n > \alpha \ln n\}$ is also mutually independent. Apply the second Borel–Cantelli lemma. The second Borel–Cantelli lemma states: for a sequence of mutually independent events $A_n$ , if $\sum_{n=1}^{\infty} P(A_n) = \infty$ , then $P(A_n \text{ i.o.}) = 1$ . Therefore, when $0 < \alpha \le 1$ : $P(X_n > \alpha \ln n \text{ i.o.}) = 1$

Step 4. Conclusion. Combining the above cases, we have proved that for the two ranges of $\alpha$ : $P(X_n > \alpha \ln n \text{ i.o.}) = \begin{cases} 1, & 0 < \alpha \le 1 \\ 0, & \alpha > 1 \end{cases}$

Final answer

QED.

Marking scheme

This marking scheme is based on the official solution approach, with a total of 7 points. Grade strictly according to the following three sections.

1. Checkpoints (max 7 pts)

Note: In this section, scores within each group cannot exceed the group cap (if any).

Part I: Core Probability Computation (1 pt)

[1 pt] [additive] Correctly apply the exponential distribution tail probability formula to derive and simplify $P(X_n > \alpha \ln n) = n^{-\alpha}$ (or $\frac{1}{n^{\alpha}}$ ).
*If only the exponential distribution formula $e^{-x}$ is stated without substituting $\alpha \ln n$ and simplifying, award 0 points.*

Part II: The case $\alpha > 1$ (2 pts)

[1 pt] [additive] State that the series $\sum_{n=1}^{\infty} P(A_n) = \sum \frac{1}{n^{\alpha}}$ converges when $\alpha > 1$ (or cite the $p$ -series test).
[1 pt] [additive] Invoke the first Borel–Cantelli lemma (BC1) to conclude that in this case $P(A_n \text{ i.o.}) = 0$ .

Part III: The case $0 < \alpha \le 1$ (4 pts)

[1 pt] [additive] State that the series $\sum_{n=1}^{\infty} P(A_n) = \sum \frac{1}{n^{\alpha}}$ diverges when $0 < \alpha \le 1$ (the case $\alpha = 1$ must be explicitly included).
[1 pt] [additive] Key theoretical condition: Explicitly state the independence of the event sequence $\{A_n\}$ (derived from the independence of $X_n$ ), and present it as a necessary prerequisite for applying the second Borel–Cantelli lemma.
[2 pts] [additive] Invoke the second Borel–Cantelli lemma (BC2) to conclude that in this case $P(A_n \text{ i.o.}) = 1$ .
*Note: If independence is not mentioned, do not deduct these 2 points on that basis; only deduct the "independence statement" point from the preceding item.*

Total (max 7)

2. Zero-credit items

Merely copying the random variable definitions from the problem statement or the statement of the Borel–Cantelli lemma, without performing any problem-specific computation or substitution.
Stating only the final conclusion (e.g., writing the piecewise result directly) while omitting all intermediate derivations (such as the probability computation and the series convergence/divergence analysis).
A serious error in the probability computation that renders $P(A_n)$ a constant (not a function of $n$ ); even if the subsequent logic is correct, the entire subsequent portion generally receives no credit (unless the subsequent portion demonstrates an independently correct judgment of series convergence/divergence).

3. Deductions

*In this section, deduct at most the single largest applicable penalty; penalties are not cumulative. The total score shall not fall below 0.*

[-1] Unclear boundary analysis: When analyzing the $p$ -series or stating the conclusion, the case $\alpha = 1$ is not handled correctly (e.g., erroneously claiming the series converges at $\alpha = 1$ , or failing to explicitly state that $\alpha = 1$ falls in the divergent case).
[-1] Logical gap: In the case $0 < \alpha \le 1$ , the correct conclusion is reached but the core reason "the series diverges" is entirely omitted (jumping directly from the probability formula to the conclusion).
[-1] Notational error: Confusing set notation with probability values (e.g., writing $A_n = n^{-\alpha}$ instead of $P(A_n) = n^{-\alpha}$ ), or confusing $X_n$ with $A_n$ .

Probability Theory – Problem 34: prove that\\