Probability Theory Solution – Problem 47: prove that

Question

The random variables $X_n$ are independent and identically distributed, each following an exponential distribution with 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 the event $P(A_n)$ . Define the event $A_n = \{X_n > \alpha \ln n\}$ . We wish to determine $P(\limsup_{n \rightarrow \infty} A_n) = P(A_n \text{ i.o.})$ . Using the exponential distribution formula: Since $X_n$ follows an exponential distribution with parameter 1, its probability density function is $f(x) = e^{-x}$ for $x \ge 0$ , 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 $n \ge 1$ , so $\alpha \ln n \ge 0$ ): $P(A_n) = P(X_n > \alpha \ln n) = e^{-\alpha \ln n}$ Simplifying using 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: $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ This is a $p$ -series with $p = \alpha$ . By a standard result from calculus, when $\alpha > 1$ , this series converges: $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} < \infty$ Applying 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: $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ For the $p$ -series, when $0 < \alpha \le 1$ , the series diverges: $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} = \infty$ Verify the independence condition: The problem explicitly states that the random variable sequence $\{X_n\}$ is independent and identically distributed. Therefore, the event sequence $A_n = \{X_n > \alpha \ln n\}$ consists of mutually independent events. Applying the Second Borel--Cantelli Lemma: The Second Borel--Cantelli Lemma states that 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. In summary, depending on the range of $\alpha$ , we have proved: $P(X_n > \alpha \ln n \text{ i.o.}) = \begin{cases} 1, & 0 < \alpha \le 1 \\ 0, & \alpha > 1 \end{cases}$

Final answer

QED.

Marking scheme

This rubric is based on the official solution approach, with a total of 7 points. Please grade strictly according to the following three sections.

1. Checkpoints (max 7 pts)

Note: Within this section, points are not to exceed the cap of each group (if any).

Part 1: Core probability computation (1 point)

[1 pt] [additive] Correctly use 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 listed without substituting $\alpha \ln n$ and simplifying, award 0 points.*

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

[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 result).
[1 pt] [additive] Invoke the First Borel--Cantelli Lemma (BC1) to conclude that $P(A_n \text{ i.o.}) = 0$ in this case.

Part 3: The case $0 < \alpha \le 1$ (4 points)

[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 included in particular).
[1 pt] [additive] Key theoretical condition: Explicitly state the independence of the event sequence $\{A_n\}$ (based on 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 $P(A_n \text{ i.o.}) = 1$ in this case.
*Note: If independence is not mentioned, do not deduct these 2 points for that reason; only deduct the "independence statement" point above.*

Total (max 7)

2. Zero-credit items

Merely copying the random variable definitions from the problem or the statement of the Borel--Cantelli lemma without performing any specific computation or substitution for this problem.
Only giving the final conclusion (e.g., directly writing the piecewise function result) without any intermediate derivation (such as probability computation, series convergence analysis).
A serious error in the probability computation causing $P(A_n)$ to be a constant (not a function of $n$ ); even if the subsequent logic is correct, the entire subsequent part typically receives no credit (unless the subsequent part demonstrates an independently correct judgment of series convergence).

3. Deductions

*In this section, deduct at most the single largest applicable item; deductions are not cumulative. The total score cannot go below 0.*

[-1] Unclear boundary discussion: When analyzing the $p$ -series or the conclusion, the case $\alpha = 1$ is not handled correctly (e.g., incorrectly claiming the series converges when $\alpha = 1$ , or not explicitly stating that $\alpha = 1$ belongs to the divergent case).
[-1] Logical jump: In the case $0 < \alpha \le 1$ , although the correct conclusion is reached, the core reason "the series diverges" is completely 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 47: prove that