Find $P(\limsup A_n)$ given $P(A_{n+1}) \ge \frac{n}{n+1} P(A_n)$

Question

Consider independent events $A_1, A_2, ...$ in the same probability space s,t. $$P(A_{n+1}) \ge \frac{n}{n+1}P(A_n) \ \forall n \ge 1$$

Find $P(\limsup A_n)$.

There seem to be two cases:

1. $\exists m > 0 \ \text{s.t.} \ P(A_i) > 0 \ \forall \ i \geq m$

2. $P(A_i) = 0 \ \forall i \geq 1$

Is that right?

Answer 1

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For case 1, it seems that $P(\limsup A_n) = 1$:

Let $N-1 := \sup\{n \mid P(A_n) = 0\}$

Then $$P(A_1) + P(A_2) + P(A_3) + \cdots$$

$$= P(A_N) + P(A_{N+1}) + P(A_{N+2}) + \cdots$$

$$\ge P(A_N) + \frac{N}{N+1}P(A_N) + \frac{N}{N+2}P(A_N) + \cdots$$

$$= (N P(A_N))[\frac{1}{N} + \frac{1}{N+1} + \frac{1}{N+2} + \cdots]$$

where $$[\frac{1}{N} + \frac{1}{N+1} + \frac{1}{N+2} + \cdots] = \infty$$ by limit comparison test with harmonic series. Hence, BCL2 gives us what we want. QED

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For case 2, it seems that $P(\limsup A_n) = 0$:

Define $B_m := \bigcup_{n \ge m} A_n$. Then

$$P(\limsup A_n) = P(\bigcap_{m \ge 1} B_m)$$

$$= 1 - P(\bigcup_{m \ge 1} B_m^c)$$

$$= 1 - \lim_{m \to \infty} P(B_m^c)$$

where

$$P(B_m^c) = 1 - P(B_m) = 1 - P(\bigcup_{n \ge m} A_n) \ge 1 - \sum_{n \ge m} P(A_n) = 1$$ QED