Fix

ε > 0

. Since a rate-one exponential variable has

P (X_{n} > x) = e^{- x}

\sum_{n} P {X_{n} > (1 + ε) lo g n} = \sum_{n} n^{- (1 + ε)} < \infty.

The first Borel--Cantelli lemma gives the upper bound

lim sup X_{n} / lo g n \leq 1

. Conversely,

\sum_{n} P {X_{n} > (1 - ε) lo g n} = \sum_{n} n^{- (1 - ε)} = \infty.

These events are independent, so the second lemma makes them occur infinitely often. Hence the lower bound is

1 - ε

. Letting rational

ε ↓ 0

proves

n \to \infty lim sup \frac{X _{n}}{lo g n} = 1 almost surely.

ii

Solution

For

ε > 0

P {M_{n} \leq (1 - ε) lo g n} = (1 - n^{- (1 - ε)})^{n} \leq exp (- n^{ε}),

using

1 - x \leq e^{- x}

. The final probabilities are summable, so Borel--Cantelli implies that eventually

M_{n} > (1 - ε) lo g n .

Intersecting over positive rational

ε

n \to \infty lim inf \frac{M _{n}}{lo g n} \geq 1 almost surely.

There are no discussions about this article yet.

There are currently no matching articles.