# Using Borel-Cantelli Lemma

Let $X_1, X_2,\ldots$ be iid Geometric(p) where $p \in (0,1)$. Thus if $q=1-p$, then $P(X_n > k) = q^k$ for $k\geq 0$. Prove that for any fixed $\epsilon \in (0,1)$,

CORRECTION: k is supposed to be an integer. Anyway, proof is similar. Just apply floor/ceiling to $\frac{1-\epsilon}{\ln(1/q)} \ln n$ or $\frac{1+\epsilon}{\ln(1/q)} \ln n$

$P(X_n > \frac{1-\epsilon}{\ln(1/q)} \ln n\text{ i.o.}) = 1$ and $P(X_n > \frac{1+\epsilon}{\ln(1/q)} \ln n\text{ i.o.}) = 0$

Note: $P(C_n\text{ i.o.}) = P(\limsup C_n)$

Hint:

$k-1 < \left \lfloor k \right \rfloor \leq k$

$k \leq \left \lceil k \right \rceil < k+1$

1 Does the hint have something to do with $\epsilon$? I seem to recall something like $\left \lfloor k \right \rfloor + 1 = k + \epsilon$ for $0 < \epsilon \leq 1$ or $\exists \ \epsilon$.

2 Anyway here are my attempts.

2.1 $P(A_n) \equiv P(X_n > \frac{1-\epsilon}{\ln(1/q)} \ln n)) = (1-p)^{\frac{1-\epsilon}{\ln(1/q)} \ln n} \equiv a^{\ln n}$ where $a > 0$.

Note that $P(A_n) \geq 0$ and $\lim_{n \to \infty} P(A_n) \neq 0$

Thus, $\sum_{n=1}^{\infty} P(A_n) = \sum_{n=1}^{\infty} a^{\ln n} = \infty$ by the test for divergence.

Also, the $A_n$'s are independent (I don't feel like typing this down, but it's do with the fact that the joint pdf splits right?).

Thus, by Borel-Cantelli 2, $P(X_n > \frac{1-\epsilon}{\ln(1/q)} \ln n\text{ i.o.}) = 1$.

Is it correct?

2.2 Here I got $\infty$ so I guess I did something wrong, if the approach is even right.

$P(B_n) \equiv P(X_n > \frac{1+\epsilon}{\ln(1/q)} \ln n) = (1-p)^{\frac{1+\epsilon}{\ln(1/q)} \ln n} \equiv b^{\ln n}$ where $b > 0$...I think?

Also, the $B_n$'s are independent (I don't feel like typing this down, but it's do with the fact that the joint pdf splits right?).

By a similar argument, $\sum_{n=1}^{\infty} P(B_n) = \sum_{n=1}^{\infty} b^{\ln n} = \infty$, I think...?

## 1 Answer

The first part looks right, the second part can be done by Borel Cantelli 1.

$$P(X_n> {\frac{b }{ln(\frac 1 q)}\ln n}) \approx q^{\frac{b }{ln(\frac 1 q)}\ln n}$$ $$=e^{\frac{b\ln n}{-\ln q}\ln q}$$ $$=n^{-b}$$ with $b=1+\epsilon$.

The sum $\sum_n n^{-b}$ converges when $b=1+\epsilon$.

Thus, by Borel Cantelli 1, $$P(X_n > \frac{1+\epsilon}{\ln(1/q)} \ln n\text{ i.o.}) = 0.$$

• My first part was presumptuous but coincidentally correct. I just did what you did. I guess there's a q for a reason. Hehehehe. Thanks i707107! – BCLC Aug 24 '14 at 2:21