-3
$\begingroup$

Consider a sequence $\{p_n\}_{n \in \mathbb{N}}$ s.t. $p_n \in [0,1)$ and $\sum_{n=1}^{\infty} p_n < \infty$.

Prove $\prod_{n=1}^{\infty} (1-p_n) > 0$.


I think there's a way to do this without probability eg this or this. I mean there's no probability here originally. But I would like to try a probabilistic proof. Kinda like this or this


What I tried (I'm not sure if my lim, inf and sup statements are right):

Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space, and let $X_n$ be iid Bernoulli random variables with parameter $p_n$ and in $(\Omega, \mathscr{F}, \mathbb{P})$. Then,

$$\sum_{n=1}^\infty p_n<\infty \iff \sum_{n=1}^\infty P(X_n = 1) <\infty$$

By Borel-Cantelli Lemma 1 we have, $$P(\limsup(X_n = 1))=0$$

which is equivalent to $$P(\liminf(X_n = 0)) = 1$$

$$ \ $$

$$\to 1 =P(\bigcup_{N \ge 1} \bigcap_{n \ge N} (X_n=0))$$

By continuity of measure $\color{red}{\text{(I think?)}}$

$$= \lim_{N \to \infty}P(\bigcap_{n \ge N} (X_n=0))$$

By monotone convergence theorem $\color{red}{\text{(I think?)}}$

$$= \sup_{N \ge 1}P(\bigcap_{n \ge N} (X_n=0))$$

By independence,

$$ = \sup_{N \ge 1} \prod_{n=N}^{\infty} P(X_n=0)$$

$$= \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n)$$

$$\to \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n) = 1$$

$$\to \forall \epsilon > 0, \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1] \ \color{red}{\text{(I think?)}}$$

Since $p_n < 1 \ \forall n \in \mathbb{N}$,

$$\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1] \ \color{red}{\text{(I think?)}}$$

Hence $\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.}$

$$\prod_{n=1}^\infty(1-p_n)=\prod_{n=1}^{m-1}(1-p_n)\prod_{n=m}^{\infty}(1-p_n) > 0 \ \because$$

  1. $\prod_{n=1}^{m-1}(1-p_n) > 0 \because p_n < 1 \ \forall n \ge 1$

  2. $\prod_{n=m}^{\infty}(1-p_n) > 0 \because \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$ QED

Any mistakes? Again, I'm not sure if my lim, inf and sup statements are right.

$\endgroup$
0
-3
$\begingroup$

Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space, and let $X_n$ be iid Bernoulli random variables with parameter $p_n$ and in $(\Omega, \mathscr{F}, \mathbb{P})$. Then,

$$\sum_{n=1}^\infty p_n<\infty \iff \sum_{n=1}^\infty P(X_n = 1) <\infty$$

By Borel-Cantelli Lemma 1 we have, $$P(\limsup(X_n = 1))=0$$

which is equivalent to $$P(\liminf(X_n = 0)) = 1$$

$$ \ $$

$$\to 1 =P(\bigcup_{N \ge 1} \bigcap_{n \ge N} (X_n=0))$$

By continuity of measure

$$= \lim_{N \to \infty}P(\bigcap_{n \ge N} (X_n=0))$$

By monotone convergence theorem

$$= \sup_{N \ge 1}P(\bigcap_{n \ge N} (X_n=0))$$

By independence,

$$ = \sup_{N \ge 1} \prod_{n=N}^{\infty} P(X_n=0)$$

$$= \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n)$$

$$\to \sup_{N \ge 1} \prod_{n=N}^{\infty} (1-p_n) = 1$$

$$\to \forall \epsilon > 0, \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$$

Since $p_n < 1 \ \forall n \in \mathbb{N}$,

$$\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.} \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$$

Hence $\forall \epsilon \in (0,1), \exists m \ge 1 \text{s.t.}$

$$\prod_{n=1}^\infty(1-p_n)=\prod_{n=1}^{m-1}(1-p_n)\prod_{n=m}^{\infty}(1-p_n) > 0 \ \because$$

  1. $\prod_{n=1}^{m-1}(1-p_n) > 0 \because p_n < 1 \ \forall n \ge 1$

  2. $\prod_{n=m}^{\infty}(1-p_n) > 0 \because \prod_{n=m}^{\infty} (1-p_n) \in (1-\epsilon, 1]$ QED


'Verified by Landon Carter'

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.