# Probabilistic proof for $\prod_{n} (1-p_n) > 0$ (Considering the $p_n$'s as parameters of random variables and using Borel-Cantelli and continuity)

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.

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'