1
$\begingroup$

$E_1,\ldots,E_n$ are independent. We want to show that $$P\left(\bigcap_{i=1}^n E_i^c\right) \le e^{-\sum\limits_{i=1}^n P(E_i)}$$ More generally, for a countable sequence $(E_i)_{1}^\infty$ of independent events, show that $$P\left(\bigcap_{i=1}^\infty E_i^c\right) \le e^{-\sum\limits_{i=1}^\infty P(E_i)}$$

I'm trying to use Boole's inequality and the fact that $e^{-x} \ge 1-x$ for all $x\in\mathbb R$.

To use Boole's inequality, I first invoked De-Morgan's Law to do the following: $$P\left(\bigcap_{i=1}^n E_i^c\right) = P\left(\left(\bigcup_{i=1}^n E_i\right)^c\right) = 1 - P\left(\bigcup_{i=1}^n E_i\right)$$

Then I tried to put $x = P\left(\bigcup_{i=1}^n E_i\right)$ and use $e^{-x} \ge 1-x$, but sadly I didn't get to the desired inequality. What am I missing? I did note that $$P\left(\bigcup_{i=1}^n E_i\right) \le \sum_{i=1}^nP\left(E_i\right)$$ but it didn't produce the right inequality.

Once I'm able to prove it for $n$, putting $n\to\infty$ should do the rest of the job. Since probability measures are continuous, we know we can switch $P$ with $\lim_{n\to\infty}$.

Thanks for the help!

$\endgroup$

1 Answer 1

2
$\begingroup$

Independence tells us that $$P\left(\bigcap_{i=1}^n E_i^c\right) = \prod_{i=1}^n P(E_i^c) = \prod_{i=1}^n (1 - P(E_i)),$$ which is the expression to which you should start applying all your inequalities.

$\endgroup$
4
  • $\begingroup$ To use the result of the finite case and move to the infinite one, would it be correct to do: $$\lim_{n\to\infty} P\left(\bigcap_{i=1}^n E_i^c\right) \le \lim_{n\to\infty} e^{-\sum\limits_{i=1}^n P(E_i)} = e^{-\sum\limits_{i=1}^\infty P(E_i)}$$ Also $$\lim_{n\to\infty} P\left(\bigcap_{i=1}^n E_i^c\right) = \lim_{n\to\infty} \prod_{i=1}^n P\left(E_i^c\right) = \prod_{i=1}^\infty P\left(E_i^c\right) = P\left(\bigcap_{i=1}^\infty E_i^c\right)$$ $\endgroup$ Commented Jan 25, 2021 at 13:22
  • 1
    $\begingroup$ Thanks a lot! I'm not very comfortable with multiplying infinite probabilities like that and putting the intersection inside, so I came up with an alternative that you could help verify: $B_n = \cap_{i=1}^n E_i^c$ is a decreasing sequence. $$\lim_{n\to\infty} P\left(\bigcap_{i=1}^n E_i^c\right) = P\left(\bigcap_{n=1}^\infty\bigcap_{i=1}^n E_i^c\right) = P\left(\bigcap_{i=1}^\infty E_i^c\right)$$ I used nothing more than the equality which says that if $B_n$ is a decreasing sequence of events, then $$\lim_{n\to\infty} P(B_n) = P\left(\bigcap_{n=1}^\infty B_n\right)$$ Sounds good too? $\endgroup$ Commented Jan 25, 2021 at 13:33
  • 1
    $\begingroup$ Yes, I think that one's actually better. Now that I think about it, I overlooked the final step where you go from $\prod_{i=1}^\infty$ to $\bigcap_{i=1}^\infty$, and you're right that without your decreasing-sequence-of-events property, that's a little dodgy (though I think it's still true). $\endgroup$ Commented Jan 25, 2021 at 15:07
  • $\begingroup$ Alright, thanks a lot Misha! $\endgroup$ Commented Jan 25, 2021 at 18:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .