If we have sets $A_1, A_2, \ldots$, then when we write

$$ P\left(\bigcap_{n=1}^{\infty}A_i\right) $$

what does this actually mean? Does it mean the probability of the intersection of infinite sets? Or does it mean that IF this infinite intersection does indeed, then what is the probability that an element is in the set? To be more explicit, is the above the same thing as:

$$ P\left(\bigcap_{n=1}^{\infty}A_i\right) = P\left(\omega \in \Omega:\bigcap_{n=1}^{\infty}A_i(\omega)\right) $$ ?

I often times get confused exactly what I am seeking the probability of, it is very easy to think of if there is some equality inequality inside the probability measure but would you know what it means usually when the intersection is left by itself inside? Thanks

  • $\begingroup$ Think of $P$ as a function taking sets to $[0,1]$. So $\bigcap_{n=1}^{\infty} A_i$ is the set and $P(\bigcap_{n=1}^{\infty} A_i)$ gives the probability that 'your result' $\omega$ is in $\bigcap_{n=1}^{\infty} A_i$. $\endgroup$ – E.Lim Nov 29 '15 at 11:14
  • $\begingroup$ E.Lim is right: $P\left(\bigcap_{n=1}^{\infty}A_i\right)$=$P\left(\left\{\omega \in\Omega \mid \omega \in \bigcap_{n=1}^{\infty}A_i\right\}\right)$ (be aware it is not a conditional probability, it is the probability measure of the set $\left\{\omega \in\Omega \mid \omega \in \bigcap_{n=1}^{\infty}A_i\right\}$ ) $\endgroup$ – MoebiusCorzer Nov 29 '15 at 12:16
  • 3
    $\begingroup$ The notation $A_i(\omega)$ sems to point at a serious misconception in the OP's mind. There is no such thing as $A_i(\omega)$. $\endgroup$ – Did Nov 29 '15 at 13:00

If you meant

$$ P\left(\bigcap_{n=1}^{\infty}A_n\right) $$

then that is the probability that given $\omega \in \Omega$,

$$\omega \in A_n \ \forall n \ge 1$$


$$\omega \in A_1, A_2, A_3, ...$$

That is, almost surely, all the events $A_1, A_2, A_3, ...$ occur.

Sometimes this arises in the case of independent events where we have

$$P\left(\bigcap_{n=1}^{\infty}A_n\right) = \prod_{n=1}^{\infty} P(A_n)$$

Sometimes we want to compute

$$ P\left(\bigcup_{n=1}^{\infty}A_n\right) $$

We can do this by noting that:

$$ P\left(\bigcup_{n=1}^{\infty}A_n\right) = 1 - P\left(\bigcap_{n=1}^{\infty}A_n^C\right) $$

So, for example, if we have a bunch of random variables $X_1, X_2, ...$ in $(\Omega, \mathscr F, \mathbb P)$

If we want to compute the probability that at least 1 is greater than 5, we can compute the probability that all are less than 5:

$$ P\left(\bigcup_{n=1}^{\infty}(X_n \ge 5)\right) = 1 - P\left(\bigcap_{n=1}^{\infty}(X_n < 5)\right) $$

If the random variables are independent, we have

$$1 - P\left(\bigcap_{n=1}^{\infty}(X_n < 5)\right) = 1 - \prod_{n=1}^{\infty}P(X_n < 5) = 1 - \prod_{n=1}^{\infty}F_{X_n}(5)$$

Edit: You said

$$ P\left(\bigcap_{n=1}^{\infty}A_i\right) = P\left(\omega \in \Omega:\bigcap_{n=1}^{\infty}A_i(\omega)\right) $$

Actually it's supposed to be

$$ P\left(\bigcap_{n=1}^{\infty}A_n\right) = P\left(\omega \in \Omega:\omega \in \bigcap_{n=1}^{\infty}A_n\right) $$

'$A_i(\omega)$' is a notation we apply to random variables. For example if we have random variables $X_1, X_2, ...$,

define $X := \limsup X_n$ s.t. for a given $\omega \in \Omega$

$$X(\omega) := [\limsup X_n](\omega) := \limsup [X_n(\omega)]$$

Perhaps you meant to say '$\omega \in A_i$' as in

$$ P\left(\bigcap_{n=1}^{\infty}A_n\right) = P\left(\omega \in \Omega: \bigcap_{n=1}^{\infty}(\omega \in A_n)\right) $$


I'm not sure, but maybe that could work. I don't think I've seen that notation a lot.


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