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I was taught that the Strong Law of Large Numbers states that for some iid random variables $X_i$ with $\overline{X_n}=\frac{1}{n}\sum_{i=1}^n(X_i)$, it follows that $P(\lim_{n\to\infty}\overline{X}=\mu)=1$. We are stating that the probability of this equality is 1, so why would it be improper to just say $\lim_{n\to\infty}\overline{X}=\mu$ ?

I know that intuitively, this latter claim is not true. For example, by flipping a bunch of coins, although the ratio between heads and tails converges to $1:1$, the difference between heads and tails is unbounded. But I am still confused as to why we can state that the probability of this equality being true is equal to 1 ,when in reality, it is not true.

June Richardson
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    A random variable is a (nice) function $X: \Omega \to \mathbb R$. What $\lim_{n\to \infty} \bar X_n = \mu$ means is that for each $\omega \in \Omega$ we have $\lim_{n\to \infty} \bar X_n(\omega) = \mu$. But it could be the case there is some $\omega$ such that $\bar X_n(\omega) \not \to \mu$. If the set of $\omega$ that the convergence does not occur has probability 0, then we say $\bar X_n \to \mu$ almost surely. – Andrew Dec 29 '22 at 18:54
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    For example, take $X$ to be i.i.id uniform on $[0,1]$. Put $X_n = X^n$. Then $\mathbb P(\lim_{n\to \infty} X_n = 0) = \mathbb P(X\neq 1) = 1$. So $X_n \to 0$ almost surely. Note, however, that we do not have $\lim X_n = 0$. For example, if $X: [0,1]\to [0,1]$ is just the identity map, then $X(1) = 1$ and $\lim X_n(1) = X(1)$. But $\mathbb P(X = 1) = 0$... – Andrew Dec 29 '22 at 18:59
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    Because an event having probability 1 does not imply that said event always happens. – Sergey Guminov Dec 29 '22 at 20:17
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    https://math.stackexchange.com/questions/41107/zero-probability-and-impossibility – user51547 Dec 29 '22 at 21:04

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