Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d. $\mathbb{R}$-valued r.v.s such that for some $M \geq 0$ it follows that $P[|X_n| \leq M] = 1$ for all $n$. Define $Y_n = \frac{1}{n}\sum_{i=1}^nX_i$, and $L = \lim\sup_{n}Y_n$, and $A = \{\lim_nY_n \text{ exists}\}$. Then my reading material makes the following claim

There exists some $c \in [-M, M]$ s.t. $P[L = c] = 1$, and $P[A] \in \{0, 1\}$

The proof for the claim is as follows

For any $q \in \mathbb{Q}$ define the tail event $A_q = \{L\geq q\}$. By Kolmogorov's 0-1 law, the probability of each of these tail events is either 0 or 1. Thus, there exists some $c = \sup\{q:P[A_q] = 1\} = \inf\{q:P[A_q] = 0\}$. Since $\mathbb{Q}$ is countable, $P[L\geq c] = P[L\leq c] = 1$, and so $P[L = c] = 1$. Finally $c \in [-M, M]$, since $P[L\in [-M, M]] = 1$.

The main points I'm struggling to understand here are 1.) How do we know from the Kolmogorov's 0-1 law that there exists some constant $c$ s.t. $c = \sup\{q:P[A_q] = 1\} = \inf\{q:P[A_q] = 0\}$, 2.) why does it follow from the countability of $\mathbb{Q}$ that $P[L\geq c] = P[L\leq c] = 1$?


1 Answer 1


I will provide a similar argument. Kolmogorov's 0-1 law tells use that the tail $\sigma$-algebra is trivial in the sense that $\forall A\in \mathscr{F}_\infty,\,P(A)\in \{0,1\}$. In general, if a $\sigma$-algebra $\mathscr{A}$ is in that sense trivial, then any $\mathscr{A}$-measurable random variable $Y$ is then equal to a constant a.s., that is $P(Y=c)=1$ for some $c$. To see this, notice that $F_Y(y)=P(Y\leq y)\in \{0,1\}$ since $\{Y\leq y\}\in \mathscr{A}$. As $F_Y$ is nondecreasing, there exists one $c$ s.t. $F_Y$ 'jumps' to $1$ at $c$. This implies that $P(Y=c)=1$.

  • $\begingroup$ For $c \in \mathbb{R}$ to exist makes total sense now that I realized that the sequence is bounded! (I had worked with a similar problem before, where there was no assumption of boundedness, hence the slight confusion regarding that). The only thing I'm quite not understanding yet is why are we using the rational numbers in the proof? Could the proof be made without them? $\endgroup$ Commented Feb 16, 2022 at 16:59
  • 1
    $\begingroup$ @EpsilonAway Yes it can. You know that $L$ is measurable wrt the tail $\sigma$-algebra. Since the tail $\sigma$-algebra is $P$-trivial, it follows by the theorem Snoop just proved that $L = c$ for some $c \in \overline{\mathbb{R}} = [-\infty, \infty]$. Now $$|X_n| \leq M \implies |Y_n| \leq M \implies |L| \leq M.$$ Hence $L \in [-M, M]$. $\endgroup$
    – Mason
    Commented Feb 16, 2022 at 18:13

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