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Any random variable measurable with respect to the tail $\sigma$-field of an infinite independent sequence of $\sigma$-fields is equal to some constant a.s.

My attempt:

Since $\alpha_n \rightarrow \infty$, we have $\limsup \frac{S_n}{\alpha_n} = \limsup \frac{X_j,\ldots,X_n}{\alpha_n}$, thus the variable $\limsup \frac{S_n}{\alpha_n}$ is measurable with respect to the tail σ-algebra of (X_n) and since the (Xn) are all independent, by the 0 − 1 law, this random variable is almost-surely constant.

Is it correct the attempt ?

Could someone help me pls? Thanks for your time and help

Rosa Maria Gtz.
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2 Answers2

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Not sure I get your attempt... the key is that we know that the event $\{X\le x\}$ is a tail event so $F(x)=P(X\le x)$ is zero or one for all $x.$ The intuition is that since $F$ is monotonic in $x,$ and (provided $X$ is not almost surely $-\infty)$ we have $F(-\infty) = 0$ and $F(\infty) = 1,$ there must be some crossover point $c$ at which $F$ jumps from $0$ to $1,$ so that $X=c$ a.s. Naturally, this will be given by $$ c=\sup\{x \mid F(x)=0\}.$$

Then you must show that $P(X=c) =1,$ which is easiest to show here by showing $P(X<c) = 0$ and $P(X\le c) =1.$ For instance, we have $$ 0=P\left(\bigcup_n \{X<c-1/n\}\right) = P(X<c).$$

spaceisdarkgreen
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  • I have to prove that $P(X \geq c)$=1 and $ P(X < c)=0$ , isn't it? or why $ P(X \leq c)=1$ – Rosa Maria Gtz. Nov 27 '17 at 01:39
  • $P(X< c) = 0$ is synonymous with $P(X\ge c) = 1,$ so proving both of them is redundant and not enough. Proving $P(X<c) = 0$ and $P(X\le c) = 1$ (my recommendation) amounts to showing $P(X\le c) = P(X\ge c) = 1$ which proves $P(X=c) =1.$ – spaceisdarkgreen Nov 27 '17 at 01:42
  • $ 1=P\left(\bigcup_n {X \leq c+1/n}\right) = P(X \leq c).$ Is it correct? – Rosa Maria Gtz. Nov 27 '17 at 02:15
  • needs to be an intersection – spaceisdarkgreen Nov 27 '17 at 02:34
  • Great answer, admittedly simpler than mine. +1. @Knight To conclude that $\mathbb P(X\leq c)=1$, just show that $$\mathbb P(X>c)=\mathbb P\left(\bigcup_{n\in\mathbb N}\left{X>c+\frac{1}{n}\right}\right)=0.$$ – triple_sec Nov 27 '17 at 05:22
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Let $\mathcal T$ denote the $\sigma$-algebra which satisfies, by Kolmogorov’s zero–one law, $\mathbb P(A)\in\{0,1\}$ for any $A\in\mathcal T$.

If $X$ is a random variable measurable with respect to $\mathcal T$, then one has, for any $n\in\mathbb Z$, $\{X\in[n,n+1]\}\in\mathcal T$, so that $\mathbb P\{X\in[n,n+1]\}\in\{0,1\}$. But since $\bigcup_{n\in\mathbb Z}\{X\in[n,n+1]\}$ is the whole state space, it must be the case that $\mathbb P\{X\in[n,n+1]\}=1$ for at least one $n\in\mathbb Z$.

Now cut that particular interval $[n,n+1]$ in two halves, $[n,n+1/2]$ and $[n+1/2,n]$. By a similar argument as above, one has either $\mathbb P\{X\in[n,n+1/2]\}=1$ or $\mathbb P\{X\in[n+1/2,n]\}=1$ (or possibly both, but that is not important for the argument). Take the interval for which the probability is $1$, cut it in half again, and choose the half for which the probability of $X$ falling into it is $1$. And so forth, keep cutting in half. Then, use Cantor’s intersection theorem to conclude that the intersection of these ever smaller nested intervals is a singleton: $\{c\}$ for some $c\in\mathbb R$. Complete the proof by showing that $\mathbb P(X=c)=1$ (use that probability measures are continuous from above).

triple_sec
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