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We want to prove the strong law of large numbers with Birkhoff's ergodic theorem.

Let $X_k$ be an i.i.d. sequence of $\mathcal{L}^1$ random variables. This is a stochastic process with measure-preserving operation $\theta$ (the shift operator). From Birkhoff's ergodic theorem, we obtain $\frac{X_0 + \dotsb + X_{n-1}}{n} \to Y$ a.s., with $Y=\mathbb{E}[X_1 \mid \mathcal{J}_{\theta}]$ a.s.

Now, if $Y$ constant a.s., $Y= \mathbb{E}[X_1]$ a.s., and we would have the desired result. But why is $Y$ constant a.s.?

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    $\begingroup$ $\theta$ is ergodic, so any $\theta$-invariant r.v. is a.s. constant. $\endgroup$
    – Zhen Lin
    Jul 26, 2012 at 8:16
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    $\begingroup$ Because the sigma-algebra $\mathcal J_\theta$ is trivial, in the sense that every $A$ in $\mathcal J_\theta$ has probability $0$ or $1$. Hence every random variable measurable for $\mathcal J_\theta$ is almost surely constant. $\endgroup$
    – Did
    Jul 26, 2012 at 8:24

1 Answer 1

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The transformation $\theta$ on $\Omega^{\Bbb N}$ is ergodic. Indeed, it's enough to show that for each cylinder $A$ and $B$, we have $$\frac 1n\sum_{k=0}^{n-1}\mu(\theta^{-k}A\cap B)\to \mu(A)\mu(B),$$ where $\mu$ is the measure on the product $\sigma$-algebra. If $A=\prod_{j=0}^NA_j\times \Omega\times\dots$ and $B=\prod_{j=0}^NB_j\times \Omega\times\dots$, we have for $k>N$ \begin{align} \theta^{-k}A\cap B&=\{(x_j)_{j\geq 0}, (x_{j+k})_{j\geq 0}\in A, (x_j)_{j\geq 0}\in B\}\\ &=\{(x_j)_{j\geq 0},x_{j+k}\in A_j, 0\leq j\leq N, x_j\in B_j,0\leq j\leq N\}\\ &=B_0\times \dots\times B_N\times \Omega\times\dots\times \Omega\times A_0\times\dots\times A_n\times \Omega\times\dots, \end{align} and we use the definition of product measure $\mu$ on cylinders (the $N$ first terms doesn't matter).

Since $\theta$ is ergodic, $\mathcal J_{\theta}$ consists only of events of measure $0$ or $1$. The conditional expectation with respect such a $\sigma$-algebra is necessarily constant.

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  • $\begingroup$ A stationary process $X$ can be charakterized by the fact that $\theta$ is invariant with respect to the measure $\mathbb{P} \circ X$.. So this means that Birkhoff's ergodic theorem includes that any stationary process converges to something almost surely constant? $\endgroup$
    – Suedklee
    Jul 26, 2012 at 9:09
  • $\begingroup$ It means that the arithmetic means converge almost surely. $\endgroup$ Jul 26, 2012 at 9:43
  • $\begingroup$ Yes, I wanted to ask if this not already implies that the arithmetic mean converges a.s. but also that it converges to a a.a. constant r.v. $\endgroup$
    – Suedklee
    Jul 26, 2012 at 10:05

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