$X_1,X_2,...$ real independent random variables $\not\implies S_n = \frac{1}{n}(X_1+ \cdots + X_n)$ are independent

Question

Is it possible to show the following Lemma:

Given independent real random variables $(X_i)_{i\in \mathbb{N}}$,

then $(S_n := \frac{1}{n}(X_1+\cdots + X_n))_{n \in \mathbb{N}}$ are independent as well.

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If so, any hints?

Using this lemma I would be able to solve my homework. But so far I didn't manage to proof mentioned lemma -- is it possible at all?

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So as this is not possible, I can't use Kolmogoroff ($S_n$ are not independent) to show the following:

$X_1,...$ independent, $S_n := \frac{1}{n}(X_1+\cdots + X_n)$ and $A := \{x: \lim_{n \to \infty} S_n(x) \text{ exists in } \mathrm{R}\cup \{\pm \infty\}\}$ Then: $P(A) \in \{0,1\}$.

See here: Why is the probability that $(X_1+\ldots+X_n)/n$ converges either $0$ or $1$?

Answer 1

This is clearly false. When trying to find counter-examples, you often just have to think simple. For example, let $X_1=X$ be some random variable, and let $X_2=x$ be some constant $x\in\mathbb{R}$. Then $X_1$ and $X_2$ are independent but $S_1=X$ and $S_2=\frac{1}{2}X+\frac{1}{2}x$ which clearly aren't independent.

Answer 2

Your lemma is false. Consider $S_1 = X_1$ and $S_2 = \frac{X_1+X_2}{2}$. Intuitively, it must be the case that $S_1$ and $S_2$ are dependent as knowledge of the value of $S_1$ gives us some information about the value of $S_2$.