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Let $\{X_n\}$ be a independent random values with $P(X_n = 1) = P(X_n = -1) = 1/2$. Let $Y_n = \prod_{i=1}^{n} X_i$. My book states, it is clear that since the $X_n$ are independent, that $Y_{n-1} $ and $X_n$ is independent - why is this "clear"?

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  • $\begingroup$ What does $Y_n=X_1 \ldots X_n$ mean ? I think that some puntuation marks are missing. $\endgroup$ – callculus Apr 17 '16 at 18:10
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    $\begingroup$ @callculus It's safe to assume that $x_1x_2$ implies multiplication between reals. $\endgroup$ – Axoren Apr 17 '16 at 18:17
  • $\begingroup$ @Axoren For me it is not obvious. Where in the exercise do you read $x_1x_2$ ? $\endgroup$ – callculus Apr 17 '16 at 18:22
  • $\begingroup$ It means multiplication. $\endgroup$ – coffeeman Apr 17 '16 at 18:41
  • $\begingroup$ Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) $\endgroup$ – Martin Sleziak Apr 18 '16 at 13:37
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$Y_{n-1}$ is independent of $X_n$ for two reasons:

  • $Y_{n-1}$ is not explicitly defined in terms of $X_n$ (being explicitly defined in terms of $X_n$ would make them clearly dependent).
  • Each $X_i$ for $i < n$ is independent of $X_n$, so their individual realizations of values do not convey the state of $X_n$ to $Y_{n-1}$.

For these two reasons, you will see that the probability of $Y_{n-1}$ taking any one value does not change with observances of the value of $X_n$. Therefore, $Y_{n-1}$ is independent of $X_n$.

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The random variables $X_1,\ldots,X_n$ are said to be independent when each $X_i$ is independent from all other variables $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n$. More precisely: $$ \begin{array}{l} \mathbb{P}(X_i \in A \ \land\ \langle X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n \rangle \in B ) = \\ \mathbb{P}(X_i \in A) \cdot \mathbb{P}(\langle X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n \rangle \in B ) \end{array} $$ for every (measurable) sets $A,B$.

This implies that any variable $X_i$ is independent of any (measurable) function of the other variables $f(X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n)$. For any (measurable) $C$ we obtain:

$$ \begin{array}{l} \mathbb{P}(X_i \in A \ \land\ f( X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n ) \in C ) = \\ \mathbb{P}(X_i \in A \ \land\ \langle X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n \rangle \in f^{-1}(C) ) = \\ \mathbb{P}(X_i \in A) \cdot \mathbb{P}(\langle X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n \rangle \in f^{-1}(C) ) \end{array} $$

Note that the above is stronger than only saying that each pair of distinct variables $X_i,X_j$ are independent.

To see why, take $X_1,X_2$ to be independent variables returning $1$ or $-1$ with probability $1/2$, and $X_3 = X_1 \cdot X_2$. (Note that this is basically the same example in the question.) Then, every variable is independent from any other variable: e.g., $X_3$ is independent from $X_1$ (alone) and from $X_2$ (alone). However, $X_3$ is not independent of $X_1,X_2$ taken together.

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One way to see why it is clear that the two random variables $Y_{n-1}$ and $X_n$ are independent is to consider the following theorem:

Let $X$ and $Y$ be independent random variables. Then, $U = g(X)$ and $V = h(Y )$ are also independent for any function $g$ and $h$.

Here you can write $Y_{n-1} = g(X_1,...,X_{n-1})$. Since the $X_n$'s are mutually independent, $g(X_1,...,X_{n-1})$ is independent of $X_n$.

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