# explanation of independence [closed]

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"?

• What does $Y_n=X_1 \ldots X_n$ mean ? I think that some puntuation marks are missing. – callculus Apr 17 '16 at 18:10
• @callculus It's safe to assume that $x_1x_2$ implies multiplication between reals. – Axoren Apr 17 '16 at 18:17
• @Axoren For me it is not obvious. Where in the exercise do you read $x_1x_2$ ? – callculus Apr 17 '16 at 18:22
• It means multiplication. – coffeeman Apr 17 '16 at 18:41
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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 $X_n$.

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.

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 $X_n$.