Showing that $\{X_n, n\geq 1 \}$ are independent if for $n\geq 2$ we have $\sigma(X_1,...,X_{n-1}) \perp \sigma(X_n)$

Question

Looking for hints to proceed or to corroborate my solution (I found it still weak).

So, as the title says:

We want to show that $\{X_n, n\geq 1 \}$ are independent random variables if for $n\geq 2$ we have: $\sigma(X_1,...,X_{n-1}) \perp\sigma(X_n)$.

My approach:

So, I can say, assume this is true for n=2.

then we have, from the "if condition" that if we know that: $\sigma(X_1) \perp \sigma(X_2)$

Then, this implies that $X_1 \perp X_2$, since their induced sigma-fields are independent, the random variables are independent.

Now, take an induction approach (still not sure if approaching it the right way though), but this is my shot:

Start by checking $n=3$:

From the "if condition", if we know that: $\sigma(X_1,X_2) \perp \sigma(X_2)$

This implies that $X_1 \perp X_2 \perp X_3$. Since the induced sigma-fields are independent, the random variables are independent.

A question is:: What is the relation between $\sigma(X_1) \perp \sigma(X_2)$ and $ \sigma(X_1,X_2) $

I little confused now. Would appreciate any help.

Thanks.!

Answer 1

I believe your perp stands fro independence, not orthogonality. By definition a sequence is independent if each finite subset is. So we have to show that $\{X_1,X_2,...,X_n\}$ is independent for each N. This means $P\{X_1^{-1}(A_1) \cap ...\cap X_N^{-1} A_N\}$ is the product of $P\{X_i^{-1} A_i\}$. Just note that $\{X_1^{-1}(A_1) \cap ...\cap X_{(N-1)}^{-1} A_{(N-1)}\}$ belongs to $\ sigma \{X_1,X-2,...,X_{(N-1)}\}$ and use induction.