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Are the running products of iid RVs independent?

Let $Y_0, Y_1, ...$ be independent random variables with $P(Y_n = 1) = P(Y_n = -1) = 1/2 \ \forall n = 0, 1, 2, ...$ (*)

Define $X_n = Y_0 Y_1 ... Y_n \ \forall n = 0, 1, 2, ...$

Prove that $X_1, X_2, ...$ are independent (strangely $X_0$ is omitted).

I tried 2 ways.


Way 1:

We must show that $P(X_1 \in B_1, ..., X_n \in B_n) = \prod_{i=1}^{n} P(X_i \in B_i)$.

$LHS = P(X_n \in B_n | X_1 \in B_1, ..., X_{n-1} \in B_{n-1}) P(X_{n-1} \in B_{n-1} | X_1 \in B_1, ..., X_{n-2} \in B_{n-2})...P(X_2 \in B_2 | X_1 \in B_1) P(X_1 \in B_1)$

From the definition of the RVs, if you know $X_{n-2}$, you don't need to know $X_1, ..., X_{n-3}$ for $P(X_{n-1} \in B_{n-1} | X_1 \in B_1, ..., X_{n-2} \in B_{n-2})$, I think.

Hence, $P(X_{n-1} \in B_{n-1} | X_1 \in B_1, ..., X_{n-2} \in B_{n-2}) = P(X_{n-1} \in B_{n-1} | X_{n-2} \in B_{n-2})$. Hence, $X_1, X_2, ...$ is a Markov process.

Now, given $X_n$, what is the probability $X_{n+1} = 1$? Well that is the same thing as asking what is the probability that sign($Y_{n+1}$) = sign($X_n$).

If $X_n > 0$, then P(sign($Y_{n+1}$) = sign($X_n$)) = P($Y_{n+1} > 0$) = 1/2

If $X_n < 0$, then P(sign($Y_{n+1}$) = sign($X_n$)) = P($Y_{n+1} < 0$) = 1/2

Same for $X_{n+1} = -1$.

One can deduce (as below) that $P(X_{n+1} = 1) = P(X_{n+1} = -1) = 1/2$.

Hence $P(X_{n+1} \in B | X_{n}) = P(X_{n+1} \in B)$.

Thus, LHS = RHS, QED.


Way 2:

  1. I got $\sigma(X_1)$ and $\sigma(X_2)$ and inferred $\sigma(X_n) = (\emptyset, \Omega, X_n = 1, X_n = -1)$.

  2. I deduced that $P(X_1 = 1) = P(X_1 = -1) = 1/2$.

  3. I tried checking if $X_1$ and $X_2$ are independent. WOLOG, I check $P(X_2 = 1, X_1 = 1) = P(X_2 = 1) P(X_1 = 1)$:

3a. I deduce that $Y_1$ and $X_2$ are independent due to some property whose proof I don't understand.

3b. I deduce that $P(X_2 = 1) = 1/2$ from 3a:

$P(X_2 = 1) = P(X_1 = -1, Y_2 = -1) + P(X_1 = 1, Y_2 = 1) = P(X_1 = -1)P(Y_2 = -1) + P(X_1 = 1) P(Y_2 = 1) = 1/2$.

3c. $P(X_2 = 1, X_1 = 1) \doteq P(X_2 = 1 | X_1 = 1) P(X_1 = 1) = 1/4$

Hence, $P(X_2 = 1, X_1 = 1) = P(X_2 = 1) P(X_1 = 1)$. WOLOG, $X_1$ and $X_2$ are independent.

  1. From 3, $X_n$ and $X_{n+1}$ are independent, $\forall \ n = 1, 2, ...$ (and I guess 0). (I think?)

  2. From 4, $X_n$ and $X_m$ are independent where $n \neq m$ and $\forall \ n, m = 1, 2, ...$ (and I guess 0). (I think?)

  3. From 5, $X_{i_1}, ..., X_{i_j}$ are independent for indices $j = 1, 2, ...$ and $(i_1, ..., i_j) \subseteq (1, 2, ...)$. (I think?)

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  1. If relevant, I deduced that $P(X_n = 1) = P(X_n = -1) = 1/2 \ \forall n = 1, 2, ...$ (and 0 I guess).

In case my question is the same as this one, and at least one of my proofs is/are wrong, pls tell me anyway why it is/they are wrong.


(*) Is it correct to describe these random variables as identically distributed?

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If $e_j\in\{-1,+1\}$ for $0\leq j\leq n$, then $$\mathbb{P}(\cap_{j=0}^n [X_j=e_j]) =\mathbb{P}(\cap_{j=0}^n [Y_j=e_{j+1}/e_{j}])=(1/2)^{n+1}. $$

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  • $\begingroup$ Thanks. Why not $e_{j+1}/e_j$ ? $\endgroup$ – BCLC Jul 6 '15 at 14:54
  • $\begingroup$ @BCLC Because $Y_j=X_j/X_{j-1}$. $\endgroup$ – user940 Jul 6 '15 at 15:13
  • $\begingroup$ Why not $Y_j = X_{j+1}/X_j$ ? $\endgroup$ – BCLC Jul 6 '15 at 15:14
  • $\begingroup$ Look at your formula for $X_n$ $\endgroup$ – user940 Jul 6 '15 at 15:15

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