Sequence of product of Rademacher variables

Question

I am trying to figure out what a sequence of product of Radenmacher variables looks like.

The sequence is defined as: Let $\{\nu_j\}$ be IID fair ±1 coin-flips (Rademacher variables). Let $Y_1 = \nu_1$, let $Y_2 = \nu_1 \nu_2$ and for $m = 2^{n−1} + j$ with $0 < j ≤2^{n−1}$ let $Y_m = \nu_{n+1}Y_j$.

Given the definition above, I can't figure out what is the form of $Y_3, Y_4, Y_5$ and so on. Would appreciate any help.

The question asks me to prove that $Y_i$ are independent which seems to be true since it is the product of coin flips. But I am finding the notation for how $Y_i$ is defined very confusing.

Answer 1

The procedure for writing $Y_i$ in terms of the $\nu_j$ is to take $i-1$ in binary, add a $1$ to the end of the binary string, and take the product of the $\nu_j$ over all $j$ where bit $j$ (counting right to left, starting with $1$) is a $1$.

For example, to find $Y_6$, note that $6-1 = 5$, the binary representation of $5$ is $101$, adding a $1$ to the end gives $1011$, which has the first, second, and fourth bits from the right equal to $1$. So $Y_6 = \nu_1 \nu_2 \nu_4$.

The proof for this is just an induction; the recurrence relation makes it clear that the recurrence can be defined in terms of the binary expansion and all you need to do is verify the base cases.

The other answer and comment tell you how to show that these variables are not independent, but are pairwise independent.

(I am curious about the context for why the $Y_i$ are defined that way. Was it to enumerate all possible products of the $\nu_i$ that contain $\nu_1$?)

Answer 2

Let's find a few more:

• $m=3$: Since $3 = 2^{2-1} + 1$, we conclude $Y_3 = \nu_3Y_1$.
• $m=4$: Since $4 = 2^{2-1} + 2$, we conclude $Y_4 = \nu_3Y_2$.
• $m=5$: Since $5 = 2^{3-1} + 1$, we conclude $Y_5 = \nu_4Y_1$.
• $m=6$: Since $6 = 2^{3-1} + 2$, we conclude $Y_6 = \nu_4Y_2$.

From this, we see $Y_6 = Y_3Y_4Y_5$, so in particular, the $Y_i$ aren't independent.