Distribution of signed random variables Define a sequence of random variables $\{Y_i\}$ where $P(Y_i =1)=\frac{1}{2}=P(Y_i = -1)$. Let $X_1 = Y_1$ and $X_{2^k +j}= X_j Y_{k+2}$ where $j=1,2,...,2^k$ and $k \in \mathbb{N}\cup\{0\}$. I am struggling to find the distribution of $X_i$.
I can see that they are pairwise independent, but I am not sure how to get the distribution of $X_i$.
I tried for $n=2^m$, it becomes:
$X_{n} = Y_1 Y_2 Y_3 ^2 Y_4...Y_{m+1}$.
So, $X_n$ can take values $+1$ or $-1$ depending on the parity of $+1$'s or $-1$'s in the product.
 A: Notice a pattern?
$${X_1=Y_1\\X_2=X_{2^0+1 }= X_1Y_{0+2}=Y_1Y_2\\X_3=X_{2^1+1}=X_1Y_{1+2}=Y_1~~~~Y_3\\X_4=X_{2^1+2}=X_2Y_{1+2} = Y_1Y_2Y_3\\X_5=X_{2^2+1}=X_1Y_{2+2}=Y_1\qquad Y_4\\X_6=X_{2^2+2}=X_2Y_{2+2}=Y_1Y_2\quad Y_4\\X_7=X_{2^2+3}=X_3Y_{2+2}=Y_1\quad Y_3Y_4\\X_8=X_{2^2+4}=X_4Y_{2+2}=Y_1Y_2Y_3Y_4\\X_9=X_{2^3+1}=X_1Y_{3+2}=Y_1 \qquad\quad Y_5\\\!\!X_{10}=X_{2^3+2}=X_2Y_{3+2}=Y_1Y_2\qquad Y_5\\\!\!X_{11}=X_{2^3+3}=X_3Y_{3+2}=Y_1\quad Y_3\quad Y_5\\~\vdots\\\!\!X_{16}=\hspace{18ex}=Y_1Y_2Y_3Y_4Y_5}$$

So, Xn can take values +1 or −1 depending on the parity of +1's or −1's in the product.

Yes, not quite.   $X_n=1$ when there are an even number of $^-1$ in the product (of $Y_m$ that comprise it).
$\small\begin{align}&\mathsf P(X_4{=}1)\\=~&\mathsf P(Y_1Y_2Y_3{=}1) \\ =~&\mathsf P(Y_1{=}1,Y_2{=}1,Y_3{=}1)+\mathsf P(Y_1{=}1,Y_2{=}{^-}1,Y_3{=}{^-}1)+\mathsf P(Y_1{=}{^-}1,Y_2{=}1,Y_3{=}{^-}1)+\mathsf P(Y_1{=}{^-}1,Y_2{=}{^-}1,Y_3{=}1)\\=~&\tfrac 4{2^3}\end{align}$
A: If we don't know the joint distribution of $Y_i$ then it doesn't have sense.
Suppose that $Y_i$ are i.i.d.
Hence $X_{2k+j} = Z \cdot Y_{k+2}$, where $Z$ is a function of $Y_1, \ldots, Y_{k+1}$ (a product of some of them) and takes only two values: value $1$ with some probability $p$ and takes value $-1$ with probability $1-p$. Hence $X_{2k+j} = Z \cdot Y_{k+2}$ where $Z$ and $Y_{k+2}$ are independent.
We have
$$P(X_{2k+j} = 1) = P(Z = Y_{k+2} = 1) + P(Z=Y_{k+2 }= -1) = p \cdot \frac12 + (1-p) \cdot \frac12 = \frac12,$$
$$P(X_{2k+j} = 1)  = 1 - P(X_{2k+j} = 0) = 1 - \frac12 = \frac12.$$
Hence $X_i$ are $Bern(\frac12)$.
