# 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.

• Do you suppose that $Y_i$ are independent? Commented Nov 2, 2021 at 21:22
• Yes we assume that $Y_i$'s are independent. Commented Nov 3, 2021 at 1:52

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}

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 $$Bern(\frac12)$$.