About joint distribution of random variables Let $W_1,W_2,W_3,W_4$ be iid random variables s.t.
$$W_i = \begin{cases}0 \text{ w.p. 1/2}\\1 \text{ w.p. 1/2} \end{cases}$$
for $i=1,2,3,4$

Is the joint distribution of $Y_1:=(W_1+W_4) \bmod 2, Y_2;=(W_1+W_2)\bmod 2,Y_3:=(W_2+W_3)\bmod 2, Y_4:=(W_3+W_4)\bmod 2$ uniform over $2^4$ possibilities?

I would think I have to look at conditional probabilities of $Y_2$ given $Y_1$, etc to check if that is correct but I'm not sure?
 A: TL;DR The distribution is uniform, but not over the $16$ cases, because some of those cases cannot occur. Conditional probabilities are not necessary.
To save writing, let $Y$ denote the binary string $Y_1Y_2Y_3Y_4$, and similarly let $W$ denote the binary string $W_1W_2W_3W_4$.
Note that for binary variables $a,b$, we have $a+b\bmod 2=\begin{cases}1&\text{if }a\ne b\\ 0&\text{if }a= 
 b\end{cases}.$
Now consider $Y_1Y_2Y_3Y_4=0000$. That can only happen if every $W_i=0$ or every $W_i=1$, so $P(Y=0000)=P(W=0000)+P(W=1111)=2(1/2^4)=1/8.$ On the other hand, consider $Y_1Y_2Y_3Y_4=1000.$ Here $Y_1=1$ implies $W=1ab0$ or $W=0ab1$, leaving no way to assign $ab$ to make $Y_2=Y_3=Y_4=0$ -- i.e. this case cannot occur, so $P(Y=1000)=0.$
Thus,
$$\begin{align}
P(Y=0000)&= P(W\in\{0000, 1111 \})=2(1/2^4)=1/8\\
P(Y=1000)&= 0\\
P(Y=1100)&= P(W\in\{1000, 0111 \})=2(1/2^4)=1/8\\
P(Y=1110)&= 0\\
P(Y=1111)&= P(W\in\{1010, 0101 \})=2(1/2^4)=1/8\\
& \,\,\,\vdots
\end{align}$$
That is, in the eight cases where $Y_1Y_2Y_3Y_4$ has oddly many $1$s, the probability is $0$, and the probability is uniform ($=1/8$) over the other eight cases of evenly many $1$s.
