# 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?

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.