Suppose I have a geometric distribution $X$ with probability of success $1/2$, and a distribution $Y$ which counts the number of successes in three subsequent trials ($X+1$, $X+2$, $X+3$). I want to show that these are independent by demonstrating that their joint pmf satisfies $p(i, j) = p_X(i)p_Y(j)$.
I found the marginal distributions to be $p_X(i) = 2^{-i}$ and $p_Y(j) = {3 \choose j}/8$, and tried to condition on $X$ while making use of the properties of conditional probability, but I can't seem to find a solution without assuming independence at some point throughout.
Let $O_i$ be the indicator variable of outcome of the $i$-th toss. The $O_i$'s are independent.
\begin{align} P(X=i, Y=j) &= P(X=i)P(Y=j|X=i)\\ &=P(X=i)P(\sum_{p=i+1}^{i+3}O_p=j|O_i=1)\\ &=P(X=i)P(\sum_{p=i+1}^{i+3}O_p=j)\\ &= P(X=i)P(Y=j) \end{align}