Pairwise Independence vs. Independence

For collections of independent random variables, the Chernoff bound and related bounds give us very sharp concentration inequalities — if $X_1,\ldots,X_n$ are independent, then their sum has a distribution that decays like $e^{-x^2}$. For random variables that are only pairwise independent, the strongest bound we have is Chebyshev’s inequality, which says that their sum decays like $\frac{1}{x^2}$.
The point of this post is to construct an equality case for Chebyshev: a collection of pairwise independent random variables whose sum does not satisfy the concentration bound of Chernoff, and instead decays like $\frac{1}{x^2}$.
The construction is as follows: let $X_1,\ldots,X_d$ be independent binary random variables, and for any $S \subset \{1,\ldots,d\}$, let $Y_S = \sum_{i \in S} X_i$, where the sum is taken mod 2. Then we can easily check that the $Y_S$ are pairwise independent. Now consider  the random variable $Z = \sum_{S} Y_S$. If any of the $X_i$ is equal to 1, then we can pair up the $Y_S$ by either adding or removing $i$ from $S$ to get the other element of the pair. If we do this, we see that $Z = 2^{d-1}$ in this case. On the other hand, if all of the $X_i$ are equal to 0, then $Z = 0$ as well. Thus, with probability $\frac{1}{2^d}$, $Z$ deviates from its mean by $2^{d-1}-\frac{1}{2}$, whereas the variance of $Z$ is $2^{d-2}-\frac{1}{4}$. The bound on this probability form Chebyshev is $\frac{2^{d-2}-1/4}{(2^{d-1}-1/2)^2}$, which is very close to $\frac{1}{2^d}$, so this constitutes something very close to the Chebyshev equality case.