Let $Z$ be some finite abelian group. Suppose that we pick a random subset $X$ of $Z$ of size $d$, and define the set $$X^+ = \{\epsilon_1 x_1 + \epsilon_2x_2 + \ldots+\epsilon_dx_d | \forall i\in [d]: \epsilon_i \in \{0,1\} \}. $$ I am trying to prove that if $d = O(\log |Z|)$, then with positive probability $|X^+| = \Omega(|Z|)$.
My general idea: First, estimate the probability $p$ that some $z \in Z$ is in $X^+$, and hopefully $p$ would be some constant. Then, using the linearity of expectation, we have $E[|X^+|] = \sum_z p_z = p|Z|$. This would end the proof, since there must be some choice of $X$ which would obtain the expected value.
My problem is with estimating $p$. I tried the following: Let $z \in Z$. For each non-empty subset $I \subseteq [d]$, define the event: $$G_I := \{z = \sum_{i \in I} x_i\}, $$ then $\Pr[G_I] = \frac{1}{|Z|}$, and $\Pr[z \in X^+] = \Pr[\bigvee_{I} G_I]$. Now the issue is that I want a lower bound, not an upper bound, and so the union bound will not help here.
I tried also to define the bad events, $B_I := \{z \neq \sum_{i\in I} x_i \}$, and then somehow argue about the probability that non of them hold simultaneously. The thing is that I am pretty sure those events are dependent, and so it seems that it would be better to somehow argue about the expectation of the sum of their indicators, i.e. count the expected number of times $z$ appears in $X^+$, which is $\frac{2^d}{|Z|}$. The problem here as well that the bounds are for the converse direction: I can say something about the probability that $z$ appears too many times in $X^+$, but I don't see a way to argue that it appears too few times.