Conditional expectation of linear combination of Rademacher RVs Let $u, v \in \mathbb{S}^{d-1}$ be two unit vectors with $u \cdot v \geq c_1$.
Let $Z \in \{-1, +1\}^d$ be a random sign vector where each coordinate is +1 or -1 independently with probability 1/2.
I would like to compute a lower bound, if one exists, on the conditional expectation
$$\mathbb{E}[\langle Z, v\rangle|\langle Z, u\rangle \geq c_2]$$
in terms of $c_1$ and $c_2$. It seems like some kind of coupling argument would work here since the conditioning makes $Z$ more likely to be close to $v$ overall, but I haven't been able to work out the details.
Edit: The specific regime I am studying is that of $c_1$ and $c_2$ both being relatively small but positive, and am looking for a positive lower bound (given that the unconditional expectation is zero).

Edit: If $u$ is a normalized sign vector, I've been able to solve the problem as follows: define $Z' = Z \odot u$ (elementwise product), and let $v' = v \odot u$. By symmetry, $Z'$ is also a random Rademacher vector.
Then, our condition translates to $\sum_i v'_i \geq c_1$, and the expectation can be written as $\mathbb{E}[\langle Z', v'\rangle|\sum_i Z_i' \geq c_2]$. For every integer $k > c_2$, the expectation $\mathbb{E}[\langle Z', v'\rangle|\sum_i Z_i' = k]$ is just a rescaling of $\sum_i v'_i$, which is positive by assumption.
However, I haven't been able to extend this beyond $u$ being a sign vector.
 A: I give a crude lower bound, which does not use the distribution of $Z$, but only that $|Z|=\sqrt{d}$. It relies on the triangle inequality for the angular distance on the unit sphere.
I assume $c_1$ and $c_2$ to be non-negative.
The angular distance between $u$ and $v$ is $\arccos \langle u,v \rangle \le \arccos c_1 \le \pi/2$.
On the event $[\langle u,Z \rangle \ge c_2]$, the angular distance between $u$ and $Z$ is $\arccos (\langle u,Z \rangle/|Z|) \le \arccos(c_2/\sqrt{d})\le \pi/2$.
Hence on the event $[\langle u,Z \rangle \ge c_2]$, the angular distance between $v$ and $Z$ is at most $\arccos c_1 + \arccos(c_2/\sqrt{d})$, which is less than $\pi$.
Since $\cos$ decreases on $[0,\pi]$, we derive
$$\langle v,Z \rangle \ge \cos\big(\arccos c_1 + \arccos(c_2/\sqrt{d})\big) = c_1c_2/\sqrt{d} - \sqrt{1-c_1^2}\sqrt{1-c_2^2/d}$$
on the event $[\langle u,Z \rangle \ge c_2]$, so
$$E\big[\langle v,Z \rangle \big| \langle u,Z \rangle \ge c_2 \big] \ge c_1c_2/\sqrt{d} - \sqrt{1-c_1^2}\sqrt{1-c_2^2/d}.$$
