# 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.

• Are you looking for an estimate for large $d$? Commented Nov 22, 2022 at 14:02
• Yes, I expect $d$ to be large (and am happy to make additional assumptions, if necessary) Commented Nov 22, 2022 at 14:04

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 $$E\big[\langle v,Z \rangle \big| \langle u,Z \rangle \ge c_2 \big] \ge c_1c_2/\sqrt{d} - \sqrt{1-c_1^2}\sqrt{1-c_2^2/d}.$$

• Thank you! I had realized this lower bound, but it has the unfortunate property that for even moderately positive values of $c_1, c_2 > 0$, the lower bound is negative (which is worse than what you get from not conditioning at all). Commented Nov 22, 2022 at 15:40