# Expectation of $\langle s,x \rangle^2$

I'm studying dimensionality reduction (SVD in particular), and I saw the following question:

Assume we have a vector $$x \in \mathbb R^d$$, and consider $$F(x)=s^t x$$ , where $$s$$ is a $$d$$-dimensional random vector with entries drawn uniformly independently from $$[-1,1]$$.

What is the value of $$\mathbb E[F(x)^2]$$?

I'm starting now from zero, so I need to study more. The exercise asks for a formal proof, but I wish to just understand the philosophy.

I see many questions in which valued are drawn uniformly and independently from $$[-1,1]$$. Is this distribution used due to its symmetrical range? Is the expectation in this range often $$1/2$$?

Let us assume that each entry of $$s$$ is drawn independently from some distribution $$\mathcal{D}$$ whose expectation is $$0$$. Then $$\mathbb{E}\left[\left(\sum_{i=1}^d s_i x_i\right)^2\right] = \sum_{i=1}^d x_i^2 \mathbb{E}[s_i^2] + \sum_{i \neq j} x_i x_j \mathbb{E}[s_i s_j] = \|x\|^2 \mathbb{V}[\mathcal{D}],$$ since $$\mathbb{E}[s_is_j] = \mathbb{E}[s_i]\mathbb{E}[s_j] = 0$$ due to independence.
In your particular case, $$\mathcal{D}$$ is the uniform distribution over $$[-1,1]$$, whose variance is $$\frac{1}{2} \int_{-1}^1 x^2 \, \mathrm{d}x = \left. \frac{x^3}{6} \right|_{-1}^1 = \frac{1}{3}.$$
• Expectaction of $F(X)$is 0 ...right? instead of $[F(x)]^2$ is the $|| v ||^2$ . – theantomc Jan 8 at 11:42
• The expectation of $F(x)$ is zero. I didn’t understand the rest of your question. – Yuval Filmus Jan 8 at 11:45