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 $1/2$?


1 Answer 1


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 $$ \frac{1}{2} \int_{-1}^1 x^2 \, \mathrm{d}x = \left. \frac{x^3}{6} \right|_{-1}^1 = \frac{1}{3}. $$

  • $\begingroup$ Expectaction of $ F(X) $is 0 ...right? instead of $[F(x)]^2$ is the $|| v ||^2$ . $\endgroup$
    – theantomc
    Commented Jan 8, 2019 at 11:42
  • $\begingroup$ The expectation of $F(x)$ is zero. I didn’t understand the rest of your question. $\endgroup$ Commented Jan 8, 2019 at 11:45

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