1
$\begingroup$

I have another question with dimensionality reduction. I have a matrix $S \in R^{k \times d}$ and S is in {$- \frac{1}{\sqrt k}, \frac{1}{\sqrt k}$} and i have two vector $u,v \in R^d $.

I need to understand why $E[u'^T v' ]= u^Tv$ where $u'=Su$.

I just have a intution, that maybe i just prove the left(or right span vector), but the hint is compute $E[S^TS]$ that for me this expectation is expectation of E[ I ] that if i m wrong will be 0...

I m a bit lost with this stuff.

i m thinking to do this

$E[u^t v] = E[u^T \sum vi ]$

another question is about the interval ... using $1 /\sqrt k $ is the same if i use a -1,1 with scaled factor of $\sqrt k $?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The idea is that $$ \mathbb{E}[u^{\prime T} v'] = \mathbb{E}[u^T S^T S v] = u^T \mathbb{E}[S^T S] v, $$ due to linearity of expectation. You can take it from here.

Improve this answer
$\endgroup$
3
  • $\begingroup$ So the $E[S^T S] = 1$ because the variables are independent ? $\endgroup$
    – theantomc
    Commented Jan 16, 2019 at 8:59
  • $\begingroup$ Note $S^T S$ is a matrix. $\endgroup$
    – Yuval Filmus
    Commented Jan 16, 2019 at 9:02
  • $\begingroup$ Yes, for me $S^T S = I $ the identity matrix, and S have entries $S_{ij}$ $\endgroup$
    – theantomc
    Commented Jan 16, 2019 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.