Consider we have samples $\mathbf{X} \in \mathcal{R}^{n\times p}$ and we aim to find a "regression" coefficients $\beta \in \mathcal{R}^{q \times 1}$ ($q>p$), but the regression is defined as a convolution function as follows:

For every $\mathbf{x}$, we have:

$$ \mathbf{x}\odot\beta = \begin{cases} \mathbf{x}_1\beta_1+\mathbf{x}_2\beta_2 + \dots + \mathbf{x}_p\beta_p \\ \mathbf{x}_1\beta_2+\mathbf{x}_2\beta_3 + \dots + \mathbf{x}_p\beta_{p+1} \\ \vdots\\ \mathbf{x}_1\beta_{q-p+1}+\mathbf{x}_2\beta_{q-p+2} + \dots + \mathbf{x}_p\beta_q \end{cases} $$ therefore, $\mathbf{x}\odot\beta$ leads to a vector.

Given a data set $\mathbf{X}$, can we solve a non-trivial $\beta$ so that for every $\mathbf{x} \in \mathbf{X}$, $\mathbf{x}\odot\beta = \mathbf{0}$?

  • $\begingroup$ There's a trivial solution: set all $\beta$ to zero. $\endgroup$ – user20160 Jul 24 '19 at 10:49
  • $\begingroup$ Thanks, I edited the question a little bit to specify that we need a non-trivial solution @user20160 $\endgroup$ – Haohan Wang Jul 24 '19 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.