1
$\begingroup$

Let random matrix $\mathbf{X} \in \mathbb{C^{\mathrm{m} \times \mathrm{n}}}$ and random vector $\mathbf{y} \in \mathbb{C^{\mathrm{m} \times 1}}$ are unknown distributed, but their covariance and correlation are known, $\mathbf{C}_{\mathbf{X}}$, $\mathbf{C}_{\mathbf{y}}$, and $\mathbf{C}_{\mathbf{Xy}}$. The question is whether there is any bound for the following expression, (like Cauchy-Schwarz or Jensen's)

$?\le \mathbb{E}[(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}] \le ?$

Also, I have to mention $\le$ here means an element-wise inequality since the corresponding expectation is an $\mathrm{n} \times 1$ vector.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

No, of course not.

Indeed, consider a simplest case when $m=n=1$, and $X:=\mathbf X$ and $y:=\mathbf y$ are iid standard normal random variables. Then $$E(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}=E\frac yX$$ does not even exist (because here $\frac yX$ has the standard Cauchy distribution).

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks a lot for your answer. Yes what you say is true, but I should add something. Here all random variables in both matrix $\mathbf{X}$ or $\mathbf{y}$ are from passing gaussian variable through sign function. As you know the resulting variable doesn't have known distribution, but at least the amplitude of each element (of matrix $\mathbf{X}$ or $\mathbf{y}$ ) can not be zero, so that fraction would be bounded. $\endgroup$
    – A. R.
    Commented Feb 18, 2023 at 21:02
  • $\begingroup$ @AminRadbord : I don't understand your comment. What do you mean by "Here"? by "are from passing gaussian variable through sign function"? by " the resulting variable"? by "the amplitude"? by "fraction"? $\endgroup$
    – Iosif Pinelis
    Commented Feb 19, 2023 at 0:16
  • $\begingroup$ @ Iosif Pinelis: I mean, assume in my question we have this assumption, each element in $\mathbf X$ or $\mathbf y$ is a random variable which is the output of the sign function where the input is a Gaussian random variable. For example, if m=1 and n=1, then X = sgn(z) and y = sgn(t), in which both z and t are gaussian random variables. so as a result, both X and y can not be zero so $\mathbb{E}[\frac{y}{X}]$ is bounded. $\endgroup$
    – A. R.
    Commented Feb 19, 2023 at 11:59
  • 1
    $\begingroup$ @AminRadbord : This is a completely different question, which should be posted separately. There is nothing like this $\text{sgn}$ condition in your post -- which only had "unknown distributed". $\endgroup$
    – Iosif Pinelis
    Commented Feb 19, 2023 at 20:18
  • 2
    $\begingroup$ @AminRadbord : One should not change the question so as to invalidate a valid answer. Such actions on your part may make people reluctant to spend their time to answer your questions -- nobody wants to waste their time. So, please restore the original question and finalize this matter appropriately. $\endgroup$
    – Iosif Pinelis
    Commented Feb 20, 2023 at 3:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .