1
$\begingroup$

Given a positive-semidefinite matrix $A$, for a random vector $x\in\mathcal{R}^n$, does there exist an upper bound for the following probability $$\mathcal{P}(\|Ax\|_{\infty}\geq t),$$ where $\|x\|_{\infty}=\max\{|x_i|\}$, $t>0$ and each entry $x_i\in x$ is an independent random variable with the expectation of $0$, $\mathcal{E}(x_i)=0$.

If necessary we can add an assumption of i.i.d. and boundedness, that means $x_i$ is an i.i.d. bounded random variable, i.e. $|x_i|<c$ for some $0<c<\infty$ $a.s.$

Thank you very much!

$\endgroup$
2
$\begingroup$

I'm not sure if I understand your question correctly. An upper bound for the probability itself is $\mathcal P = 1$ (as for any probability), attained here for $t=0$, as $\|Ax\|_\infty$ is nonnegative. So I'm guessing you want to known an upper bound $t = t_\max$ such that $$ P(\|Ax\|_\infty \geqslant t) = 0 \quad \text{for }t > t_\max, $$ so that $t>t_\max$ are impossible to obtain from the expression.

If that is what you need, consider $$ \|Ax\|_\infty \leqslant \|A\|_\infty \|x\|_\infty $$ so if indeed $|x_i| < c$ then $\|Ax\|_\infty < c \|A\|_\infty =: t_\max$, as specified above. Note that this does not use the positive-semidefiniteness of $A$, nor the expected value of $x$. I don't see any obvious connection of these properties to (my interpretation of) the question. And in general, if $x$ is unbounded and $A$ is nonzero, then $\mathcal P$ does not vanish at any finite $t$.

$\endgroup$

Your Answer

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

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