# Is there an upper bound about the probability $\mathcal{P}(\|Ax\|_{\infty}\geq t)$

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!

## 1 Answer

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 $t$.