# Lower bound on norm of vector given estimation vector

I have the following problem:

The are two numbers unknown real $A_1, A_2$, and two known drawn approximations $B_1, B_2$ such that $$A_i-B_i\sim N(0, \sigma).$$

Does there exist a function $f:\mathbb{R}^2\rightarrow\mathbb{R}_{\ge0}$, independent of $A_1, A_2$, such that $$Pr\big(f(B_1, B_2)\le A_1^2+A_2^2\big)=0.5.$$

Or, more generally:

For $\delta>0$, does there exists a function $f_\delta:\mathbb{R}^n\rightarrow\mathbb{R}_{\ge0}$ such that for all $A_1, ..., A_n\in\mathbb{R}^n$ we have: $$Pr\big(f_\delta(A_1+\varepsilon_1, ..., A_n+\varepsilon_n)\le A_1^2+...+A_n^2\big)=1-\delta,$$ where $$\varepsilon_i\sim N(0,\sigma).$$

At first I thought that maybe $f_\delta$ should be the ppf of the noncentral chi-squared distribution with $n$ degrees of freedom and $\lambda=A_1^2+...+A_n^2$ at $\delta$, but that has a problem with the case $A_1=...=A_n=0$, since then we need $$Pr\big(f_\delta(\varepsilon_1, ...,\varepsilon_n)=0\big)=1-\delta,$$

which no ppf meets.

Googling for "estimation of vector norm" and similar wordings didn't come up with something helpful.

Does this problem have a name? Maybe a solution?

If it is known that no such solution exists, what is known about the problem when we relax equality to inequality, i.e. $$Pr\big(f_\delta(A_1+\varepsilon_1, ..., A_n+\varepsilon_n)\le A_1^2+...+A_n^2\big)\ge 1-\delta?$$