Concentration inequality for square roots

Question

Given a sequence of (not-necessarily-iid) real-valued random variables $X_n$ that converge to $a\in\mathbb{R}$ in probability, suppose we have an exponential concentration inequality of the form $$ P(|X_n-a|>t)\le Ce^{-cnt^2} $$ for come constants $C,c>0$. What (if anything) can be said about $$ P(|\sqrt{X_n}-\sqrt{a}|>t)? \quad (*) $$

Just to be clear, here, $a$ is a constant that may not be equal to $\mathbb{E} X_n$ (which themselves need not be the same).

My suspicion is that "not much" can be said, so I am either looking for a counterexample to show how bad this gets, or a proof of a nontrivial concentration inequality for $(*)$.

Answer 1

Let $X:=X_n$. For your probability $$p_{a,t}:=P(|\sqrt X-\sqrt a|>t)$$ to make sense, we need to assume that $X\ge0$ and $a\ge0$. Also, if $t<0$, then $p_{a,t}=1$. So, without loss of generality $t\ge0$. Then, solving the inequality $|\sqrt X-\sqrt a|>t$ for $X$, we get $$p_{a,t}=P(X-a>b^+_{a,t})+P(-a\le X-a<-b^-_{a,t}), \tag{1}\label{1}$$ where $$b^+_{a,t}:=(\sqrt a+t)^2-a,\quad b^-_{a,t}:=a-(\sqrt a-t)_+^2,$$ and $u_+:=\max(0,u)$. Note also that $P(-a\le X-a<-b^-_{a,t})=0$ if $-a\ge-b^-_{a,t}$, that is, if $t\ge\sqrt a$.

Comparing the right-hand side of \eqref{1} with the right-hand side of the identity $$P(|X-a|>b)=P(X-a>b)+P(X-a<-b)$$ and noting that $0\le b^-_{a,t}\le b^+_{a,t}$ for $a\ge0$ and $t\ge0$, we see that the best possible upper bound on $p_{a,t}=P(|\sqrt X-\sqrt a|>t)$ that can be obtained from the given upper bound on $P(|X-a|>t)$ (for all $t\ge0$) is given by the following: $$P(|\sqrt X-\sqrt a|>t)\le P(|X-a|>b^-_{a,t}) \le C\exp(-cn(b^-_{a,t})^2)$$ if $t<\sqrt a$ and $$P(|\sqrt X-\sqrt a|>t)\le P(|X-a|>b^+_{a,t}) \le C\exp(-cn(b^+_{a,t})^2)$$ if $t\ge\sqrt a$.