Concentration inequality for sum of squares of i.i.d. sub-exponential random variables?
Solution 1:
No. For non-negative i.i.d. $Y_i$, $$P(Y_1\ge (\mu+t)n)\le P\Bigg(\sum_{i=1}^n Y_i\ge (\mu+t)n\Bigg)\le\exp(-nf(t))$$ implies that $Y_1$ is sub-exponential and a square of a sub-exponential is not guaranteed to be sub-exponential.
However you can obtain $$P(X_1^2+\dots+X_n^2>nt)\sim nP(X_1^2>nt)=n\exp(-\lambda\sqrt {nt})$$ for $X_i\sim \exp(\lambda)$ which you can extend to all subexponential $X_i$ that have exponential tails.