WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that

$$ P(|\|X\|^2-n| \le n\epsilon) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0. $$

Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ has good (exponential / Gaussian) concentration around the value $n$.

WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that

$$ \mathbb P\left(\left|\frac{\|X\|^2}{n}-1\right| \le \epsilon\right) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0. $$

Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ has good (exponential / Gaussian) concentration around the value $n$.

WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that

$$ P(|\|X\|^2-n| \le n\epsilon) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0. $$

Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ as good (exponential / Gaussian) concentration around the value $n$.

WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that

$$ P(|\|X\|^2-n| \le n\epsilon) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0. $$

Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ has good (exponential / Gaussian) concentration around the value $n$.