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Let $X$ be an $n$-dimensional Gaussian random vector. Namely its $n$-components are i.i.d standard normal random variables. Consider its $2$-norm $\|X\|_2$. It is intuitively very clear to me that there exists a positive absolute constant $c$ (independent of $n$) such that

$$\text{Var}(\|X\|_2) \ge c$$

But how do we give this a rigorous proof? I don't really need to know what $c$ is, as long as it is an absolute constant.

Just in case someone would expect $\text{Var}(\|X\|_2) \ge \text{Var}(|X_1|)$ to hold, but $\text{Var}$ doesn't have such monotonicity condition.

No One
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1 Answers1

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Your question is related to this post. I do not give a rigorous proof here, however note that, \begin{align} \text{Var}(\|X\|_2) = \text{Var}\left(\sqrt{\sum_{i=1}^{n}X_i^2}\right). \end{align} The square root of the sum, i.e, $\sqrt{\sum_{i=1}^{n}X_i^2}$ follows a $\chi(n)$ distribution. The variance is given by,

\begin{align} \text{Var}\left(\sqrt{\sum_{i=1}^{n}X_i^2}\right) &= n - \mu^2, \\ \mu &= \sqrt{2}\frac{\Gamma((n+1)/2)}{\Gamma(n/2)}. \end{align}

Here $\Gamma(\cdot)$ is the gamma function. Depending on your taste for rigorous, here is a proof that the sum of squared standard normal variables follows a chi-square distribution with $n$ degrees of freedom and here is a proof that the square root of a chi-squared distribution, follows a chi distribution. I hope this is indeed what you are looking for.

vshas
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