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Suppose $X$ is a random vector denoted as $(X_1,\cdots,X_n)$, where $X_1,\cdots,X_n$ are iid random variables with sub-Gaussian distributions. For all $i$, suppose $E[X_i^2]=1$ for simplicity and $\|X_i\|_{\psi_2}=K$ where$\|\cdot\|_{\psi_2}$ is the sub-Gaussian norm.

Let $Y=\|X\|$ be the 2-norm of $X$. A known fact is $E[Y]-\sqrt n$ can be const bounded. On the other hand, one may ask when $n\rightarrow \infty$, will $E[Y]-\sqrt n \rightarrow 0$?

For example, consider the standard normal distribution. Then $Y^2$ is $\chi^2(n)$ distribution, and $$E[Y]=\sqrt 2 \frac {\Gamma\left(\frac {n+1} 2\right)} {\Gamma\left(\frac {n} 2\right)}$$ so $E[Y]-\sqrt n\rightarrow 0$ when $n\rightarrow \infty$. So for arbitary random variable $X_i$ which satisfy the above condition, does $E[Y]-\sqrt n\rightarrow 0$ always hold?

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$\newcommand{\si}{\sigma}$ Let us prove a stronger estimate of $EY$, and let us do that under less restrictive conditions. Namely, let us prove that \begin{equation*} EY-\sqrt n=O(1/\sqrt n) \tag{1} \end{equation*} assuming only that $EX_1^4<\infty$ (instead of the $X_i$'s being sub-Gaussian).

Substituting $$U:=\frac{Y^2}n=\frac1n\,\sum_1^n X_i^2$$ for $u$ in the inequalities $$\frac{1+u-(u-1)^2}2\le\sqrt u\le\frac{1+u}2$$ for $u\ge0$, taking the expectations, and using that $EU=1$ and $E(U-1)^2=Var\,U=\sigma^2/n$, where $\si^2:=Var(X_1^2)<\infty$,
we have $$1-\frac{\sigma^2}{2n}\le\frac{EY}{\sqrt n}\le1,$$ so that (1) indeed follows.

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  • $\begingroup$ This proof is great! $\endgroup$ – zbh2047 Mar 24 at 2:21

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