I want to establish an asymptotic rate for the quantity $\|\bar{x} - \mu\|_2^2$. Here, $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ where $x_i$ are iid ${\rm N}_p(\mu, \Sigma)$ for $i=1,\dots, n$. Here, I assume that the maximum eigenvalue of $\Sigma$ is bounded above by come constant $C < \infty$.

In my first attempt, I (believe) showed: $$ P\left( \|\bar{x} - \mu\|_2^2 \geq \epsilon\right) \leq P\left(\sum_{j=1}^p Y_j \geq n \epsilon/C\right)$$ where $Y_j \sim \chi^2_1$ for $j=1, \dots, p$ and $Y_1,\dots, Y_p$ are independent. I am wondering whether I can bound the right hand side in such a way that I can make a statement of the form: $$\|\bar{x} - \mu \|_2^2 = O(f\left\{p, n\right\})$$ for some function $f$.

EDIT: For example, using the union bound and the fact that $\chi^2_1$ is subgaussian, I know that:
$$P\left(\sum_{j=1}^p Y_j \geq t\right) = 2p \text{exp}\left(-\frac{t}{p}\right),$$ so that using $t = p\log p$, I have that $\chi^2_p = O(p\log p)$ and consequently, $$ \|\bar{x} - \mu \|_2^2 = O\left(\frac{p\log p}{n}\right).$$ However, I suspect that this rate can be improved since the union bound does not use the independence of $Y_1, \dots, Y_p$.


On the one hand, the rhs has a chi squared p distribution, its density is known and you can look it up. On the the other you can probably do better. Using the known distribution of $\bar x - \mu $ the lhs is a single (p dimensional) integral and if you pick the point of greatest density on the sphere of radius $\sqrt \epsilon$, which is related to the largest eigenvalue, it probably gives you the right asymptotic rate.

  • $\begingroup$ I've added an edit to make clearer what I'm asking about. $\endgroup$ – user23658 Sep 13 '16 at 14:33

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