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Let $X_1, \dots, X_n \stackrel{\text{i.i.d}}{\sim} \mathcal{N}(0, 1)$ and $Y_i = |X_i|^q$. The very first thing I thought is that the power of normal distribution should be Gamma (but I guess it is true for $q=2$ only). I derived the p.d.f. of $Y = |X|^q$: $$ Y \sim f_Y(y) := \sqrt{\frac{2}{q^2\pi}} y^{1/q - 1} \exp\left\{- \frac{x^{2/q}} 2 \right\} \quad \text{ for } y > 0. $$

I tried to compute the convolution of above mentioned distribution with itself, but did not succeed. I was trying to compute the distribution of $\sum_{i=1}^n Y_i$ after which try to obtain a concentration inequality.

The problem can also be formulated in terms of $\mathbf{X} = (X_1, \dots, X_n)^T$ and we want to obtain some kind of concentration bound for $\| \mathbf{X} \|_{q}$. It can be shown that using the union bound one has the following inequality $$ \| \mathbf{X}\|_q \le n^{1/q} \cdot \sqrt{2 \log \frac{2n} {\delta}} $$ with probability of at least $1 - \delta$.

Clearly, for $q = 2$ we get something of order $n \log n$, while the optimal bound for $\chi^2_n$ is of order $n + \sqrt{n}$, but I didn't use any of the properties of $\chi^2_n$ distribution. So, I wonder whether this inequality can be sharpened in some way, say getting rid of the additional $\log n$ factor.

tortue
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1 Answers1

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Let $q\geq 2$ and $f:x\mapsto \|x\|_q$. By the reverse triangle inequality and this, $$|f(x)-f(y)|\leq \|x-y\|_q \leq \|x-y\|_2$$

Note that Jensen's inequality implies $E(\|X\|_q) \leq [E(\|X\|_q^q)]^{1/q}\leq [nE(|X_1|^q)]^{1/q}$

Concentration of the Gaussian measure (see Concentration in Gauss space) yields

$$P(\|X\|_q - n^{1/q}E(|X_1|^q)^{1/q} \geq t)\leq \exp\left(-\frac{t^2}2\right)$$ hence $$P\left(\|X\|_q \leq n^{1/q}E(|X_1|^q)^{1/q} + \sqrt{2\log\left(\frac 1{\delta} \right)}\right)\geq 1-\delta$$

It can be shown that $\displaystyle E(|X_1|^q) = \sqrt{\frac{2^q}{\pi}}\Gamma\bigg(\frac{q+1}2\bigg)$ so the bound can be made more explicit, but I'll leave that to you.

tortue
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Gabriel Romon
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