# The power of standard normal distribution

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.

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 $$E(\|X\|_q) \leq [E(\|X\|_q^q)]^{1/q}\leq [nE(|X_1|^q)]^{1/q}$$
$$P(\|X\|_q - n^{1/q}E(|X_1|^q)^{1/q} \geq t)\leq \exp\left(-\frac{t^2}2\right)$$ 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 $$\displaystyle E(|X_1|^q) = \sqrt{\frac{2^q}{\pi}}\Gamma(\frac{q+1}2)$$ so the bound can be made more explicit, but I'll leave that to you.