If $X_1, \ldots, X_n \sim t_\nu$, a t-distribution with $\nu >1$, how to show $E\left(\max_{1 \leq i \leq n}|X_i|\right) = O\left(n^{1/\nu}\right)$? If $X_1, \ldots, X_n \sim t_\nu$, a t-distribution with $\nu >1$ degrees of free, with each of them independent, then a  result from probability theory is that: 
$$
E\left(\max_{1 \leq i \leq n}|X_i|\right) = O\left(n^{1/\nu}\right)
$$
$X_n = O(Y_n)$ means that there exists a constant $a > 0$ such that $|X_n| \leq a|Y_n|$ for all $n$.
The common trick is to use moment generating functions, for which it doesnt exist for a t-distribution. Does anyone have any ideas how to approach this?
 A: The most common trick I know is not to use the moment generating function, but the cumulative distribution function. From Wikipedia, the density of the distribution of $|X_i|$ is:
$$\frac{2\Gamma \left( \frac{1+\nu}{2}\right)}{\sqrt{\nu \pi} \Gamma \left( \frac{\nu}{2}\right)} \frac{1}{\left( 1+\frac{x^2}{\nu} \right)^{\frac{1+\nu}{2}}}.$$
Hence, for $t > 0$,
$$\mathbb{P} (|X_i| \geq t) = \frac{2\Gamma \left( \frac{1+\nu}{2}\right)}{\sqrt{\nu \pi} \Gamma \left( \frac{\nu}{2}\right)} \int_t^{+\infty}\frac{1}{\left( 1+\frac{x^2}{\nu} \right)^{\frac{1+\nu}{2}}} dx \leq \frac{2\Gamma \left( \frac{1+\nu}{2}\right) \nu^{\frac{1+\nu}{2}}}{\sqrt{\nu \pi} \Gamma \left( \frac{\nu}{2}\right)} \int_t^{+\infty}\frac{1}{x^{1+\nu}} dx = C_\nu t^{-\nu}.$$
For the maximum, we get a crude upper bound:
$$\mathbb{P} (\max_{1\leq i\leq n} |X_i| \geq t) \leq C_\nu n t^{-\nu}.$$
Now, we only need to integrate of $t$:
$$\mathbb{E} (\max_{1\leq i\leq n} |X_i|) = \int_0^{+ \infty} \mathbb{P} (\max_{1\leq i\leq n} |X_i| \geq t) dt \leq \int_0^{+ \infty} \min \{1, C_\nu n t^{-\nu}\} dt = \int_0^{(C_\nu n)^{1/\nu}} 1 dt + \int_{(C_\nu n)^{1/\nu}}^{+\infty} C_\nu n t^{-\nu} dt = \frac{\nu(C_\nu n)^{\frac{1}{\nu}}}{\nu-1} = O(n^{\frac{1}{\nu}}).$$
Note that the bounds we used are so crude that this result holds even if the $X_i$ are not independent.
