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Let $X_1, X_2, \dots, X_n$ be identically distributed but not necessarily independent random variables with $E|X_i|^{\alpha} < \infty, \alpha >1$.

In part a) we are required to show that for $\lambda >0$:

$\max_{1 \leq i \leq n} |X_i|^{\alpha} \leq \lambda^{\alpha} + \sum_{i=1}^{n} |X_i|^{\alpha} I(|X_i|>\lambda).$

This bit is not a problem. However I am stuck showing the next part:

b) Hence, show that $n^{-1}E \left(\max_{1 \leq i \leq n} |X_i|^{\alpha}\right) \rightarrow 0$, as $n \rightarrow \infty$.

My working so far:

\begin{align*} n^{-1}E \left(\max_{1 \leq i \leq n} |X_i|^{\alpha}\right) & \leq n^{-1}E \left( \lambda^{\alpha} + \sum_{i=1}^{n} |X_i|^{\alpha} I(|X_i|>\lambda) \right) \\ & = \frac{\lambda^{\alpha}}{n} + \frac{E \sum_{i=1}^{n} |X_i|^{\alpha}I(|X_i|>\lambda)}{n} \\ &= \frac{\lambda^{\alpha}}{n} + \frac{ \sum_{i=1}^{n} E(|X_i|^{\alpha}I(|X_i|>\lambda))}{n} \\ &= \frac{\lambda^{\alpha}}{n} + \frac{ \sum_{i=1}^{n} E(|X_1|^{\alpha}I(|X_1|>\lambda))}{n} \quad \text{(identically distributed)}\\ &= \frac{\lambda^{\alpha}}{n} + \frac{ \sum_{i=1}^{n} \beta}{n} \quad \text{for some finite $\beta$} \\ &= \frac{\lambda^{\alpha}}{n} + \beta \end{align*}

Which clearly doesn't go to 0. Can anyone help me spot any errors or provide some insight?

PS this question is for self study - thanks!

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    $\begingroup$ Please add tag "self-study" $\endgroup$ – Learner Nov 11 '15 at 3:48
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You got the bound $$\frac 1n\mathbb E\left[\max_{1\leqslant i\leqslant n}\left|X_i\right|^\alpha \right]\leqslant \frac{\lambda^\alpha}{n}+\mathbb E\left[\left|X_1\right|^\alpha I\left(\left|X_1\right| \gt \lambda\right)\right],$$ which is valid for any $n$ and $\lambda$. Taking the $\limsup$ as $n$ goes to infinity, we derive that for each $\lambda$, $$\tag{*} \limsup_{n\to +\infty}\frac 1n\mathbb E\left[\max_{1\leqslant i\leqslant n}\left|X_i\right|^\alpha \right]\leqslant \mathbb E\left[\left|X_1\right|^\alpha I\left(\left|X_1\right| \gt \lambda\right)\right].$$ Now, by monotone convergence, we have $\lim_{\lambda\to +\infty}\mathbb E\left[\left|X_1\right|^\alpha I\left(\left|X_1\right| \gt \lambda\right)\right]=0$ and since $(*)$ is valid for any positive $\lambda$, we get the wanted result.

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