Ive got a short question, I wasn't able to solve. Theres the following inequality which I`m trying to understand: Let $\varepsilon>0$ be a constant, then $$ P_{\eta}\bigg(\frac{1}{m}\sum\limits_{j=1}^m \delta^2_{n,j}>\varepsilon\bigg)\leq m\sup\limits_{j=1,\ldots,m}P_\eta\Big(\delta_{n,j}^2>\varepsilon / 2\Big),$$ where $P_\eta$ is a probability measure and $\delta_{n,j}$ random variables (assume $n$ is fixed so $j$ is the index). Further, I only know, that $\delta$ isn't centered and has finite second moment.
My idea was, that one might start with $$ P_{\eta}\bigg(\frac{1}{m}\sum\limits_{j=1}^m \delta^2_{n,j}>\varepsilon\bigg)\leq P_\eta\bigg(\sup\limits_{j=1,\ldots,m}\delta_{n,j}^2>\epsilon\bigg). $$ But beyond this obvious step, I've had no clue what to do. Maybe some of you can help me out.
Thanks!
As Kolmin said we can omit the $n$ and the $\eta$. Your idea is correct, then you can use the union bound (Bonferroni's inequality): $$ %P\left(\frac1m\sum_{j=1}^m \delta_j^2>\epsilon\right) %\le P\left(\sup_{j=1,\dots,m}\delta_j^2>\epsilon\right) = P\left(\bigcup_{j=1}^m\{\delta_j^2>\epsilon\}\right) \le \sum_{j=1}^m P\left(\delta_j^2>\epsilon\right) \le m\sup_{j=1,\dots,m}P\left(\delta_j^2>\epsilon\right). $$ The last inequality is bcecause a sum of $m$ numbers is smaller than $m$ times the maximum of these real numbers. Of course, $P(\delta_j^2>\epsilon)\le P(\delta_j^2>\epsilon/2)$, which completes the proof.