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Suppose $X_1,X_2,...,X_n$ are i.i.d. strictly positive random variables.

Is there any way to upper bound the following probability?

$$P\left[ \sum_{1\leq i\leq n} X_i \geq \alpha ,\ \max_{1\leq i\leq n} X_i \leq \beta \right]$$

I am not able to proceed at all with this. The only bound I could come up with is the trivial one $$P\left[ \sum_{1\leq i\leq n} X_i \geq \alpha ,\ \max_{1\leq i\leq n} X_i \leq \beta \right]\leq P\left[ \sum_{1\leq i\leq n} X_i \geq \alpha \right]$$ and then use some concentration inequality. This means the final expression is independent of $\beta$, which is not a good bound.

Does any one have any other idea on how to proceed?

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  • $\begingroup$ Well, you have trivially also $\sum X_i\leq n\beta$ (or that $X_i\leq\beta$ for all $i$) so there is a $\beta$-dependence if you want. That is still not a good bound though. $\endgroup$ – user10354138 Oct 18 '18 at 9:44
  • $\begingroup$ You also have $$P\left[ \sum_{1\leq i\leq n} X_i \geq \alpha ,\ \max_{1\leq i\leq n} X_i \leq \beta \right]\leq P\left[ \max_{1\leq i\leq n} X_i \leq \beta \right] = P(X_1 \leq \beta)^n$$ since the random variables are i.i.d. $\endgroup$ – elexhobby Oct 20 '18 at 1:25

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