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Let $X$ and $X_1,X_2, X_3$ be non-negative random variables, and $a_i, b_i$ be also nonnegative scalars for $i=1,2,3$.

Suppose

$\text{Prob}(X_1 \geq a_1) \leq b_1$

$\text{Prob}(X_2 \geq a_2) \leq b_2$

$\text{Prob}(X_3 \geq a_3) \leq b_3$

and $X \leq X_1+X_2+X_3$.

Why is $\text{Prob}(X \geq a_1 + a_2 + a_3) \leq b_1 + b_2 + b_3$ true?

The reason I am asking is this paper Stochastic First-and Zeroth-order Methods for Nonconvex, where on pages 12 and 13, formulas (2.29-31) are the first three inequalities and (2.28) corresponds to the forth one. Using these formulas he shows (2.23).

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If $X \geq a_1+a_2+a_3$ then $X_1 \geq a_1$ or $X_2 \geq a_2$ or $X_3 \geq a_3$. [Prove this by contradiction]. Probability of a union of events is less than or equal to the sum of the probabilities. Can you finish?

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  • $\begingroup$ can you elaborate on contradiction part? and union of three events? $\endgroup$
    – user494522
    Apr 8, 2020 at 8:34
  • $\begingroup$ @Saeed If $X_1 <a_1, X_2<a_2$ and $X_3 <a_3$ then $X \leq X_1+X_2+X_3 <a_1+a_2+a_3$. $\endgroup$
    – Kavi Rama Murthy
    Apr 8, 2020 at 8:37
  • $\begingroup$ got it. Can you also help me to understand (2.30) in the paper where he finds the probability of maximum when he has each individual in one line above it? $\endgroup$
    – user494522
    Apr 8, 2020 at 8:41

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