How to prove $$P(a\geq b+c)\leq P(a\geq b)+P(c\leq0)?$$
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1$\begingroup$ $a\ge b+c$ happens when either (i) $a\ge b+c$ and $c\ge 0$ or (ii) $a\ge b+c$ and $c\le 0$, which are incompatible events. $\endgroup$– Xi'anCommented Apr 9, 2019 at 7:51
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1 Answer
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The critical relation is: $$P(a\geq b+c)\leq P(a\geq b \cup c \leq 0)$$ because RHS is more general, i.e. when it doesn't happen, we have $a<b\cap c>0$, in which we cannot have $a\geq b+c$.
Then, we just apply set rules: $$P(a\geq b+c)\leq P(a\geq b)+P(c\leq 0)-P(a\leq b\cap c\leq 0)\leq P(a\geq b)+P(c\leq 0)$$