3
$\begingroup$

A text I'm using passes off the following statement as obvious, but I can't see where their logic came from,

As $X_n \le x - c - \epsilon$ and $Y_n \le c + \epsilon$ imply $X_n + Y_n \le x$, it is apparent that $P(X_n \le x-c-\epsilon) - P(Y_n > c+\epsilon) \le P(X_n + Y_n \le x) \le P(X_n \le x-c+\epsilon) + P(Y_n < c-\epsilon)$

Where does this statement come from? It seems entirely out of the blue to me.

  • 0
    Did you mean $Y_n\leq c+\varepsilon$?2011-09-26

1 Answers 1

4

I guess the $e$ in the second equation is $\varepsilon$. Put $c'=c+\varepsilon$. The implication shows that $$P(X_n+Y_n\leq x)\geq P(\{X_n\leq x+c'\}\cap\{Y_n\leq c'\}).$$ Now, we use the fact that for events $A$ and $B$ we have $$P(A \cap B)=P(A)-P(A\cap B^c)\geq P(A)-P(B^c),$$ since $P(A\cap B^c)\leq P(B^c)$.

For the second inequality, let $c'' = c-\varepsilon$ and note that $\{X_n+Y_n\leq x\}\subset \{X_n\leq x-c''\}\cup \{Y_n

  • 0
    That's exactly it! I didn't realize $P(A \cap B) = P(A) - P(A \cap B^c)$, though it's obvious now that I see it.2011-09-26
  • 0
    Davide: Excellent! The second inequality uses $c''=c-\varepsilon$ rather than $c'$.2011-09-26