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Consider $P_1$ and $P_2$ as two prob measures on $(R,B)$. Does knowing that $P_1([a,\infty)) \ge P_2([a,\infty)) \forall a \in R$ mean that for any open set in R (call it A), $P_1(A) \ge P_2(A)$ ?

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up vote 2 down vote accepted

Consider what happens if $P_1$ is a point mass at 2 and $P_2$ is a point mass at 0. Think about $A = (-1,1)$.

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If $P_{1}=P_{2}$, then what you wish to prove follows trivially.

If $P_{1} \neq P_{2}$, then there exists $a$ such that $P_{1}([a,\infty)) > P_{2}([a,\infty))$. Take the open set $A = (-\infty,a)$.

$P_{1}(A) = 1-P_{1}([a,\infty)) < 1-P_{2}([a,\infty)) = P_{2}(A)$

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Thanks madprob. Very helpful –  Eli Dec 12 '12 at 20:45
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