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Suppose there are two CDF's $F$ and $G$ over the same support, $[0,1]$, and assume that one first order stochastically dominates the other: $$\tag{FOSD 1}F\succsim_{FOSD}G$$ meaning that $F(x)\leq G(x),~\forall x\in[0,1]$.

Can we say this fact implies the dominance relation of the conditional distributions of the following? $$\tag{FOSD 2}F(x|x\geq y)\succsim_{FOSD} G(x|x\geq y),~\forall y\in[0,1]$$

Clearly, (FOSD $2$) implies (FOSD $1$) as we can just take $y=0$. However, can we guarantee that the converse also hold? or do we need something more to guarantee (FOSD $2$)?

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  • $\begingroup$ $F(x|x\ge y)$ does not make any sense. Maybe you mean $P(X\le x|X\ge y)$ where $X$ is a random variable having distribution $F$? $\endgroup$
    – user52227
    Feb 15 '19 at 10:17
  • $\begingroup$ That's correct. I meant the probability. $\endgroup$
    – Andeanlll
    Feb 15 '19 at 11:05
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It is not true. Look at Example 1.3.1. in "Comparison Methods for Stochastic Models and Risks", Alfred Müller, Dietrich Stoyan, 2002.

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  • $\begingroup$ Thanks. I couldn't find any online access to the article so I'll visit a library and check it. $\endgroup$
    – Andeanlll
    Feb 15 '19 at 12:27
  • $\begingroup$ Would there be any condition that guarantee the stochastic order of the conditional probabilities? $\endgroup$
    – Andeanlll
    Feb 15 '19 at 12:27
  • $\begingroup$ Yes, that's called Hazard Rate order, that you can find in the book. $\endgroup$
    – user52227
    Feb 15 '19 at 12:42
  • $\begingroup$ I see. Thanks a lot! $\endgroup$
    – Andeanlll
    Feb 15 '19 at 12:43

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