Suppose we have two distributions $F$ and $G$ over $\left[0,1\right]$. Suppose $F(x) \leq G(x)$ for all $x$, i.e. $F$ first-order stochastically dominates $G$. Is it true that $F(x|x\leq k) \leq G(x|x\leq k)$ for all $k$ and for all $x \in \left[0,k\right]$? Put in another way, does first-order stochastic dominance survive half-truncations (including when $x \geq k$).



1 Answer 1


No. In fact, simple examples show that they may exist $k$ such that conditioning on $x\in [0,k]$ may reverse, and not only destroy, the original ordering.

Hence the property in your question defines a stronger relation than first order stochastic dominance (also known as the usual stochastic order), and it is called the reversed hazard rate order.

Two good references containing the above (at least as far as I can remember) are the books M\"uller and Stoyan (2002), Comparison Methods for Stochastic Models and Risks, and Shaked and Shanthikumar (2007), Stochastic Orders. Neither is completely free of errors, though.

  • $\begingroup$ Thank you. Could you recommend any reading for truncation-proofness of this reversed hazard rate order? $\endgroup$
    – firemind
    Nov 2, 2014 at 12:03
  • $\begingroup$ I expanded the answer to include two general references which should contain what you want. $\endgroup$ Nov 2, 2014 at 16:17

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