Issue in understanding stochastic ordering

[pakkanen]

This is not actually a problem but I need clarification for stochastic ordering

From Wikipedia, there is stated: (http://en.wikipedia.org/wiki/Stochastic_ordering)

Usual stochastic order:
a real random variable A is less than a random variable B in the "usual stochastic order" if
Pr(A>x) ≤ Pr(B>x) ,where Pr(°) denotes the probability of an event.

Here comes the issue that I do not understand:
Characterizations:
The following rules describe cases when one random variable is stochastically less than or equal to another. Strict version of some of these rules also exist.

1. A ≤ B if and only if for all non-decreasing functions u, E[u(A)] ≤ E[u(B)].

Why it is not possible to attain Pr(A>x) ≥ Pr(B>x) for some x even if E[u(A)] ≤ E[u(B)] ?

 

[Ray Vickson]

This is not actually a problem but I need clarification for stochastic ordering

From Wikipedia, there is stated: (http://en.wikipedia.org/wiki/Stochastic_ordering)

Usual stochastic order:
a real random variable A is less than a random variable B in the "usual stochastic order" if
Pr(A>x) ≤ Pr(B>x) ,where Pr(°) denotes the probability of an event.

Here comes the issue that I do not understand:
Characterizations:
The following rules describe cases when one random variable is stochastically less than or equal to another. Strict version of some of these rules also exist.

1. A ≤ B if and only if for all non-decreasing functions u, E[u(A)] ≤ E[u(B)].

Why it is not possible to attain Pr(A>x) ≥ Pr(B>x) for some x even if E[u(A)] ≤ E[u(B)] ?

The inequality E[u(A)] ≤ E[u(B)] is assumed to hold for ALL non-decreasing functions. For the special case where u(t) = 0 if t ≤ x and u(t) = 1 if t > x (a non-decreasing function) we have E[u(A)] = Pr(A>x), etc.

Note: we can assume the seemingly-weaker hypothesis that E[u(A)] ≤ E[u(B)] holds for all strictly increasing u; then, by a limiting argument we can prove it for all non-decreasing u. Therefore, the special u above is in no sense "artificial".

RGV

 

[pakkanen]

Late gratitudes! I believe I get it.