Does P(X>t) Differ from P(Y>t) When X>Y>t?

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The discussion centers on the probability inequality between two random variables, X and Y, given the condition X > Y > t. It is established that if X and Y are random variables, then P(X > t) is greater than or equal to P(Y > t), but not necessarily larger. The concept of stochastic dominance is introduced, where Y is considered stochastically larger than X if P(Y > t) is greater than or equal to P(X > t) for all t. This forms a basis for analyzing two-sample location problems.

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I am confused on a basic probability inequality, could anyone help me on this:
If X>Y>t, then is P(X>t) larger or smaller than P(Y>t)?

Thanks
 
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I am confused about this. You start by asserting that "X> Y> t". Given that, P(x> t) and P(Y> t) are both 1.0!

If you just mean that X and Y are random variables, requiring that X> Y, and t is some number, then, yex, P(X> t) is greater than or equal to P(Y> t) (not necessarily larger- they might be equal). If Y> t, then X> t follows from X> Y. But X> t may be true even if Y> t is not.
 
This may not be the point of the question about the r.vs, but: we say that the random variable [tex]Y[/tex] is stochastically larger than the random variable [tex]X[/tex] provided that

[tex] P(Y > t) \ge P(X > t), \quad \forall t[/tex]

This idea is one way of discussing two-sample location problems.
 

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