Re: Dynamical Systems and Expansion-Contraction

>>From Osher Doctorow

Often readers will find "X stochastically dominates Y" defined this
way:

1) P(Y > = x) < = P(X > = x) for all x

Let's prove that this implies FX(x) < = FY(x), all x.

Since P(A' ) = 1 - P(A) where A' is "not A" (the complement of A), we
have from (1):

2) 1 - P(Y < x) < = 1 - P(X < x)

and adding P(Y < x) + P(X < x) - 1 to both sides yields:

3) P(X < x) < = P(Y < x)

Since P(X = x) = P(Y = x) = 0 for continuous random variables (for
similar reasons to Lebesgue measure being 0 at a point), it follows
that P(X < = x) < = P(Y < = x), which by definition says FX(x) <
FY(x). Q.E.D.

Osher Doctorow

.

Relevant Pages