F and G are cumulative probability distributions with identical support. G first order stochastically dominates F, i.e., for every X on the support, F(x) > G(x). Prove (or disprove) the proposition that argmax [X(1-G(X))] > argmax [X(1-F(X))], where argmax is the value of x that maximizes the expression in brackets.
See attached file for full problem description.
Here is an example demonstrating that the proposition is WRONG.
(Proof of wrong can be done by a single example to the contrary of a proposition)
Let G(x) = x for all x in [0,1] ;
Then arg(max(x(1-G(x)))) = 1/2
Two probability distributions are compared.