# Mean and Variance: first-order stochastic dominance

Suppose $$X$$ and $$Y$$ follow the same distribution, with same mean. And $$Var(X). Then, does $$X$$ first-order stochastically dominate $$Y$$? Intuitively, I think this will hold. But I do not know how to prove this more formally.

Using the definition that $$A$$ first order stochastically dominates $$B$$ if $$P(A \geq x) \geq P(B \geq x)$$ for all $$x$$, and for at least one $$x$$ we have a strict inequality, the answer is no.
Consider two normal distributions centered at $$0$$. Then if $$Y$$ has larger variance than $$X$$, the survival function's $$(1-F(x))$$ will cross at $$0$$. Meaning that neither first order scholastically dominates the other.